Cambridge
International AS and A Level Mathematics
Statistics
Sophie Goldie
Series Editor: Roger Porkess
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Much of the material in this book was published originally as part of the MEI Structured
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Mathematics syllabus.
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Bob Francis, Bill Gibson, Gerald Goddall, Alan Graham, Nigel Green and Roger Porkess.
Copyright in this format © Roger Porkess and Sophie Goldie, 2012
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ISBN 978 1444 14650 9
Contents
Key to symbols in this book
vi
Introduction
vii
The Cambridge International AS and A Level Mathematics syllabus viii
S1 Statistics 1
1
Chapter 1
Exploring data
Looking at the data
Stem-and-leaf diagrams
Categorical or qualitative data
Numerical or quantitative data
Measures of central tendency
Frequency distributions
Grouped data
Measures of spread (variation)
Working with an assumed mean
2
4
7
13
13
14
19
24
34
45
Chapter 2
Representing and interpreting data
Histograms
Measures of central tendency and of spread using quartiles
Cumulative frequency curves
52
53
62
65
Chapter 3
Probability
Measuring probability
Estimating probability
Expectation
The probability of either one event or another
Independent and dependent events
Conditional probability
77
78
79
81
82
87
94
Chapter 4
Discrete random variables
Discrete random variables
Expectation and variance
105
106
114
iii
Chapter 5
Permutations and combinations
Factorials
Permutations
Combinations
The binomial coefficients
Using binomial coefficients to calculate probabilities
123
124
129
130
132
133
Chapter 6
The binomial distribution
The binomial distribution
The expectation and variance of B(n, p)
Using the binomial distribution
141
143
146
147
Chapter 7
The normal distribution
Using normal distribution tables
The normal curve
Modelling discrete situations
Using the normal distribution as an approximation for the
binomial distribution
154
156
161
172
S2 Statistics 2
iv
173
179
Chapter 8
Hypothesis testing using the binomial distribution
Defining terms
Hypothesis testing checklist
Choosing the significance level
Critical values and critical (rejection) regions
One-tail and two-tail tests
Type I and Type II errors
180
182
183
184
189
193
196
Chapter 9
The Poisson distribution
The Poisson distribution
Modelling with a Poisson distribution
The sum of two or more Poisson distributions
The Poisson approximation to the binomial distribution
Using the normal distribution as an approximation for the
Poisson distribution
202
204
207
210
216
224
Chapter 10
Continuous random variables
Probability density function
Mean and variance
The median
The mode
The uniform (rectangular) distribution
233
235
244
246
247
249
Chapter 11
Linear combinations of random variables
The expectation (mean) of a function of X, E(g[X])
Expectation: algebraic results
The sums and differences of independent random variables
More than two independent random variables
256
256
258
262
269
Chapter 12
Sampling
Terms and notation
Sampling
Sampling techniques
277
277
278
281
Chapter 13
Hypothesis testing and confidence intervals using
the normal distribution
Interpreting sample data using the normal distribution
The Central Limit Theorem
Confidence intervals
How large a sample do you need?
Confidence intervals for a proportion
285
285
298
300
304
306
Answers
Index
312
342
v
Key to symbols in this book
?
●
This symbol means that you may want to discuss a point with your teacher. If
you are working on your own there are answers in the back of the book. It is
important, however, that you have a go at answering the questions before looking
up the answers if you are to understand the mathematics fully.
!
This is a warning sign. It is used where a common mistake, misunderstanding or
tricky point is being described.
This is the ICT icon. It indicates where you could use a graphic calculator or a
computer. Graphic calculators and computers are not permitted in any of the
examinations for the Cambridge International AS and A Level Mathematics 9709
syllabus, however, so these activities are optional.
This symbol and a dotted line down the right-hand side of the page indicate
material which is beyond the syllabus for the unit but which is included for
completeness.
vi
Introduction
This is part of a series of books for the University of Cambridge International
Examinations syllabus for Cambridge International AS and A Level Mathematics
9709. There are thirteen chapters in this book; the first seven cover Statistics 1
and the remaining six Statistics 2. The series also includes two books for pure
mathematics and one for mechanics.
These books are based on the highly successful series for the Mathematics in
Education and Industry (MEI) syllabus in the UK but they have been redesigned
for Cambridge international students; where appropriate, new material has been
written and the exercises contain many past Cambridge examination questions.
An overview of the units making up the Cambridge international syllabus is given
in the diagram on the next page.
Throughout the series the emphasis is on understanding the mathematics as well
as routine calculations. The various exercises provide plenty of scope for practising
basic techniques; they also contain many typical examination questions.
An important feature of this series is the electronic support. There is an
accompanying disc containing two types of Personal Tutor presentation:
examination-style questions, in which the solutions are written out, step by step,
with an accompanying verbal explanation, and test-yourself questions; these are
multiple-choice with explanations of the mistakes that lead to the wrong answers
as well as full solutions for the correct ones. In addition, extensive online support
is available via the MEI website, www.mei.org.uk.
The books are written on the assumption that students have covered and
understood the work in the Cambridge IGCSE® syllabus. However, some
of the early material is designed to provide an overlap and this is designated
‘Background’. There are also places where the books show how the ideas can be
taken further or where fundamental underpinning work is explored and such
work is marked as ‘Extension’.
The original MEI author team would like to thank Sophie Goldie who has carried
out the extensive task of presenting their work in a suitable form for Cambridge
international students and for her original contributions. They would also like to
thank University of Cambridge International Examinations for their detailed advice
in preparing the books and for permission to use many past examination questions.
Roger Porkess
Series Editor
vii
The Cambridge International AS
and A Level Mathematics syllabus
P2
Cambridge
IGCSE
Mathematics
P1
S1
AS Level
Mathematics
M1
S1
M1
S2
P3
M1
viii
S1
M2
A Level
Mathematics
Statistics 1
S1
Exploring data
S1
1
2
1
Exploring data
A judicious man looks at statistics, not to get knowledge but to save
himself from having ignorance foisted on him.
Carlyle
Source: The Times 2012
The cuttings on page 2 all appeared in one newspaper on one day. Some of them
give data as figures, others display them as diagrams.
How do you interpret this information? Which data do you take seriously and
which do you dismiss as being insignificant or even misleading?
Exploring data
To answer these questions fully you need to understand how data are collected
and analysed before they are presented to you, and how you should evaluate what
you are given to read (or see on the television). This is an important part of the
subject of statistics.
S1
1
In this book, many of the examples are set as stories from fictional websites.
Some of them are written as articles or blogs; others are presented from the
journalists’ viewpoint as they sort through data trying to write an interesting
story. As you work through the book, look too at the ways you are given such
information in your everyday life.
bikingtoday.com
Another cyclist seriously hurt. Will you be next?
On her way back home from school on
Wednesday afternoon, little Rita Roy
was knocked off her bicycle and taken to
hospital with suspected concussion.
Rita was struck by a Ford Transit van, only
50 metres from her own house.
Rita is the fourth child from the Nelson
Mandela estate to be involved in a serious
cycling accident this year.
The busy road where Rita Roy was
knocked off her bicycle yesterday.
After reading the blog, the editor of a local newspaper commissioned one of the
paper’s reporters to investigate the situation and write a leading article for the
paper on it. She explained to the reporter that there was growing concern locally
about cycling accidents involving children. She emphasised the need to collect
good quality data to support presentations to the paper’s readers.
?
●
Is the aim of the investigation clear?
Is the investigation worth carrying out?
What makes good quality data?
The reporter started by collecting data from two sources. He went through back
numbers of the newspaper for the previous two years, finding all the reports of
cycling accidents. He also asked an assistant to carry out a survey of the ages of
3
local cyclists; he wanted to know whether most cyclists were children, young
adults or whatever.
Exploring data
S1
1
?
●
Are the reporter’s data sources appropriate?
Before starting to write his article, the reporter needed to make sense of the data
for himself. He then had to decide how he was going to present the information
to his readers. These are the sorts of data he had to work with.
Name
Age
Distance
from home
Cause
Injuries
Treatment
Rahim Khan
45
3 km
skid
Concussion
Hospital
outpatient
Debbie Lane
5
75 km
hit kerb
Broken arm
Hospital
outpatient
Arvinder Sethi
12
1200 m
lorry
Multiple
fractures
Hospital
3 weeks
Husna Mahar
8
300 m
Bruising
Hospital
outpatient
David Huker
8
50 m
hit
each
other
Concussion
Hospital
outpatient
}
There were 92 accidents listed in the reporter’s table.
Ages of cyclists (from survey)
66
35
64
37
9
18
18
6
26
11
18
23
20
15
62
61
39
138
12
11
19
13
22
16
9
25
20
61
9
67
37
7
15
28
13
45
7
42
21 8 21 63
21 7 10 52
9 17 64 32
10 55 14 66
36 9 88 46
29 6 60 60
44
13
8
67
12
16
10
52
9
14
59
50
44
20
31
62
61
16
34
17
19
28
22
34
18
26
22
36
49
14
This information is described as raw data, which means that no attempt has yet
been made to organise it in order to look for any patterns.
Looking at the data
4
At the moment the arrangement of the ages of the 92 cyclists tells you very little
at all. Clearly these data must be organised so as to reveal the underlying shape,
the distribution. The figures need to be ranked according to size and preferably
grouped as well. The reporter had asked an assistant to collect the information
and this was the order in which she presented it.
Tally
Tallying is a quick, straightforward way of grouping data into suitable intervals.
You have probably met it already.
Tally
Frequency
13
10–19
26
20–29
16
30–39
10
40–49
6
50–59
5
60–69
0–9
70–79
80–89
Looking at the data
Stated age
(years)
S1
1
14
0
1
1
130–139
Total
92
Extreme values
A tally immediately shows up any extreme values, that is values which are far
away from the rest. In this case there are two extreme values, usually referred to
as outliers: 88 and 138. Before doing anything else you must investigate these.
In this case the 88 is genuine, the age of Millie Smith, who is a familiar sight
cycling to the shops.
The 138 needless to say is not genuine. It was the written response of a man who
was insulted at being asked his age. Since no other information about him is
available, this figure is best ignored and the sample size reduced from 92 to 91.
You should always try to understand an outlier before deciding to ignore it; it
may be giving you important information.
!
Practical statisticians are frequently faced with the problem of outlying
observations, observations that depart in some way from the general pattern of
a data set. What they, and you, have to decide is whether any such observations
belong to the data set or not. In the above example the data value 88 is a genuine
member of the data set and is retained. The data value 138 is not a member of the
data set and is therefore rejected.
5
Describing the shape of a distribution
An obvious benefit of using a tally is that it shows the overall shape of the
distribution.
30
frequency density (people/10 years)
Exploring data
S1
1
20
10
0
10
20
30
40
50
60
70
80
90
age (years)
Figure 1.1 Histogram to show the ages of people involved in cycling accidents
You can now see that a large proportion (more than a quarter) of the sample are
in the 10 to 19 year age range. This is the modal group as it is the one with the
most members. The single value with the most members is called the mode, in
this case age 9.
You will also see that there is a second peak among those in their sixties; so this
distribution is called bimodal, even though the frequency in the interval 10–19 is
greater than the frequency in the interval 60–69.
Different types of distribution are described in terms of the position of their
modes or modal groups, see figure 1.2.
(a)
(b)
(c)
Figure 1.2 Distribution shapes: (a) unimodal and symmetrical (b) uniform (no
mode but symmetrical) (c) bimodal
6
When the mode is off to one side the distribution is said to be skewed. If the
mode is to the left with a long tail to the right the distribution has positive (or
right) skewness; if the long tail is to the left the distribution has negative (or left)
skewness. These two cases are shown in figure 1.3.
S1
1
Stem-and-leaf diagrams
(a)
(b)
Figure 1.3 Skewness: (a) positive (b) negative
Stem-and-leaf diagrams
The quick and easy view of the distribution from the tally has been achieved at
the cost of losing information. You can no longer see the original figures which
went into the various groups and so cannot, for example, tell from looking at the
tally whether Millie Smith is 80, 81, 82, or any age up to 89. This problem of the
loss of information can be solved by using a stem-and-leaf diagram (or stemplot).
This is a quick way of grouping the data so that you can see their distribution
and still have access to the original figures. The one below shows the ages of the
91 cyclists surveyed.
n = 91
6
7 represents 67 years
0
1
2
3
4
5
6
7
8
6
9
0
4
4
2
6
HIGH
8
5
1
5
4
2
2
7
0
1
9
5
5
3
9
8
6
2
6
9
1
9
3
8
1
9
0
1
8
0
1
7
2
9
3
0
6
This is the scale.
These are branches.
9
7
6
7
9
1
2
6
7 9 7 6
3 7 9 8 6 0 4 4 2 2 8 1 6 6 4 8 5
2 8 3 2 0 5 9
4
4 4 7 6 7 2 1 0 0
8
138
Individual numbers
are called leaves.
Extreme values are placed on a separate
HIGH or LOW branch. These values are given
in full as they may not fit in with the scale
being used for more central values.
This is the stem.
Value 138 is ignored
Figure 1.4 Stem-and-leaf diagram showing the ages of a sample of
91 cyclists (unsorted)
?
●
Do all the branches have leaves?
7
Exploring data
S1
1
The column of figures on the left (going from 0 to 8) corresponds to the tens
digits of the ages. This is called the stem and in this example it consists of
9 branches. On each branch on the stem are the leaves and these represent the
units digits of the data values.
In figure 1.4, the leaves for a particular branch have been placed in the order in
which the numbers appeared in the original raw data. This is fine for showing the
general shape of the distribution, but it is usually worthwhile sorting the leaves,
as shown in figure 1.5.
n = 91
6
7 represents 67 years
0
1
2
3
4
5
6
7
8
6
0
0
1
2
0
0
6
0
0
2
4
2
0
7
0
0
4
4
2
1
7
1
1
4
5
5
1
7
1
1
5
6
9
1
8
2
1
6
9
8
2
2
6
9
3
2
7
9
3
2
7
9 9 9 9
3 4 4 4 5 5 6 6 6 7 7 8 8 8 8 9 9
3 5 6 6 8 8 9
9
2 2 3 4 4 6 6 7 7
8
Note that the value 138 is left out as it has been identified as not belonging to this set of
data.
Figure 1.5 Stem-and-leaf diagram showing the ages of a sample of
91 cyclists (sorted)
The stem-and-leaf diagram gives you a lot of information at a glance:
●
The youngest cyclist is 6 and the oldest is 88 years of age
●
More people are in the 10–19 year age range than in any other 10 year age
range
●
There are three 61 year olds
●
The modal age (i.e. the age with the most people) is 9
●
The 17th oldest cyclist in the survey is 55 years of age.
If the values on the basic stem-and-leaf diagram are too cramped, that is, if there
are so many leaves on a line that the diagram is not clear, you may stretch it. To
do this you put values 0, 1, 2, 3, 4 on one line and 5, 6, 7, 8, 9 on another. Doing
this to the example results in the diagram shown in figure 1.6.
When stretched, this stem-and-leaf diagram reveals the skewed nature of the
distribution.
8
n = 91
6
6
0
5
0
5
1
5
2
5
0
5
0
6
6
0
5
0
6
2
6
4
6
2
9
0
6
7
0
6
0
6
4
6
4
9
2
7
1
6
1
8
4
7
7
1
6
1
8
8
2
7
1
9
8
2
7
2
9
3
8
2
9
3
8
2
9 9 9 9
3 4 4 4
8 8 9 9
3
Stem-and-leaf diagrams
0
0
1*
1
2*
2
3*
3
4*
4
5*
5
6*
6
7*
7
8*
8
S1
1
7 represents 67 years
7 9
1 1 1 2 2 3 4 4
7 7
8
Figure 1.6 Stem-and-leaf diagram showing the ages of a sample of
91 cyclists (sorted)
?
●
How would you squeeze a stem-and-leaf diagram? What would you do if the data
have more significant figures than can be shown on a stem-and-leaf diagram?
Stem-and-leaf diagrams are particularly useful for comparing data sets. With two
data sets a back-to-back stem-and-leaf diagram can be used, as shown in figure 1.7.
represents 590
Figure 1.7
?
●
9
5
2
9
2
5 3 0
9 7 5 1 1
8 6 2 1
5
6
7
8
9
1
0
1
3
2
represents 520
7
2 3 5 8
2 5 6 6 7
5
Note the numbers on the left
of the stem still have the smallest
number next to the stem.
How would you represent positive and negative data on a stem-and-leaf diagram?
9
EXERCISE 1A
1
Write down the numbers which are represented by this stem-and-leaf diagram.
n = 15
Exploring data
S1
1
2
32
1 represents 3.21 cm
32
33
34
35
36
37
7
2
3
0
1
2
6
5 9
2 6 6 8
1 4
Write down the numbers which are represented by this stem-and-leaf diagram.
n = 19
8
9 represents 0.089 mm
8
9
10
11
12
13
3
7
4 8
5 8 9 9
4
0.223
0.248
0.226
0.253
0.230
0.253
0.233
0.259
82.00
79.04
78.01
81.03
80.08
79.06
0.241
82.05
80.04
Write down the numbers which are represented by this stem-and-leaf diagram.
n = 21
34
5 represents 3.45 m
LOW 0.013, 0.089, 1.79
34
35
36
37
38
39
3
1
0
1
0
4
7 9
4 6 8
1 3 8 9
5
HIGH 7.45, 10.87
10
0.237
0.262
Show the following numbers on a sorted stem-and-leaf diagram with five
branches, remembering to include the appropriate scale.
81.07
81.09
5
6
1
3
1
5
Show the following numbers on a sorted stem-and-leaf diagram with six
branches, remembering to include the appropriate scale.
0.212
0.242
4
3
0
2
0
3
1
6
The information was copied from the forms and the ages listed as:
28
19
61
37
(i)
(ii)
7
52
55
38
41
44
34
26
39
28
35
29
81
38
66
63
35
46
37
38
35
62
22
29
32
59
26
36
36
37 60
45 5
45 33
39 33
S1
1
Exercise 1A
Forty motorists entered for a driving competition. The organisers were
anxious to know if the contestants had enjoyed the event and also to know
their ages, so that they could plan and promote future events effectively. They
therefore asked entrants to fill in a form on which they commented on the
various tests and gave their ages.
Plot these data as a sorted stem-and-leaf diagram.
Describe the shape of the distribution.
The unsorted stem-and-leaf diagram below gives the ages of males whose
marriages were reported in a local newspaper one week.
n = 42
1
0
1
2
3
4
5
6
7
8
(i)
(ii)
(iii)
(iv)
8
9 represents 19
9
5
0
8
2
6
6
0
4
2
9
8
5
7
1
8
9 1 1 0 3 6 8 4 1 2 7
2 3 9 1 2 0
9 6 5 3 3 5 6
7
3
What was the age of the oldest person whose marriage is included?
Redraw the stem-and-leaf diagram with the leaves sorted.
Stretch the stem-and-leaf diagram by using steps of five years between the
levels rather than ten.
Describe and comment on the distribution.
On 1 January the average daily temperature was recorded for 30 cities around
the world. The temperatures, in °C, were as follows.
21
32
35
(i)
(ii)
3
2
14
18
–9
23
–4
29
19
10
11
–15
27
26
8
14
–7
8
7
–11
–2
19
15
3
–14
4
1
Illustrate the distribution of temperatures on a stem-and-leaf diagram.
Describe the shape of the distribution.
11
9
The following marks were obtained on an A Level mathematics paper by the
candidates at one centre.
26
29
54
77
73
35
Exploring data
S1
1
54
52
17
76
87
56
50
43
26
30
49
36
37
66
40
100
90
74
54
59
69
98
53
25
34
22
80
44
45
70
34
74
90
60
40
69
66
51
95
46
61
67
44
49
96
97
66
48
76
39
95
75
94
65
45
32
70
52
62
55
71
37
68
82
39
64
51
57
97
92
100
75
37
87
51
91
30
18
68
44
66
Draw a sorted stem-and-leaf diagram to illustrate these marks and comment
on their distribution.
10
The ages of a sample of 40 hang-gliders (in years) are given below.
28 19 24 20 28
35 69 65 26 17
72 23 21 30 28
(i)
(ii)
11
26 22 19 37 40
22 26 45 58 30
65 21 67 23 57
19 25 65 34 66
31 58 26 29 23
Using intervals of ten years, draw a sorted stem-and-leaf diagram to
illustrate these figures.
Comment on and give a possible explanation for the shape of the
distribution.
An experimental fertiliser called GRO was applied to 50 lime trees, chosen at
random, in a plantation. Another 50 trees were left untreated. The yields in
kilograms were as follows.
Treated
59 25 52
23 61 35
20 21 41
50 44 25
19
38
33
42
32
44
35
18
26 33 24 35 30
27 24 30 62 23
38 61 63 44 18
23 54 33 31 25
47 42 41 53 31
53 38 33 49 54
Untreated
8 11 22
55 30 30
29 40 61
63 43 61
22
25
53
12
20
29
22
42
5 31 40 14 45
12 48 17 12 52
33 41 62 51 56
10 16 14 20 51
58 61 14 32 5
10 48 50 14 8
Draw a sorted back-to-back stem-and-leaf diagrams to compare the two sets
of data and comment on the effects of GRO.
12
A group of 25 people were asked to estimate the length of a line which they
were told was between 1 and 2 metres long. Here are their estimates, in metres.
1.15 1.33 1.42 1.26 1.29
1.21 1.30 1.32 1.33 1.29
1.41 1.28 1.65 1.54 1.14
12
1.30 1.30 1.46 1.18 1.24
1.30 1.40 1.26 1.32 1.30
(i)
(ii)
(iii)
Categorical or qualitative data
Chapter 2 will deal in more detail with ways of displaying data. The remainder of
this chapter looks at types of data and the basic analysis of numerical data.
Some data come to you in classes or categories. Such data, like these for the sizes
of sweatshirts, are called categorical or qualitative.
XL,
S,
S,
L,
M,
S,
M,
M,
XL,
L,
S1
1
Numerical or quantitative data
(iv)
Represent these data in a sorted stem-and-leaf diagram.
From the stem-and-leaf diagram which you drew, read off the third
highest and third lowest length estimates.
Find the middle of the 25 estimates.
On the evidence that you have, could you make an estimate of the length
of the line? Justify your answer.
XS
XS = extra small; S = small; M = Medium; L = Large; XL = extra large
Most of the data you encounter, however, will be numerical data (also called
quantitative data).
Numerical or quantitative data
Variables
The score you get when you throw an ordinary die is one of the values 1, 2, 3, 4, 5
or 6. Rather than repeatedly using the phrase ‘The score you get when you throw
an ordinary die’, statisticians find it convenient to use a capital letter, X, say. They
let X stand for ‘The score you get when you throw an ordinary die’ and because
this varies, X is referred to as a variable.
Similarly, if you are collecting data and this involves, for example, noting
the temperature in classrooms at noon, then you could let T stand for ‘the
temperature in a classroom at noon’. So T is another example of a variable.
Values of the variable X are denoted by the lower case letter x, e.g. x = 1, 2, 3, 4, 5
or 6.
Values of the variable T are denoted by the lower case letter t, e.g. t = 18, 21, 20,
19, 23, ... .
Discrete and continuous variables
The scores on a die, 1, 2, 3, 4, 5 and 6, the number of goals a football team scores,
0, 1, 2, 3, ... and amounts of money, $0.01, $0.02, ... are all examples of discrete
variables. What they have in common is that all possible values can be listed.
13
Exploring data
S1
1
Distance, mass, temperature and speed are all examples of continuous variables.
Continuous variables, if measured accurately enough, can take any appropriate
value. You cannot list all possible values.
You have already seen the example of age. This is rather a special case. It is
nearly always given rounded down (i.e. truncated). Although your age changes
continuously every moment of your life, you actually state it in steps of one year,
in completed years, and not to the nearest whole year. So a man who is a few days
short of his 20th birthday will still say he is 19.
In practice, data for a continuous variable are always given in a rounded form.
●
A person’s height, h, given as 168 cm, measured to the nearest centimetre;
167.5 h 168.5
●
A temperature, t, given as 21.8 °C, measured to the nearest tenth of a degree;
21.75 t 21.85
●
The depth of an ocean, d, given as 9200 m, measured to the nearest 100 m;
9150 d 9250
Notice the rounding convention here: if a figure is on the borderline it is rounded
up. There are other rounding conventions.
Measures of central tendency
When describing a typical value to represent a data set most people think of a
value at the centre and use the word average. When using the word average they
are often referring to the arithmetic mean, which is usually just called the mean
and when asked to explain how to get the mean most people respond by saying
‘add up the data values and divide by the total number of data values’.
There are actually several different averages and so, in statistics, it is important
for you to be more precise about the average to which you are referring. Before
looking at the different types of average or measure of central tendency, you need
to be familiar with some notation.
Σ notation and the mean, x–
A sample of size n taken from a population can be identified as follows.
The first item can be called x1, the second item x2 and so on up to xn.
The sum of these n items of data is given by x1 + x2 + x3 + ... + xn.
i =n
n
i =1
i =1
A shorthand for this is ∑ xi or ∑ xi . This is read as ‘the sum of all the terms xi
when i equals 1 to n’.
n
14
So
∑ xi
i =1
= x1 + x2 + x3 + + xn.
Σ is the Greek capital letter, sigma.
If there is no ambiguity about the number of items of data, the subscripts i can be
n
dropped and
∑ xi becomes ∑ x .
i =1
x1 + x 2 + x3 + + xn
n
where x is the symbol for the mean, referred to as ‘x-bar’.
∑ x or 1 x .
It is usual to write x =
n
n∑
The mean of these n items of data is written as x =
This is a formal way of writing ‘To get the mean you add up all the data values
and divide by the total number of data values’.
Measures of central tendency
∑ x is read as ‘sigma x’ meaning ‘the sum of all the x items’.
S1
1
The mean from a frequency table
Often data is presented in a frequency table. The notation for the mean is slightly
different in such cases.
Alex is a member of the local bird-watching group. The group are concerned about
the effect of pollution and climatic change on the well-being of birds. One spring
Alex surveyed the nests of a type of owl. Healthy owls usually lay up to 6 eggs. Alex
collected data from 50 nests. His data are shown in the following frequency table.
Number of eggs, x
Frequency, f
1
4
2
12
3
9
4
18
5
7
6
0
Total
Σf = 50
This represents ‘the sum of the
separate frequencies is 50’. That is,
4 + 12 + 9 + 18 + 7 = 50
It would be possible to write out the data set in full as 1, 1, 1, … , 5, 5 and then
calculate the mean as before. However, it would not be sensible and in practice
the mean is calculated as follows:
x = 1 × 4 + 2 × 12 + 3 × 9 + 4 × 18 + 5 × 7
50
= 162
50
= 3. 24
∑ xf
In general, this is written as x =
n
This represents the sum of
each of the x terms multiplied by
its frequency
n = Σf
15
Exploring data
S1
1
!
In the survey at the beginning of this chapter the mean of the cyclists’ ages,
2717
x =
= 29.9 years.
91
However, a mean of the ages needs to be adjusted because age is always rounded
down. For example, Rahim Khan gave his age as 45. He could be exactly 45 years
old or possibly his 46th birthday may be one day away. So, each of the people
in the sample could be from 0 to almost a year older than their quoted age. To
adjust for this discrepancy you need to add 0.5 years on to the average of 29.9 to
give 30.4 years.
Note
The mean is the most commonly used average in statistics. The mean described
here is correctly called the arithmetic mean; there are other forms, for example, the
geometric mean, harmonic mean and weighted mean, all of which have particular
applications.
The mean is used when the total quantity is also of interest. For example, the
staff at the water treatment works for a city would be interested in the mean
amount of water used per household (x ) but would also need to know the total
amount of water used in the city (Σx). The mean can give a misleading result if
exceptionally large or exceptionally small values occur in the data set.
There are two other commonly used statistical measures of a typical (or
representative) value of a data set. These are the median and the mode.
Median
The median is the value of the middle item when all the data items are ranked in
order. If there are n items of data then the median is the value of the n + 1 th item.
2
If n is odd then there is a middle value and this is the median. In the survey of the
cyclists we have
The 46th item of data is 22 years.
6, 6, 7, 7, 7, 8, …, 20, 21, 21, 21, 22, 22, 22, …
So for the ages of the 91 cyclists, the median is the age of the 91 + 1 = 46th person
2
and this is 22 years.
a +b
If n is even and the two middle values are a and b then the median is
.
2
16
For example, if the reporter had not noticed that 138 was invalid there would
have been 92 items of data. Then the median age for the cyclists would be found
as follows.
}
The 46th and 47th items of data are the
two middle values and are both 22.
So the median age for the cyclists is given as the mean of the 46th and 47th items
of data. That is, 22 + 22 = 22 .
2
It is a coincidence that the median turns out to be the same. However, what is
important to notice is that an extreme value has little or no effect on the value of
the median. The median is said to be resistant to outliers.
The median is easy to work out if the data are already ranked, otherwise it can
be tedious. However, with the increased availability of computers, it is easier to
sort data and so the use of the median is increasing. Illustrating data on a stemand-leaf diagram orders the data and makes it easy to identify the median. The
median usually provides a good representative value and, as seen above, it is not
affected by extreme values. It is particularly useful if some values are missing;
for example, if 50 people took part in a marathon then the median is halfway
between the 25th and 26th values. If some people failed to complete the course
the mean would be impossible to calculate, but the median is easy to find.
Measures of central tendency
6, 6, 7, 7, 7, 8, …, 20, 21, 21, 21, 22, 22, 22, …
S1
1
In finding an average salary the median is often a more appropriate measure than
the mean since a few people earning very large salaries may have a big effect on
the mean but not on the median.
Mode
The mode is the value which occurs most frequently. If two non-adjacent values
occur more frequently than the rest, the distribution is said to be bimodal, even if
the frequencies are not the same for both modes.
Bimodal data usually indicates that the sample has been taken from two
populations. For example, a sample of students’ heights (male and female) would
probably be bimodal reflecting the different average heights of males and females.
For the cyclists’ ages, the mode is 9 years (the frequency is 6).
For a small set of discrete data the mode can often be misleading, especially if
there are many values the data can take. Several items of data can happen to
fall on a particular value. The mode is used when the most probable or most
frequently occurring value is of interest. For example, a dress shop manager who
is considering stocking a new style would first buy dresses of the new style in the
modal size, as she would be most likely to sell those ones.
Which average you use will depend on the particular data you have and on what
you are trying to find out.
17
The measures for the cyclists’ ages are summarised below.
S1
1
Exploring data
Mean
Mode
Median
?
●
EXAMPLE 1.1
29.9 years
9 years
22 years
(adjusted = 30.4 years)
Which do you think is most representative?
These are the times, in minutes, that a group of people took to answer a
Sudoku puzzle.
5, 4, 11, 8, 4, 43, 10, 7, 12
Calculate an appropriate measure of central tendency to summarise these times.
Explain why the other measures are not suitable.
SOLUTION
First order the data.
4, 4, 5, 7, 8, 10, 11, 12, 43
One person took much longer to solve the puzzle than the others, so the mean is
not appropriate to use as it is affected by outliers.
The mode is 4 which is the lowest data value and is not representative of the
data set.
So the most appropriate measure to use is the median.
( )
There are nine data values; the median is the 9 + 1 th value, which is 8 minutes.
2
EXERCISE 1B
18
1
Find the mode, mean and median of these figures.
(i)
23 46 45 45 29
51 36 41 37 47
(ii)
110 111 116 119 129
116 132 118 122 127
(iii)
5 7 7 9 1
6 4 7 7 6
2 3 5 6 6
1 3 3 5 7
45 44 41 31 33
126 132 116 122 130
132 126 138 117 111
8 6 5 7 9
8 2 8 7 6
2 2 5 6 6
5 4 3 6 7
2
For each of these sets of data
(a)
(b)
The ages of students in a class in years and months.
14.1 14.11 14.5 14.6 14.0
14.7 14.7 14.9 14.1 14.2
14.6 14.5 14.8 14.2 14.0
14.9 14.2 14.8 14.11 14.8
15.0 14.7 14.8 14.9 14.3
14.5 14.4 14.3 14.6 14.1
(ii)
Students’ marks on an examination paper.
55 78 45 54 0
62 43 56 71 65
39 45 66 71 52
71 0 0 59 61
(iv)
The scores of a cricketer during a season’s matches.
10 23 65 0 1
24 47 2 21 53
17 34 33 21 0
10 78 1 56 3
Scores when a die is thrown 40 times.
2 4 5 5 1
3 4 6 2 5
2 4 6 1 2
3 4 6 5 5
2 3 3 1 6
5 4 2 1 3
0 67 75
56 59 64
5
2
51 100
57 63
4 23 169
0 128 12
S1
1
Frequency distributions
(i)
(iii)
3
find the mode, mean and median
state, with reasons, which you consider to be the most appropriate form of
average to describe the distribution.
21
19
5 4 4 1 1
3 2 1 6 6
The lengths of time in minutes to swim a certain distance by the members of a
class of twelve 9-year-olds and by the members of a class of eight 16-year-olds
are shown below.
9-year-olds: 13.0 16.1 16.0 14.4 15.9 15.1 14.2 13.7 16.7 16.4 15.0 13.2
16-year-olds: 14.8 13.0 11.4 11.7 16.5 13.7 12.8 12.9
(i)
(ii)
Draw a back-to-back stem-and-leaf diagram to represent the information
above.
A new pupil joined the 16-year-old class and swam the distance. The
mean time for the class of nine pupils was now 13.6 minutes. Find the new
pupil’s time to swim the distance.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q4 June 2007]
Frequency distributions
You will often have to deal with data that are presented in a frequency table.
Frequency tables summarise the data and also allow you to get an idea of the
shape of the distribution.
19
Exploring data
S1
1
EXAMPLE 1.2
Claire runs a fairground stall. She has designed a game where customers pay $1
and are given 10 marbles which they have to try to get into a container 4 metres
away. If they get more than 8 in the container they win $5. Before introducing the
game to the customers she tries it out on a sample of 50 people. The number of
successes scored by each person is noted.
5
4
6
7
5
7
8
5
7
2
8
8
5
6
1
7
9
7
3
6
5
5
6
5
8
4
6
7
5
5
0
3
5
6
4
9
2
6
9
4
10
4
9
8
3
6
4
2
7
3
The data are discrete. They
have not been organised in any
way, so they are referred to
as raw data.
Calculate the mode, median and mean scores. Comment on your results.
SOLUTION
The frequency distribution of these data can be illustrated in a table. The number
of 0s, 1s, 2s, etc. is counted to give the frequency of each mark.
Score
Frequency
0
1
1
1
2
3
3
4
4
6
5
10
6
8
7
7
8
5
9
4
10
1
Total
50
With the data presented in
this form it is easier to find or
calculate the different averages.
The mode is 5 (frequency 10).
As the number of items of data is even, the distribution has two middle values,
the 25th and 26th scores. From the distribution, by adding up the frequencies, it
can be seen that the 25th score is 5 and the 26th score is 6. Consequently the
median score is 12(5 + 6) = 5.5.
20
Representing a score by x and its frequency by f, the calculation of the mean is
shown in this table.
Frequency, f
x×f
0
1
0×1=0
1
1
1×1=1
2
3
2×3=6
3
4
12
4
6
24
5
10
50
6
8
48
7
7
49
8
5
40
9
4
36
10
1
10
Totals
50
276
x =
Frequency distributions
So
Score, x
S1
1
∑ xf
n = Σf
n
276
=
= 5.52
50
The values of the mode (5), the median (5.5) and the mean (5.52) are close. This
is because the distribution of scores does not have any extreme values and is
reasonably symmetrical.
EXAMPLE 1.3
The table shows the number of mobile phones owned by h households.
Number of mobile phones
0
1
2
3
4
5
Frequency
3
5
b
10
13
7
The mean number of mobile phones is 3. Find the values of b and h.
SOLUTION
The total number of households, h = 3 + 5 + b + 10 + 13 + 7.
So h = b + 38
The total number of mobile phones = 0 × 3 + 1 × 5 + 2 × b + 3 × 10 + 4 × 13 + 5 × 7
= 2b + 122
21
total number of mobile phones
=3
total frequency
So h = b + 38
2b + 122
So, b + 38 = 3
Mean =
Exploring data
S1
1
2b + 122 = 3(b + 38)
2b + 122 = 3b + 114
So b = 8 and h = 8 + 38 = 46.
EXERCISE 1C
1
A bag contained six counters numbered 1, 2, 3, 4, 5 and 6. A counter was
drawn from the bag, its number was noted and then it was returned to the
bag. This was repeated 100 times. The results were recorded in a table giving
the frequency distribution shown.
(i)
(ii)
(iii)
2
Number, x
Frequency, f
1
15
2
25
3
16
4
20
5
13
6
11
A sample of 50 boxes of matches with stated contents 40 matches was taken.
The actual number of matches in each box was recorded. The resulting
frequency distribution is shown in the table.
(i)
(ii)
(iii)
22
State the mode.
Find the median.
Calculate the mean.
(iv)
Number of matches, x
Frequency, f
37
5
38
5
39
10
40
8
41
7
42
6
43
5
44
4
State the mode.
Find the median.
Calculate the mean.
State, with reasons, which you think is the most appropriate form of
average to describe the distribution.
3
A survey of the number of students in 80 classrooms in Avonford College was
carried out. The data were recorded in a table as follows.
(i)
(ii)
(iv)
4
Number of students, x
Frequency, f
5
1
11
1
15
6
16
9
17
12
18
16
19
18
20
13
21
3
22
1
Total
80
Exercise 1C
(iii)
State the mode.
Find the median.
Calculate the mean.
State, with reasons,
which you think is
the most appropriate
form of average to
describe the
distribution.
S1
1
The tally below gives the scores of the football teams in the matches of the
1982 World Cup finals.
Score
Tally
0
1
2
3
4
5
6
7
8
9
10
Find the mode, mean and median of these data.
(ii) State which of these you think is the most representative measure.
(For football enthusiasts: find out which team conceded 10 goals and why.)
(i)
23
5
The vertical line chart below shows the number of times the various members
of a school year had to take their driving test before passing it.
20
Exploring data
S1
1
frequency
15
10
5
0
(i)
(ii)
1
2
3
4
number of driving tests
5
6
Find the mode, mean and median of these data.
State which of these you think is the most representative measure.
Grouped data
Grouping means putting the data into a number of classes. The number of data
items falling into any class is called the frequency for that class.
When numerical data are grouped, each item of data falls within a class interval
lying between class boundaries.
class width
class interval
value of
variable
class boundaries
Figure 1.8
You must always be careful about the choice of class boundaries because it must be
absolutely clear to which class any item belongs. A form with the following wording:
How old are you? Please tick one box.
0–10
10–20
20–30
30–40
40–50
50+
would cause problems. A ten-year-old could tick either of the first two boxes.
24
A better form of wording would be:
How old are you (in completed years)? Please tick one box.
0–9
10–19
20–29
30–39
40–49
50+
Grouped data
Notice that this says ‘in completed years’. Otherwise a 912 -year-old might not
know which of the first two boxes to tick.
S1
1
Another way of writing this is:
0 A 10
30 A 40
10 A 20
40 A 50
20 A 30
50 A
Even somebody aged 9 years and 364 days would clearly still come in the first
group.
?
●
Another way of writing these classes, which you will sometimes see, is
0–, 10–, 20–, ... , 50–.
What is the disadvantage of this way?
Working with grouped data
There is often a good reason for grouping raw data.
●
There may be a lot of data.
●
The data may be spread over a wide range.
●
Most of the values collected may be different.
Whatever the reason, grouping data should make it easier to analyse and present
a summary of findings, whether in a table or in a diagram.
For some discrete data it may not be necessary or desirable to group them. For
example, a survey of the number of passengers in cars using a busy road is
unlikely to produce many integer values outside the range 0 to 4 (not counting
the driver). However, there are cases when grouping the data (or perhaps
constructing a stem-and-leaf diagram) is an advantage.
Discrete data
At various times during one week the number of cars passing a survey point
was noted. Each item of data relates to the number of cars passing during a
five-minute period. A hundred such periods were surveyed. The data is
summarised in the following frequency table.
25
Exploring data
S1
1
Number of cars, x
Frequency, f
0–9
5
10–19
8
20–29
13
30–39
20
40–49
22
50–59
21
60–70
11
Total
100
From the frequency table you can see there is a slight negative (or left) skew.
Estimating the mean
When data are grouped the individual values are lost. This is not often a serious
problem; as long as the data are reasonably distributed throughout each interval
it is possible to estimate statistics such as the mean, knowing that your answers
will be reasonably accurate.
To estimate the mean you first assume that all the values in an interval are
equally spaced about a mid-point. The mid-points are taken as representative
values of the intervals.
The mid-value for the interval 0–9 is 0 + 9 = 4.5.
2
The mid-value for the interval 10–19 is 10 + 19 = 14.5, and so on.
2
The x × f column can now be added to the frequency distribution table and an
estimate for the mean found.
Number of cars, x
(mid-values)
26
Frequency, f
x×f
4.5
5
4.5 × 5 = 22.5
14.5
8
14.5 × 8 = 116.0
24.5
13
318.5
34.5
20
690.0
44.5
22
979.0
54.5
21
1144.5
65.0
11
715.0
Totals
100
3985.5
The mean is given by
S1
1
∑ xf
x =
∑f
3985.5
= 39.855
100
The original raw data, summarised in the frequency table on the previous page,
are shown below.
10
9
20
35
70
40
46
38
62
18
18
46
16
40
52
48
50
39
47
30
68
53
29
45
21
45
8
53
58
32
67
57
13
48
25
38
25
45
54
45
25
30
31
54
53
51
56
42
59
49
62
63
56
50
41
25
18
42
25
28
49
34
9
34
29
52
20
61
24
31
11
21
34
32
63
55
36
55
53
27
12
68
45
47
43
47
36
30
42
54
Grouped data
=
8
31
55
60
50
46
9
38
61
38
In this form it is impossible to get an overview of the number of cars, nor would
listing every possible value in a frequency table (0 to 70) be helpful.
However, grouping the data and estimating the mean was not the only option.
Constructing a stem-and-leaf diagram and using it to find the median would
have been another possibility.
?
●
Is it possible to find estimates for the other measures of centre?
Find the mean of the original data and compare it to the estimate.
The data the reporter collected when researching his article on cycling accidents
included the distance from home, in metres, of those involved in cycling
accidents. In full these were as follows.
3000
200
1250
15
75
4500
3500
4000
1200
35
30
300
60
75
50
120
250
10
400
1200
150
2400
250
1500
140
50
250
45
250
25
5
450
It is clear that there is considerable spread in the data. It is continuous data and
the reporter is aware that they appear to have been rounded but he does not
know to what level of precision. Consequently there is no way of reflecting the
level of precision in setting the interval boundaries.
27
The reporter wants to estimate the mean and decides on the following grouping.
Exploring data
S1
1
x×f
Location
relative to
home
Distance, d,
in metres
Very close
0 d 100
50
12
600
Close
100 d 500
300
11
3 300
Not far
500 d 1500
1000
3
3 000
Quite far
1500 d 5000
3250
6
19 500
32
26 400
Distance
mid-value, x
Totals
Frequency
(number of
accidents), f
x = 26400 = 825m
32
A summary of the measures of centre for the original and grouped accident data
is given below.
Raw data
Mean
Mode
Median
?
●
Grouped data
25 785 ÷ 32 = 806 m
250 m
1(200 + 250) = 225 m
2
825 m
Modal group 0 d 100 m
Which measure of centre seems most appropriate for these data?
The reporter’s article
The reporter decided that he had enough information and wrote the article below.
A town council that does not care
28
The level of civilisation of any society can be
measured by how much it cares for its most
vulnerable members.
On that basis our town council rates
somewhere between savages and barbarians.
Every day they sit back complacently while
those least able to defend themselves, the
very old and the very young, run the gauntlet
of our treacherous streets.
I refer of course to the lack of adequate
safety measures for our cyclists, 60% of
whom are children or senior citizens.
Statistics show that they only have to put
one wheel outside their front doors to be
in mortal danger. 80% of cycling accidents
happen within 1500 metres of home.
Last week Rita Roy became the latest
unwitting addition to these statistics. Luckily
she is now on the road to recovery but that is
no thanks to the members of our unfeeling
town council who set people on the road to
death and injury without a second thought.
What, this paper asks our councillors, are
you doing about providing safe cycle tracks
from our housing estates to our schools and
shopping centres? And what are you doing
to promote safety awareness among our
cyclists, young and old?
Answer: Nothing.
?
●
S1
1
Is it a fair article? Is it justified, based on the available evidence?
For a statistics project Robert, a student at Avonford College, collected the
heights of 50 female students.
He constructed a frequency table for his project and included the calculations to
find an estimate for the mean of his data.
Height, h
Mid-value, x
Frequency, f
157 h 159
158
4
632
159 h 161
160
11
1760
161 h 163
162
19
3078
163 h 165
164
8
1312
165 h 167
166
5
830
167 h 169
168
3
504
50
8116
Totals
Grouped data
Continuous data
xf
x = 8116
50
= 162.32
Note: Class boundaries
His teacher was concerned about the class boundaries and asked Robert ‘To what
degree of accuracy have you recorded your data? ’Robert told him ‘I rounded all my
data to the nearest centimetre’. Robert showed his teacher his raw data.
163
160
167
168
166
164
166
162
163
163
165
163
163
159
159
158
162
163
163
166
164
162
164
160
161
162
162
160
169
162
163
160
167
162
158
161
162
163
165
165
163
163
168
165
165
161
160
161
161
161
Robert’s teacher said that the class boundaries should have been
157.5 h 159.5
159.5 h 161.5, and so on.
He explained that a height recorded to the nearest centimetre as 158 cm has a value
in the interval 158 ± 0.5 cm (this can be written as 157.5 h 158.5). Similarly the
actual values of those recorded as 159 cm lie in the interval 158.5 h 159.5. So,
the interval 157.5 h 159.5 covers the actual values of the data items 158 and 159.
The interval 159.5 h 161.5 covers the actual values of 160 and 161 and so on.
29
S1
1
?
●
What adjustment does Robert need to make to his estimated mean in the light of
his teacher’s comments?
Exploring data
Find the mean of the raw data. What do you notice when you compare it with
your estimate?
You are not always told the level of precision of summarised data and the class
widths are not always equal, as the reporter for the local newspaper discovered.
Also, there are different ways of representing class boundaries, as the following
example illustrates.
EXAMPLE 1.4
The frequency distribution shows the lengths of telephone calls made by Emily
during August. Choose suitable mid-class values and estimate Emily’s mean call
time for August.
SOLUTION
Time (seconds)
Mid-value, x
Frequency, f
xf
0–
30
39
1170
60–
90
15
1350
120–
150
12
1800
180–
240
8
1920
300–
400
4
1600
500–1000
750
1
750
79
8590
Totals
x = 8590
79
= 108.7 seconds
Emily’s mean call time is 109 seconds, to 3 significant figures.
Notes
1 The interval ‘0–’can be written as 0 x 60, the interval ‘60–’can be written as
60 x 120, and so on, up to ‘500–1000’which can be written as 500 x 1000.
2 There is no indication of the level of precision of the recorded data. They may
have been recorded to the nearest second.
3 The class widths vary.
30
EXERCISE 1D
1
A college nurse keeps a record of the heights, measured to the nearest
centimetre, of a group of students she treats.
Her data are summarised in the following grouped frequency table.
Number of students
110–119
1
120–129
3
130–139
10
140–149
28
150–159
65
160–169
98
170–179
55
180–189
15
Exercise 1D
Height (cm)
S1
1
Choose suitable mid-class values and calculate an estimate for the mean
height.
2
A junior school teacher noted the time to the nearest minute a group of
children spent reading during a particular day.
The data are summarised as follows.
(i)
(ii)
Time (nearest minute)
Number of children
20–29
12
30–39
21
40–49
36
50–59
24
60–69
12
70–89
9
90–119
2
Choose suitable mid-class values and calculate an estimate for the mean
time spent reading by the pupils.
Some time later, the teacher collected similar data from a group of
25 children from a neigbouring school. She calculated the mean to be
75.5 minutes. Compare the estimate you obtained in part (i) with this value.
What assumptions must you make for the comparison to be meaningful?
31
3
The stated ages of the 91 cyclists considered earlier are summarised by the
following grouped frequency distribution.
Exploring data
S1
1
(i)
(ii)
(iii)
4
Stated age (years)
Frequency
0–9
13
10–19
26
20–29
16
30–39
10
40–49
6
50–59
5
60–69
14
70–79
0
80–89
1
Total
91
Choose suitable mid-interval values and calculate an estimate of the mean
stated age.
Make a suitable error adjustment to your answer to part (i) to give an
estimate of the mean age of the cyclists.
The adjusted mean of the actual data was 30.4 years. Compare this with
your answer to part (ii) and comment.
In an agricultural experiment, 320 plants were grown on a plot. The lengths of
the stems were measured, to the nearest centimetre, 10 weeks after planting.
The lengths were found to be distributed as in the following table.
Length, x (cm)
Frequency (number of plants)
20.5 x 32.5
30
32.5 x 38.5
80
38.5 x 44.5
90
44.5 x 50.5
60
50.5 x 68.5
60
Calculate an estimate of the mean of the stem lengths from this experiment.
32
5
The reporter for the local newspaper considered choosing different classes for
the data dealing with the cyclists who were involved in accidents.
He summarised the distances from home of 32 cyclists as follows.
(ii)
6
Frequency
0 d 50
7
50 d 100
5
100 d 150
2
150 d 200
1
200 d 300
5
300 d 500
3
500 d 1000
0
1000 d 5000
9
Total
32
Exercise 1D
(i)
Distance, d (metres)
S1
1
Choose suitable mid-class values and estimate the mean.
The mean of the raw data is 806 m and his previous grouping gave an
estimate for the mean of 825 m. Compare your answer to this value and
comment.
A crate containing 270 oranges was opened and each orange was weighed.
The masses, given to the nearest gram, were grouped and the resulting
distribution is as follows.
(i)
(ii)
Mass, x
(grams)
Frequency
(number of oranges)
60–99
20
100–119
60
120–139
80
140–159
50
160–220
60
State the class boundaries for the interval 60–99.
Calculate an estimate for the mean mass of the oranges from the crate.
33
Measures of spread (variation)
In the last section you saw how an estimate for the mean can be found from
grouped data. The mean is just one example of a typical value of a data set. You
also saw how the mode and the median can be found from small data sets. The
next chapter considers the use of the median as a typical value when dealing
with grouped data and also the interquartile range as a measure of spread. In this
chapter we will consider the range, the mean absolute deviation, the variance and
the standard deviation as measures of spread.
Exploring data
S1
1
Range
The simplest measure of spread is the range. This is just the difference between
the largest value in the data set (the upper extreme) and the smallest value (the
lower extreme).
●
Range = largest − smallest
The figures below are the prices, in cents, of a 100 g jar of Nesko coffee in ten
different shops.
161
161
163
163
167
168
170
172
172
172
The range for this data is
Range = 172 − 161 = 11 cents.
EXAMPLE 1.5
Ruth is investigating the amount of money, in dollars, students at Avonford
College earn from part-time work on one particular weekend. She collects and
orders data from two classes and this is shown below.
Class 1
Class 2
10 10 10 10 10 10 12 15 15 15
16 16 16 16 18 18 20 25 38 90
10 10 10 10 10 10 12 12 12 12
15 15 15 15 16 17 18 19 20 20
25 35 35
She calculates the mean amount earned for each class. Her results are
Class 1:
Class 2:
x1 = $19.50
x 2 = $16.22
She concludes that the students in Class 1 each earn about $3 more, on average,
than do the students in Class 2.
Her teacher suggests she look at the spread of the data. What further information
does this reveal?
SOLUTION
Ruth calculates the range for each class:
34
Range (Class 1) = $80
Range (Class 2) = $25
She concludes that the part-time earnings in Class 1 are much more spread out.
However, when Ruth looks again at the raw data she notices that one student in
Class 1 earned $90, considerably more than anybody else. If that item of data is
ignored then the spread of data for the two classes is similar.
?
●
Calculate the mean earnings of Class 1 with the item $90 removed.
What can you conclude about the effect of extreme values on the mean?
The range does not use all of the available information; only the extreme values
are used. In quality control this can be an advantage as it is very sensitive to
something going wrong on a production line. Also the range is easy to calculate.
However, usually we want a measure of spread that uses all the available data and
that relates to a central value.
Measures of spread (variation)
! One of the problems with the range is that it is prone to the effect of extreme values.
S1
1
The mean absolute deviation
Kim and Joe play as strikers for two local football teams. They are being considered
for the state team. The team manager is considering their scoring records.
Kim’s scoring record over ten matches looks like this:
0
1
0
3
0
2
0
0
0
4
2
1
1
2
2
Joe’s record looks like this:
1
1
1
0
0
The mean scores are, for Kim, x1 = 1 and, for Joe, x 2 = 1.1.
Looking first at Kim’s data consider the differences, or deviations, of his scores
from the mean.
Number of goals scored, x
0
1
0
3
0
2
0
0
0
4
Deviations (x – x )
–1
0
–1
2
–1
1
–1
–1
–1
3
To find a summary measure you need to combine the deviations in some way.
If you just add them together they total zero.
?
●
Why does the sum of the deviations always total zero?
The mean absolute deviation ignores the signs and adds together the absolute
deviations. The symbol d tells you to take the positive, or absolute, value of d.
35
Exploring data
S1
1
For example − 2 = 2 and 2 = 2.
It is now possible to sum the deviations:
1 + 0 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 3 = 12,
the total of the absolute deviations.
It is important that any measure of spread is not linked to the sample size so you
have to average out this total by dividing by the sample size.
In this case the sample size is 10. The mean absolute deviation = 12 =1.2.
10
1
Remember
● The mean absolute deviation from the mean = ∑ x − x
n
n = Σ f.
For Joe’s data the mean absolute deviation is
1
10 (0.1
+ 0.1 + 0.1 + 1.1 + 1.1 + 0.9 + 0.1 + 0.1 + 0.9 + 0.9) = 0.54
The average numbers of goals scored by Kim and Joe are similar (1.0 and 1.1) but
Joe is less variable (or more consistent) in his goal scoring (0.54 compared to 1.2).
The mean absolute deviation is an acceptable measure of spread but is not
widely used because it is difficult to work with. The standard deviation is more
important mathematically and is more extensively used.
The variance and standard deviation
An alternative to ignoring the signs is to square the differences or deviations. This
gives rise to a measure of spread called the variance, which when square-rooted
gives the standard deviation.
Though not as easy to calculate as the absolute mean deviation, the standard
deviation has an important role in the study of more advanced statistics.
To find the variance of a data set:
For Kim’s data this is:
(x − x )
2
●
Square the deviations
●
Sum the squared deviations
∑( x − x )2
●
Find their mean
∑( x − x )2
n
(0 − 1)2, (1 − 1)2, (0 − 1)2, etc.
1+0+1+4+1+1+1+1+
1 + 9 = 20
20 = 2
10
This is known as the variance.
●
Variance =
∑( x − x )2
n
The square root of the variance is
called the standard deviation.
●
36
sd =
∑( x − x )2
n
So, for Kim’s data the variance is 2, but
what are the units? In calculating the
variance the data are squared. In order to
get a measure of spread that has the same
units as the original data it is necessary to
take the square root of the variance. The
resulting statistical measure is known as
the standard deviation.
!
In other books or on the internet, you may see this calculation carried out using
n – 1 rather than n as the divisor. In this case the answer is denoted by s.
∑( x − x )2
n −1
In Statistics 1, you should always use n as the divisor. You will meet s if you go on
to study Statistics 2.
So for Kim’s data the variance is 2, sd is 2 = 1.41 (to 3 s.f.).
This example, using Joe’s data, shows how the variance and standard deviation
are calculated when the data are given in a frequency table. We’ve already
calculated the mean; x = 1.1.
Number of
goals scored, x
Frequency, f
Deviation
(x − x–)
Deviation2
(x − x–)2
Deviation2 × f
[(x − x–)2 f ]
0
2
0 − 1.1 = −1.1
1.21
1.21 × 2 = 2.42
1
5
1 − 1.1 = −0.1
0.01
0.01 × 5 = 0.05
2
3
2 − 1.1 = 0.9
0.81
0.81 × 3 = 2.43
Totals
10
Measures of spread (variation)
s=
S1
1
4.90
For data presented in this way,
∑(x − x )2 f
standard deviation =
n
=
The standard deviation for Joe’s data is sd =
∑(x − x )2 f
∑f
Σf = n
4.90
= 0.7 goals.
10
Comparing this to the standard deviation of Kim’s data (1.41), we see that Joe’s
goal scoring is more consistent (or less variable) than Kim’s. This confirms what
was found when the mean absolute deviation was calculated for each data set. Joe
was found to be a more consistent scorer (mean absolute deviation = 0.54) than
Kim (mean absolute deviation = 1.2).
An alternative form for the standard deviation
The arithmetic involved in calculating ∑(x − x )2 f can often be very messy.
An alternative formula for calculating the standard deviation is given by
●
standard deviation =
∑x 2f
n
− x 2 or
∑x 2f
∑f
− x2
37
Consider Joe’s data one more time.
Exploring data
S1
1
Number of goals
scored, x
Frequency, f
xf
x2f
0
2
0
0
1
5
5
5
2
3
6
12
Total
10
11
17
x = 11 = 1.1 standard deviation = 17 − 1.12
10
10
= 1.7 − 1.21
=
0.49
= 0.7
This gives the same result as using
∑(x − x )2 f
∑f
. The derivation of this
alternative form for the standard deviation is given in Appendix 1 on the CD.
!
In practice you will make extensive use of your calculator’s statistical functions to
find the mean and standard deviation of sets of data.
Care should be taken as the notations S, s, sd, σ and σ̂ are used differently by
different calculator manufacturers, authors and users. You will meet σ in
Chapter 4.
The following examples involve finding or using the sample variance.
EXAMPLE 1.6
Find the mean and the standard deviation of a sample with
∑ x = 960, ∑ x 2 = 18 000, n = 60.
SOLUTION
x =
variance =
staa ndard deviation =
38
∑ x = 960 = 16
n
60
∑ x 2 − x 2 = 18 000 − 162 = 44
n
60
44 = 6.63 (to 3 s.f.)
EXAMPLE 1.7
Find the mean and the standard deviation of a sample with
∑( x − x )
2
S1
1
= 2000, ∑ x = 960, ∑ f = 60.
x = 960 = 16
60
Remember:
Σf = n
∑(x − x )2 = 2000 = 33.3 ...
variance =
60
∑f
standard deviation =
EXAMPLE 1.8
33.3 ... = 5.77 (to 3 s.f.).
As part of her job as quality controller, Stella collected data relating to the life
expectancy of a sample of 60 light bulbs produced by her company. The mean
life was 650 hours and the standard deviation was 8 hours. A second sample of
80 bulbs was taken by Sol and resulted in a mean life of 660 hours and standard
deviation 7 hours.
Measures of spread (variation)
SOLUTION
Find the overall mean and standard deviation.
SOLUTION
Mean of first sample
× first sample size
x =
Overall mean:
Mean of second sample
× second sample size
Overall Σx
x1 × n + x 2 × m
n +m
Total sample size
x = 650 × 60 + 660 × 80 = 91800 = 655.71… = 656 hours(tto 3 s.f.)
60 + 80
140
For Stella’s sample the variance is 82. Therefore 82 =
For Sol’s sample the variance is 72. Therefore 72 =
∑ x12 − 6502.
60
∑ x22 − 6602.
80
From the above Stella found that
∑ x12 = (82 + 6502) × 60 = 25 353 840 and ∑ x22 = 34 851 920.
Overall Σx2
The overall variance is
The total number of
light bulbs is 140
Do not round any numbers
until you have completed
all calculations
25 353840 + 34 851920
− 655.71 ... 2
140
= 430 041.14… – 429 961.22…
= 79.91…
The overall standard deviation is
79.91 … = 8.94 hours (to 3 s.f. ).
39
?
●
Exploring data
S1
1
Carry out the calculation in Example 1.8 using rounded numbers. That is, use
656 for the overall mean rather than 655.71…. What do you notice?
The standard deviation and outliers
Data sets may contain extreme values and when this occurs you are faced with
the problem of how to deal with them.
Many data sets are samples drawn from parent populations which are normally
distributed. You will learn more about the normal distribution in Chapter 7. In
these cases approximately:
●
68% of the values lie within 1 standard deviation of the mean
●
95% lie within 2 standard deviations of the mean
●
99.75% lie within 3 standard deviations of the mean.
If a particular value is more than two standard deviations from the mean it should
be investigated as possibly not belonging to the data set. If it is as much as three
standard deviations or more from the mean then the case to investigate it is
even stronger.
!
The 2-standard-deviation test should not be seen as a way of defining outliers.
It is only a way of identifying those values which it might be worth looking at
more closely.
In an A level Spanish class the examination marks at the end of the year are
shown below.
35
52
55
61
96
63
50
58
58
49
61
The value 96 was thought to be significantly greater than the other values. The
mean and standard deviation of the data are x = 58 and sd = 14.16… . The value
96 is more than two standard deviations above the mean:
58
2 14.16...
29.7
Figure 1.9
40
58
14.16...
43.8
mean
58
58
14.16...
72.2
58
2 14.16...
86.3
96
When investigated further it turned out that the mark of 96 was achieved by a
Spanish boy who had taken A level Spanish because he wanted to study Spanish
at university. It might be appropriate to omit this value from the data set.
Calculate the mean and standard deviation of the data with the value 96 left out.
Investigate the value using your new mean and standard deviation.
Exercise 1E
?
●
S1
1
The times taken, in minutes, for some train journeys between Kolkata and
Majilpur were recorded as shown.
56
61
57
55
58
57
5
60
61
59
It is unnecessary here to calculate the mean and standard deviation. The value
5 minutes is obviously a mistake and should be omitted unless it is possible to
correct it by referring to the original source of data.
EXERCISE 1E
1 (i)
Find the mean of the following data.
0 0 0 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 4 4 4 4 4 5 5
(ii)
2
3
Find the standard deviation using both forms of the formula.
Find the mean and standard deviation of the following data.
x
3
4
5
6
7
8
9
f
2
5
8
14
9
4
3
Mahmood and Raheem are football players. In the 30 games played so far this
season their scoring records are as follows.
Goals scored
0
1
2
3
4
Frequency (Mahmood)
12
8
8
1
1
Frequency (Raheem)
4
21
5
0
0
(i)
(ii)
Find the mean and the standard deviation of the number of goals each
player scored.
Comment on the players’ goal scoring records.
4
For a set of 20 items of data ∑ x = 22 and ∑ x 2 = 55. Find the mean and the
standard deviation of the data.
5
For a data set of 50 items of data ∑(x − x )2 f = 8 and ∑ x f = 20. Find the
mean and the standard deviation of the data.
41
6
Two thermostats were used under identical conditions. The water
temperatures, in °C, are given below.
Thermostat A:
Thermostat B:
Exploring data
S1
1
24
26
25
26
27
23
23
22
26
28
(i)
Calculate the mean and standard deviaton for each set of water
temperatures.
(ii)
Which is the better thermostat? Give a reason.
A second sample of data was collected using thermostat A.
25
(iii)
7
24
24
26
25
24
24
Find the overall mean and the overall standard deviation for the two sets
of data for thermostat A.
Ditshele has a choice of routes to work. She timed her journey along each
route on several occasions and the times in minutes are given below.
Town route:
Country route:
(i)
(ii)
8
25
15
19
16
21
20
20
28
22
21
18
Calculate the mean and standard deviation of each set of jourmey times.
Which route would you recommend? Give a reason.
In a certain district, the mean annual rainfall is 80 cm, with standard
deviation 4 cm.
(i)
(ii)
One year it was 90 cm. Was this an exceptional year?
The next year had a total of 78 cm. Was that exceptional?
Jake, a local amateur meteorologist, kept a record of the weekly rainfall
in his garden. His first data set, comprising 20 weeks of figures, resulted
in a mean weekly rainfall of 1.5 cm. The standard deviation was 0.1 cm.
His second set of data, over 32 weeks, resulted in a mean of 1.7 cm and a
standard deviation of 0.09 cm.
(iii)
(iv)
9
Calculate the overall mean and the overall standard deviation for the
whole year.
Estimate the annual rainfall in Jake’s garden.
A farmer expects to harvest a crop of 3.8 tonnes, on average, from each
hectare of his land, with standard deviation 0.2 tonnes.
One year there was much more rain than usual and he harvested 4.1 tonnes
per hectare.
(i)
(ii)
42
Was this exceptional?
Do you think the crop was affected by the unusual weather or was the
higher yield part of the variability which always occurs?
10
A machine is supposed to produce ball bearings with a mean diameter of
2.0 mm. A sample of eight ball bearings was taken from the production line
and the diameters measured. The results, in millimetres, were as follows:
2.0
(ii)
11
2.0
1.8
2.4
2.3
1.9
2.1
Exercise 1E
(i)
2.1
S1
1
Calculate the mean and standard deviation of the diameters.
Do you think the machine is correctly set?
On page 29 you saw the example about Robert, the student at Avonford
College, who collected data relating to the heights of female students. This is
his corrected frequency table and his calculations so far.
Height, h
Mid-value, x
Frequency, f
xf
157.5 h 159.5
158.5
4
634.0
159.5 h 161.5
160.5
11
1765.5
161.5 h 163.5
162.5
19
3087.5
163.5 h 165.5
164.5
8
1316.0
165.5 h 167.5
166.5
5
832.5
167.5 h 169.5
168.5
3
505.5
50
8141.0
Totals
x = 8141.0 = 162.82
50
(i)
Calculate the standard deviation.
Robert’s friend Asha collected a sample of heights from 50 male PE students.
She calculated the mean and standard deviation to be 170.4 cm and 2.50 cm.
Later on they realised they had excluded two measurements. It was not
clear to which of the two data sets, Robert’s or Asha’s, the two items of data
belonged. The values were 171 cm and 166 cm. Robert felt confident about
one of the values but not the other.
(ii)
Investigate and comment.
43
12
As part of a biology experiment Thabo caught and weighed 120 minnows. He
used his calculator to find the mean and standard deviation of their weights.
Mean
Standard deviation
Exploring data
S1
1
(i)
(ii)
26.231 g
4.023 g
Find the total weight, ∑ x , of Thabo’s 120 minnows.
∑ x 2 − x 2 to find ∑ x 2 for
Use the formula standard deviation =
Thabo’s minnows.
n
Another member of the class, Sharon, did the same experiment with
minnows caught from a different stream. Her results are summarised by:
n = 80 x = 25.214
standard deviation = 3.841
Their teacher says they should combine their results into a single set but they
have both thrown away their measurements.
13
(iii)
Find n, ∑ x and ∑ x 2 for the combined data set.
(iv)
Find the mean and standard deviation for the combined data set.
A frequency diagram for a set of data is shown below. No scale is given on the
frequency axis, but summary statistics are given for the distribution.
∑f
∑ fx = 100, ∑ fx 2 = 344.
= 50,
f
0
(i)
(ii)
(iii)
(iv)
(v)
44
1
2
3
4
5
6
7
8
x
State the mode of the data.
Identify two features of the distribution.
Calculate the mean and standard deviation of the data and explain why
the value 8, which occurs just once, may be regarded as an outlier.
Explain how you would treat the outlier if the diagram represents
(a) the difference of the scores obtained when throwing a pair of
ordinary dice
(b) the number of children per household in a neighbourhood survey.
Calculate new values for the mean and standard deviation if the single
outlier is removed.
[MEI, adapted]
14
A group of 10 married couples and 3 single men found that the mean age xw
of the 10 women was 41.2 years and the standard deviation of the women’s
ages was 15.1 years. For the 13 men, the mean age xm was 46.3 years and the
standard deviation was 12.7 years.
(ii)
Find the mean age of the whole group of 23 people.
The individual women’s ages are denoted by xw and the individual men’s
ages by xm. By first finding ∑ xw2 and ∑ xm2 , find the standard deviation
for the whole group.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q4 November 2005]
15
The numbers of rides taken by two students, Fei and Graeme, at a fairground
are shown in the following table.
Student
Roller coaster
Water slide
Revolving drum
Fei
4
2
0
Graeme
1
3
6
(i)
(ii)
Working with an assumed mean
(i)
S1
1
The mean cost of Fei’s rides is $2.50 and the standard deviation of the
costs of Fei’s rides is $0. Explain how you can tell that the roller coaster
and the water slide each cost $2.50 per ride.
The mean cost of Graeme’s rides is $3.76. Find the standard deviation of
the costs of Graeme’s rides.
[Cambridge International AS and A Level Mathematics 9709, Paper 61 Q4 June 2010]
Working with an assumed mean
Human computer has it figured
mathman.com
Schoolboy, Simon Newton, astounded his classmates and their parents at a school open
evening when he calculated the average of a set of numbers in seconds while everyone else
struggled with their adding up.
Mr Truscott, a parent of one of the other children, said, ‘I was still looking for my calculator
when Simon wrote the answer on the board’.
Simon modestly said when asked about his skill ‘It’s simply a matter of choosing a good
assumed mean’.
Mathman.com wants to know ‘What is the secret method, Simon?’
Without a calculator, see if you can match Simon’s performance. The data is repeated below.
Send your result and how you did it into Mathman.com. Don’t forget – no calculators!
Number
3510
3512
3514
3516
3518
3520
Frequency
6
4
3
1
2
4
45
Simon gave a big clue about how he calculated the mean so quickly. He said
‘It’s simply a matter of choosing a good assumed mean’. Simon noticed that
subtracting 3510 from each value simplified the data significantly. This is how he
did his calculations.
Exploring data
S1
1
Number, x
Number – 3510, y
Frequency, f
x×f
3510
0
6
0×6= 0
3512
2
4
2×4= 8
3514
4
3
4 × 3 = 12
3516
6
1
6×1= 6
3518
8
2
8 × 2 = 16
3520
10
4
10 × 4 = 40
20
82
Totals
Average (mean) = 82 = 4.1
20
(3510 is now added back) 3510 + 4.1 = 3514.1
Simon was using an assumed mean to ease his arithmetic.
Sometimes it is easier to work with an assumed mean in order to find the
standard deviation.
EXAMPLE 1.9
Using an assumed mean of 7, find the true
mean and the standard deviation of the data
set 5, 7, 9, 4, 3, 8.
It doesn’t matter if the assumed
mean is not very close to the
correct value for the mean but the
closer it is, the simpler the working
will be.
SOLUTION
Let d represent the variation from the mean. So d = x – 7.
x
d=x–7
d 2 = (x – 7)2
5
5 − 7 = −2
4
7
7−7=0
0
9
9−7=2
4
4
4 − 7 = −3
9
3
3 − 7 = −4
16
8
8−7=1
1
Totals
46
∑ d = ∑ (x – 7) = –6
∑ d2 = ∑ (x – 7)2 = 34
The mean of d is given by
d
S1
1
∑ d = −6 = −1
=
sdd =
∑d 2 − ∑d 2 =
n
34 − − 6
6 6
n
2
d2
= 2.16 to 3 s.f.
So the true mean is 7 – 1 = 6.
The true standard deviation is 216 to 3 s.f.
7 is the assumed mean.
In general,
●
●
Working with an assumed mean
n
6
The standard deviation of d is given by
x = a + d where a is the assumed mean.
the standard deviation of x is the standard deviation of d.
The following example uses summary statistics, rather than the raw data values.
EXAMPLE 1.10
For a set of 10 data items, ∑(x − 9) = 7 and ∑(x − 9)2 = 17.
Find their mean and standard deviation.
SOLUTION
Let x − 9 = d
∑(x − 9) = 7
⇒
∑d = 7
⇒ d = 7 = 0.7
10
⇒ x = 9 + 0.7 = 9.7
The mean of x is 9.7.
The standard deviation of d =
∑d 2 − ∑d 2 where d = x − 9
n
The assumed
mean is 9.
n
( )
= 17 − 7
10 10
2
= 1.7 − 0.49
= 1.21
= 1.1
Since the standard deviation of x is equal to the standard deviation of d, it follows
that the standard deviation of x is 1.1.
47
Exploring data
S1
1
The next example shows you how to use an assumed mean with grouped data.
EXAMPLE 1.11
Using 162.5 as an assumed mean, find the mean and standard deviation of the
data in this table. (These are Robert’s figures for the heights of female students.)
Height, x (cm)
mid-points
Frequency, f
158.5
4
160.5
11
162.5
19
164.5
8
166.5
5
168.5
3
Total
50
SOLUTION
The working is summarised in the table below.
Height, x (cm)
mid-points
d = x − 162.5
Frequency, f
df
d 2f
158.5
−4
4
−16
64
160.5
−2
11
−22
44
162.5
0
19
0
0
164.5
2
8
16
32
166.5
4
5
20
80
168.5
6
3
18
108
50
16
328
Totals
d = 16 = 0.32
50
(sdd)2 = 328 – 0.322 = 6.4576
50
sdd = 2.54 to 3 s .f.
So the mean and standard deviation of the original data are
x = 162.5 + 0.32 = 162.82
48
sdx = 2.54 to 3 s.f.
EXERCISE 1F
1
Calculate the mean and standard deviation of the following masses, measured
to the nearest gram, using a suitable assumed mean.
Mass (g)
245–248
249–252
253–256
257–260
261–264
4
7
14
15
7
3
Frequency
2
A production line produces steel bolts which have a nominal length of
95 mm. A sample of 50 bolts is taken and measured to the nearest 0.1 mm.
Their deviations from 95 mm are recorded in tenths of a millimetre and
summarised as ∑x = −85, ∑x2 = 734. (For example, a bolt of length 94.2 mm
would be recorded as −8.)
Exercise 1F
241–244
S1
1
Find the mean and standard deviation of the x values.
(ii) Find the mean and standard deviation of the lengths of the bolts in
millimetres.
(iii) One of the figures recorded is −18. Suggest why this can be regarded as an
outlier.
(iv) The figure of −18 is thought to be a mistake in the recording. Calculate the
new mean and standard deviation of the lengths in millimetres, with the
−18 value removed.
A system is used at a college to predict a student’s A level grade in a particular
subject using their GCSE results. The GCSE score is g and the A level score is a
and for Maths in 2011 the equation of the line of best fit relating them was a =
2.6g − 9.42.
(i)
3
This year there are 66 second-year students and their GCSE scores are
summarised as ∑g = 408.6, ∑g 2 = 2545.06.
(i)
(ii)
4 (i)
Find the mean and standard deviation of the GCSE scores.
Find the mean of the predicted A level scores using the 2011 line of best fit.
Find the mode, mean and median of:
2
8
6
5
4
5
6
3
6
4
9
1
5
6
5
Hence write down, without further working, the mode, mean and median of:
(ii)
(iii)
(iv)
5
20 80 60 50 40
12 18 16 15 14
4 16 12 10 8
50 60 30 60 40
15 16 13 16 14
10 12 6 12 8
90 10 50 60 50
19 11 15 16 15
18 2 10 12 10
A manufacturer produces electrical cable which is sold on reels. The reels are
supposed to hold 100 metres of cable. In the quality control department the
length of cable on randomly chosen reels is measured. These measurements
are recorded as deviations, in centimetres, from 100 m. (So, for example, a
length of 99.84 m is recorded as –16.)
49
Exploring data
S1
1
For a sample of 20 reels the recorded values, x, are summarised by
∑ x = − 86
∑ x 2 = 4281
Calculate the mean and standard deviation of the values of x.
(ii) Later it is noticed that one of the values of x is −47, and it is thought that
so large a value is likely to be an error. Give a reason to support this view.
(iii) Find the new mean and standard deviation of the values of x when the
value −47 is discarded.
6 On her summer holiday, Felicity recorded the temperatures at noon each day
for use in a statistics project. The values recorded, f degrees Fahrenheit, were
as follows, correct to the nearest degree.
(i)
47
59
68 62
49
67
66 73
70
68
74 84
80
72
Represent Felicity’s data on a stem-and-leaf diagram. Comment on the
shape of the distribution.
(ii) Using a suitable assumed mean, find the mean and standard deviation of
Felicity’s data.
For a set of ten data items, ∑(x – 20) = – 140 and ∑(x – 20)2 = 2050. Find
their mean and standard deviation.
(i)
7
8
9
10
11
For a set of 20 data items, ∑(x + 3) = 140 and
mean and standard deviation.
∑(x + 3)2 = 1796. Find their
For a set of 15 data items, ∑(x + a) = 156 and
of these values is 5.4.
Find the value of a and the standard deviation.
∑(x + a)2 = 1854. The mean
For a set of 10 data items, ∑(x – a) = – 11 and
of these values is 5.9.
Find the value of a and the standard deviation.
∑(x – a)2 = 75. The mean
The length of time, t minutes, taken to do the crossword in a certain
newspaper was observed on 12 occasions. The results are summarised below.
∑(t − 35) = −15 ∑(t − 35)2 = 82.23
Calculate the mean and standard deviation of these times taken to do the
crossword.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q1 June 2007]
12
A summary of 24 observations of x gave the following information:
∑(x − a) = −73.2
and
∑(x − a)2 = 2115.
The mean of these values of x is 8.95.
(i)
(ii)
Find the value of the constant a.
Find the standard deviation of these values of x.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q1 November 2007]
50
KEY POINTS
1
S1
1
An item of data x may be identifed as an outlier if
x − x 2 × standard deviation.
2
Categorical data are non-numerical; discrete data can be listed; continuous
data can be measured to any degree of accuracy and it is not possbile to list
all values.
3
Stem-and-leaf diagrams (or stemplots) are suitable for discrete or
continuous data. All data values are retained as well as indicating
properties of the distribution.
4
The mean, median and the mode or modal class are measures of central
tendency.
5
The mean, x =
∑ x . For grouped data x = ∑ xf .
n
∑f
6
The median is the mid-value when the data are presented in rank order; it
is the value of the n + 1th item of n data items.
2
7
The mode is the most common item of data. The modal class is the class
containing the most data, when the classes are of equal width.
8
The range, variance and standard deviation are measures of spread or
variation or dispersion.
9
Range = maximum data value – minimum data value.
10
The standard deviation =
11
An alternative form is
∑(x − x )2 f
standard deviation =
12
Key points
That is, if x is more than two standard deviations above or below the
sample mean.
n
∑x 2f
n
or
∑(x − x )2 f
∑f
− x 2 or
∑x 2f
∑f
− x2
Working with an assumed mean a,
x =a+d
where a is the assumed mean and d is the deviation from the assumed
mean and the standard deviation of x is the standard deviation of d.
51
Representing and interpreting data
S1
2
Representing and
interpreting data
2
A picture is worth a thousand numbers.
Anon
Latest news from Alpha High
The Psychology department at Alpha High have found that girls have more
on-line friends than boys. Mr Rama, the head of department, said ‘The results are
quite marked, girls in all age groups with the exception of the youngest age group
had significantly more on-line friends. We have several hypotheses to explain
this, but want to do more research before we draw any conclusions.’ The group
of student psychologists have won first prize in a competition run by Psychology
Now for their research.
The Psychology department are now intending to compare these results with the
number of friends that students have ‘real-life’ contact with.
200
mean number of friends
girls
150
boys
100
50
0
?
●
15 years
16 years
17 years
age group
18 years
What is the mean number of friends for 17-year-old girls?
220 students aged 17 were surveyed from Alpha High. 120 of these students were
girls. What is the overall mean number of friends for the 17-year-olds?
52
Most raw data need to be summarised to make it easier to see any patterns that
may be present. You will often want to draw a diagram too. The Psychology
department used the following table to construct the diagram for their article.
15 years
16 years
17 years
18 years
Sample size
200
170
220
310
Mean number of friends – girls
40
70
170
150
Mean number of friends – boys
50
60
110
130
Histograms
Age group
S1
2
You will often want to use a diagram to communicate statistical findings. People
find diagrams to be a very useful and easy way of presenting and understanding
statistical information.
Histograms
Histograms are used to illustrate continuous data. The columns in a histogram
may have different widths and the area of each column is proportional to the
frequency. Unlike bar charts, there are no gaps between the columns because
where one class ends, the next begins.
Continuous data with equal class widths
A sample of 60 components is taken from a production line and their
diameters, d mm, recorded. The resulting data are summarised in the
following frequency table.
Diameter (mm)
Frequency
25 d 30
1
30 d 35
3
35 d 40
7
40 d 45
15
45 d 50
17
50 d 55
10
55 d 60
5
60 d 65
2
53
S1
2
18
Representing and interpreting data
16
14
frequency density
(components/5 mm)
12
10
8
6
The vertical axis shows
the frequency density.
In this example it is
frequency per 5 mm.
4
2
0
25
30
35
40
45
50
diameter (mm)
55
60
65
Figure 2.1 Histogram to show the distribution of component diameters
The class boundaries are 25, 30, 35, 40, 45, 50, 55, 60 and 65. The width of each
class is 5.
The area of each column is proportional to the class frequency. In this example
the class widths are equal so the height of each column is also proportional to the
class frequency.
The column representing 45 d 50 is the highest and this tells you that this is
the modal class, that is, the class with highest frequency per 5 mm.
?
●
How would you identify the modal class if the intervals were not of equal width?
Labelling the frequency axis
The vertical axis tells you the frequency density. Figure 2.3 looks the same as
figure 2.2 but it is not a histogram. This type of diagram is, however, often
incorrectly referred to as a histogram. It is more correctly called a frequency
chart. A histogram shows the frequency density on the vertical axis.
54
18
3.2
16
2.8
14
2.4
12
frequency
3.6
2.0
1.6
1.2
10
8
6
0.8
4
0.4
0.0
2
0
25 30 35 40 45 50 55 60 65
diameter (mm)
S1
2
Histograms
frequency density
(components/1 mm)
The frequency density in
figure 2.2 is frequency per 1 mm.
25 30 35 40 45 50 55 60 65
diameter (mm)
Figure 2.3
Figure 2.2
Comparing Figures 2.1 and 2.2, you will see that the shape of the distribution
remains the same but the values on the vertical axes are different. This is because
different units have been used for the frequency density.
Continuous data with unequal class widths
The heights of 80 broad bean plants were measured, correct to the nearest
centimetre, ten weeks after planting. The data are summarised in the following
frequency table.
Height (cm)
Frequency
Class width (cm)
Frequency density
7.5 x 11.5
1
4
0.25
11.5 x 13.5
3
2
1.5
13.5 x 15.5
7
2
3.5
15.5 x 17.5
11
2
5.5
17.5 x 19.5
19
2
9.5
19.5 x 21.5
14
2
7
21.5 x 23.5
13
2
6.5
23.5 x 25.5
9
2
4.5
25.5 x 28.5
3
3
1
When the class widths are unequal you can use
frequency density =
frequency
class width
55
S1
2
10
Representing and interpreting data
9
frequency density
(number of beans/class width)
8
7
6
5
4
3
2
1
0
8.0
10.0
12.0
14.0
16.0
18.0 20.0
height (cm)
22.0
24.0
26.0
28.0
30.0
Figure 2.4
Discrete data
!
Histograms are occasionally used for grouped discrete data. However, you should
always first consider the alternatives.
A test was given to 100 students. The maximum mark was 70. The raw data are
shown below.
10
9
20
35
70
40
46
38
62
18
18
46
16
40
52
48
50
39
47
30
68
53
29
45
21
45
8
53
58
32
67
57
13
48
25
38
25
45
54
45
25
30
31
54
53
51
56
42
59
49
62
63
56
50
41
25
18
42
25
28
49
34
9
34
29
52
20
61
24
31
11
21
34
32
63
55
36
55
53
27
12
68
45
47
43
47
36
30
42
54
8
31
55
60
50
46
9
38
61
38
Illustrating this data using a vertical line graph results in figure 2.5.
56
frequency
6
5
4
3
2
1
S1
2
10
20
30
40
mark
50
60
Histograms
0
70
Figure 2.5
This diagram fails to give a clear picture of the overall distribution of marks.
In this case you could consider a bar chart or, as the individual marks are known,
a stem-and-leaf diagram, as follows.
n = 100
2
5 represents 25 marks
0
1
2
3
4
5
6
7
8
0
0
0
0
0
0
0
8
1
0
0
0
0
1
9
2
1
0
1
0
1
9
3
1
1
2
1
2
9
6
4
1
2
2
2
8
5
1
2
2
3
8
5
2
3
3
3
8
5
2
5
3
7
5
4
5
3
8
5
4
5
3
8
7
4
5
4
8
5
5
4
9
6
6
4
9
6 8 8 8 8 9
6 6 7 7 7 8 8 9 9
5 5 5 6 6 7 8 9
Figure 2.6
If the data have been grouped and the original data have been lost, or are
otherwise unknown, then a histogram may be considered. A grouped frequency
table and histogram illustrating the marks are shown below.
Frequency, f
0–9
5
10–19
8
20–29
14
30–39
19
40–49
22
50–59
21
60–70
11
2.5
2.0
frequency density
(students per mark)
Marks, x
1.5
1.0
0.5
0
Figure 2.7
10
20
30
40
marks (x)
50
60
70
57
Note
S1
2
The class boundary 10–19 becomes 9.5 x 19.5 for the purpose of drawing the
histogram. You must give careful consideration to class boundaries, particularly if
Representing and interpreting data
you are using rounded data.
?
●
Look at the intervals for the first and last classes. How do they differ from the
others? Why is this the case?
Grouped discrete data are illustrated well by a histogram if the distribution is
particularly skewed as is the case in the next example.
The first 50 positive integers squared are:
Number, n
Frequency, f
0 n 250
15
250 n 500
7
500 n 750
5
750 n 1000
4
1000 n 1250
4
1250 n 1500
3
1500 n 1750
3
1750 n 2000
3
2000 n 2250
3
2250 n 2500
3
4
100
324
676
1156
1764
2500
9
121
361
729
1225
1849
16
144
400
784
1296
1936
36
196
484
900
1444
2116
49
225
529
961
1521
2209
64
256
576
1024
1600
2304
12
8
4
0
0
Figure 2.8
58
25
169
441
841
1369
2025
16
frequency density
(square numbers/250)
1
81
289
625
1089
1681
2401
500
1000
n
1500
2000
2500
The main points to remember when drawing a histogram are:
Histograms are usually used for illustrating continuous data. For discrete data
it is better to draw a stem-and-leaf diagram, line graph or bar chart.
●
Since the data are continuous, or treated as if they were continuous, adjacent
columns of the histogram should touch (unlike a bar chart where the columns
should be drawn with gaps between them).
●
It is the areas and not the heights of the columns that are proportional to the
frequency of each class.
●
The vertical axis should be marked with the appropriate frequency density
(frequency per 5 mm for example), rather than frequency.
1
A number of trees in two woods were measured. Their diameters, correct to
the nearest centimetre, are summarised in the table below.
Diameter (cm)
1–10
11–15
16–20
21–30
31–50
Total
Mensah’s Wood
10
5
3
11
1
30
6
8
20
5
1
40
Ashanti Forest
S1
2
Exercise 2A
EXERCISE 2A
●
(Trees less than 12 cm in diameter are not included.)
(i)
(ii)
(iii)
(iv)
2
Write down the actual class boundaries.
Draw two separate histograms to illustrate this information.
State the modal class for each wood.
Describe the main features of the distributions for the two woods.
Listed below are the prime numbers, p, from 1 up to 1000. (1 itself is not
usually defined as a prime.)
Primes up to 1000
2
47
3
5
7
11
13
17
19
23
29
31
37
41
43
53
59
61
67
71
73
79
83
89
97 101 103 107
109
113 127 131 137 139 149 151 157 163 167 173 179 181
191
193 197 199 211 223 227 229 233 239 241 251 257 263
269
271 277 281 283 293 307 311 313 317 331 337 347 349
353
359 367 373 379 383 389 397 401 409 419 421 431 433
439
443 449 457 461 463 467 479 487 491 499 503 509 521
523
541 547 557 563 569 571 577 587 593 599 601 607 613
617
619 631 641 643 647 653 659 661 673 677 683 691 701
709
719 727 733 739 743 751 757 761 769 773 787 797 809
811
821 823 827 829 839 853 857 859 863 877 881 883 887
907
911 919 929 937 941 947 953 967 971 977 983 991 997
59
(i)
Representing and interpreting data
S1
2
(ii)
3
A crate containing 270 oranges was opened
and each orange was weighed to the
nearest gram. The masses were found
to be distributed as in this table.
(i)
(ii)
4
(ii)
Number of
oranges
60–99
20
100–119
60
120–139
80
140–159
50
160–219
60
Length (cm)
Number of
plants
20–31
30
32–37
80
38–43
90
44–49
60
50–67
60
Draw a histogram to illustrate the data.
From the table, calculate an estimate
of the mean length of stem of a plant
from this experiment.
The lengths of time of sixty songs recorded by a certain group of singers are
summarised in the table below.
Song length in seconds
(x)
(i)
(ii)
60
Mass (grams)
Draw a histogram to illustrate the data.
From the table, calculate an estimate
of the mean mass of an orange from
this crate.
In an agricultural experiment,
320 plants were grown on a plot, and the
lengths of the stems were measured to the
nearest centimetre ten weeks after
planting. The lengths were found to be
distributed as in this table.
(i)
5
Draw a histogram to illustrate these data with the following class intervals:
1 p 20
20 p 50
50 p 100
100 p 200
200 p 300
300 p 500 and 500 p 1000.
Comment on the shape of the distribution.
Number of songs
0 x 120
1
120 x 180
9
180 x 240
15
240 x 300
17
300 x 360
13
360 x 600
5
Display the data on a histogram.
Determine the mean song length.
6
A random sample of 200 batteries, of nominal potential 6 V, was taken from a
very large batch of batteries. The potential difference between the terminals of
each battery was measured, resulting in the table of data below.
Number of batteries
5.80
1
5.85
4
5.90
22
5.95
42
6.00
60
6.05
44
6.10
24
6.15
2
6.20
1
Exercise 2A
Potential difference in volts (mid-interval value)
S1
2
Calculate the mean and standard deviation of these voltages and illustrate the
data on a histogram. Mark clearly on the histogram the mean voltage and the
voltages which are two standard deviations either side of the mean.
[MEI]
7
After completing a long assignment, a student was told by his tutor that it was
more like a book than an essay. He decided to investigate how many pages
there are in a typical book and started by writing down the numbers of pages
in the books on one of his shelves, as follows.
256 128 160 128 192
464 128 96 96 556
940 676 128 196 640
(i)
(ii)
8
64 356 96 64 160
148 64 192 96 512
44 64 144 256 72
Look carefully at the data and state, giving your reasons, whether they are
continuous or discrete. Give an explanation for your answer.
Decide on the most helpful method of displaying the data and draw the
appropriate diagram.
As part of a data collection exercise, members of a certain school year group
were asked how long they spent on their Mathematics homework during one
particular week. The times are given to the nearest 0.1 hour. The results are
displayed in the following table.
Time spent
0.1 t 0.5 0.6 t 1.0 1.1 t 2.0 2.1 t 3.0 3.1 t 4.5
(t hours)
Frequency
(i)
(ii)
11
15
18
30
21
Draw, on graph paper, a histogram to illustrate this information.
Calculate an estimate of the mean time spent on their Mathematics
homework by members of this year group.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q5 June 2008]
61
Representing and interpreting data
S1
2
Measures of central tendency and of spread using quartiles
You saw in Chapter 1 how to find the median of a set of discrete data. As a
reminder, the median is the value of the middle item when all the data items have
been ranked in order.
The median is the value of the
n +1
th item and is half-way through the data set.
2
The values one-quarter of the way through the data set and three-quarters of
the way through the data set are called the lower quartile and the upper quartile
respectively. The lower quartile, median and upper quartile are usually denoted
using Q1, Q2 and Q3.
Quartiles are used mainly with large data sets and their values found by looking
at the 14 , 12 and 43 points. So, for a data set of 1000, you would take Q1 to be the
value of the 250th data item, Q2 to be the value of the 500th data item and Q3 to
be the value of the 750th data item.
Quartiles for small data sets
For small data sets, where each data item is known (raw data), calculation of
the middle quartile Q2, the median, is straightforward. However, there are no
standard formulae for the calculation of the lower and upper quartiles, Q1 and
Q3, and you may meet different ones. The one we will use is consistent with
the output from some calculators which display the quartiles of a data set and
depends on whether the number of items, n, is even or odd.
If n is even then there will be an equal number of items in the lower half and
upper half of the data set. To calculate the lower quartile, Q1, find the median of
the lower half of the data set. To calculate the upper quartile, Q3, find the median
of the upper half of the data set.
For example, for the data set {1, 3, 6, 10, 15, 21, 28, 36, 45, 55} the median, Q2,
is 15 + 21 = 18. The lower quartile, Q1, is the median of {1, 3, 6, 10, 15}, i.e. 6.
2
The upper quartile, Q3, is the median of {21, 28, 36, 45, 55}, i.e. 36.
If n is odd then define the ‘lower half’ to be all data items below the median.
Similarly define the ‘upper half’ to be all data items above the median. Then
proceed as if n were even.
For example, for the data set {1, 3, 6, 10, 15, 21, 28, 36, 45} the median, Q2, is
15. The lower quartile, Q1, is the median of {1, 3, 6, 10}, i.e. 3 + 6 = 4.5. The
2
upper quartile, Q3, is the median of {21, 28, 36, 45}, i.e. 28 + 36 = 32.
2
62
!
Use a spreadsheet to find the median and quartiles of a small data set. Find out
the method the spreadsheet uses to determine the position of the lower and
upper quartiles.
ACTIVITY 2.1
Catherine is a junior reporter. As part of an investigation into consumer affairs
she purchases 0.5 kg of chicken from 12 shops and supermarkets in the town.
The resulting data, put into rank order, are as follows:
EXAMPLE 2.1
$1.39 $1.39 $1.46 $1.48 $1.48 $1.50 $1.52 $1.54 $1.60 $1.65 $1.68 $1.72
Find Q1, Q2 and Q3.
SOLUTION
139
139
146
148
148
150
152
154
160
165
168
172
1
2
3
4
5
6
7
8
9
10
11
12
Q1 has position
6 + 1 = 31
2.
2
Value = $1.47.
Q2 has position
12 + 1 = 6 1
2.
2
Value = $1.51.
S1
2
Measures of central tendency and of spread using quartiles
The definition of quartiles on a spreadsheet may be different from that described
above. Values of Q1 and Q3 in the even case shown above are given as 7 and 34
respectively on an Excel spreadsheet. Similarly, values of Q1 and Q3 in the odd
case shown above are given as 6 and 28 respectively.
6 + 1 = 31
2
2
from the top.Value = $1.63.
Q3 has position
In fact, the upper quartile has a value of $1.625 but this has been rounded up to
the nearest cent.
!
?
●
You may encounter different formulae for finding the lower and upper quartiles.
The ones given here are relatively easy to calculate and usually lead to values of
Q1 and Q3 which are close to the true values.
What are the true values?
Interquartile range or quartile spread
The difference between the lower and upper quartiles is known as the
interquartile range or quartile spread.
●
Interquartile range (IQR) = Q3 − Ql.
63
Representing and interpreting data
S1
2
In Example 2.1 IQR = 163 − 147 = 16 cents.
The interquartile range covers the middle 50% of the data. It is relatively easy to
calculate and is a useful measure of spread as it avoids extreme values. It is said to
be resistant to outliers.
Box-and-whisker plots (boxplots)
The three quartiles and the two extreme values of a data set may be illustrated in
a box-and-whisker plot. This is designed to give an easy-to-read representation of
the location and spread of a distribution. Figure 2.9 shows a box-and-whisker
plot for the data in Example 2.1.
139
135
140
147
145
151
150
163
155
price (cents)
160
172
165
170
175
Figure 2.9
The box represents the middle 50% of the distribution and the whiskers stretch
out to the extreme values.
Outliers
In Chapter 1 you met a definition of an outlier based on the mean and standard
deviation. A different approach gives the definition of an outlier in terms of the
median and interquartile range (IQR).
Data which are more than 1.5 × IQR beyond the lower or upper quartiles are
regarded as outliers.
The corresponding boundary values beyond which outliers may be found are
Q1 − 1.5 × (Q3 − Q1)
and
Q3 + 1.5 × (Q3 − Q1).
For the data relating to the ages of the cyclists involved in accidents discussed in
Chapter 1, for all 92 data values Q1 = 13.5 and Q3 = 45.5.
Hence Q1 − 1.5 × (Q3 − Q1) = 13.5 − 1.5 × (45.5 − 13.5)
= 13.5 − 1.5 × 32
= −34.5
and
Q3 + 1.5 × (Q3 − Q1) = 45.5 + 1.5 × (45.5 − 13.5)
= 45.5 + 1.5 × 32
= 93.5
64
From these boundary values you will see that there are no outliers at the lower
end of the range, but the value of 138 is an outlier at the upper end of the range.
Figure 2.10 shows a box-and-whisker plot for the ages of the cyclists with the
outlier removed. For the remaining 91 data items Q1 = 13 and Q3 = 45.
6 13
45
25
88
50
75
age (years)
138
100
125
150
Figure 2.10
From the diagram you can see that the distribution has positive or right skewness.
The indicates an outlier and is above the upper quartile. Outliers are usually
labelled as they are often of special interest. The whiskers are drawn to the most
extreme data points which are not outliers.
Cumulative frequency curves
0
22
S1
2
Cumulative frequency curves
When working with large data sets or grouped data, percentiles and quartiles can
be found from cumulative frequency curves as shown in the next section.
Sheuligirl
I am a student trying to live on a small allowance. I’m trying my best to allow myself a
sensible monthly budget but my lecturers have given me a long list of textbooks to buy.
If I buy just half of them I will have nothing left to live on this month. The majority of
books on my list are over $16.
I want to do well at my studies but I won’t do well without books and I won’t do well if
I am ill through not eating properly.
Please tell me what to do, and don’t say ‘go to the library’ because the books I need are
never there.
After reading this opening
post a journalist wondered if
there was a story in it. He
decided to carry out a survey
of the prices of textbooks in a
large shop. The reporter took
a large sample of 470
textbooks and the results are
summarised in the table.
Cost, C ($)
Frequency (no. of books)
0 C 10
13
10 C 15
53
15 C 20
97
20 C 25
145
25 C 30
81
30 C 35
40
35 C 40
23
40 C 45
12
45 C 50
6
65
Representing and interpreting data
S1
2
He decided to estimate the median and the upper and lower quartiles of the costs
of the books. (Without the original data you cannot find the actual values so all
calculations will be estimates.) The first step is to make a cumulative frequency
table, then to plot a cumulative frequency curve.
Cost, C ($)
Frequency
Cost
Cumulative
frequency
0 C 10
13
C 10
13
10 C 15
53
C 15
66
See note 1.
15 C 20
97
C 20
163
See note 2.
20 C 25
145
C 25
308
25 C 30
81
C 30
389
30 C 35
40
C 35
429
35 C 40
23
C 40
452
40 C 45
12
C 45
464
45 C 50
6
C 50
470
Notes
1 Notice that the interval
500
C 15 means 0 C 15
and so includes the 13 books
450
in the interval 0 C < 10 and
the 53 books in the interval
400
10 C 15, giving 66 books
in total.
your previous total, giving
you 66 + 97 = 163.
A cumulative frequency
curve is obtained by plotting
the upper boundary of each
class against the cumulative
frequency. The points are
joined by a smooth curve, as
shown in figure 2.11.
66
250
Q2
200
150
Q1
100
50
0
0
10
Figure 2.11
$27
the interval 15 C 20 to
$22.50
add the number of books in
300
$18
the interval C 20 you must
cumulative frequency
2 Similarly, to find the total for
Q3
350
20
30
cost ($)
40
50
!
The 235th ( 470
2 ) item of data identifies the median which has a value of about
$22.50. The 117.5th ( 470
4 ) item of data identifies the lower quartile, which has a
value of about $18 and the 352.5th ( 43 × 470) item of data identifies the upper
quartile, which has a value of about $27.
S1
2
Cumulative frequency curves
In this example the actual values are unknown and the median must therefore be
an estimate. It is usual in such cases to find the estimated value of the n th item.
2
This gives a better estimate of the median than is obtained by using
n + 1, which is used for ungrouped data. Similarly, estimates of the lower
2
and upper quartiles are found from the n th and 3n th items.
4
4
Notice the distinctive shape of the cumulative frequency curve. It is like a
stretched out S-shape leaning forwards.
What about Sheuligirl’s claim that the majority of textbooks cost more than $16?
Q1 = $18. By definition 75% of books are more expensive than this, so
Sheuligirl’s claim seems to be well founded. We need to check exactly how many
books are estimated to be more
expensive than $16.
500
From the cumulative frequency
curve 85 books cost $16 or less
(figure 2.12). So 385 books or
about 82% are more expensive.
450
400
!
You should be cautious about
any conclusions you draw.
This example deals with books,
many of which have prices like
$9.95 or $39.99. In using
a cumulative frequency curve
you are assuming an even
spread of data throughout
the intervals and this may not
always be the case.
cumulative frequency
350
300
250
200
150
100
85 books
50
0
0
10
20
30
cost ($)
40
50
Figure 2.12
67
Box-and-whisker plots for grouped data
S1
2
Representing and interpreting data
It is often helpful to draw a box-and-whisker plot. In cases when the extreme
values are unknown the whiskers are drawn out to the 10th and 90th percentiles.
Arrows indicate that the minimum and maximum values are further out.
150
0
0
10
20
$27
$18
50
$22.50
100
30
40
50
cost ($)
Figure 2.13
EXAMPLE 2.2
A random sample of people were asked how old they were when they first met
their partner. The histogram represents this information.
8
frequency density (people per year)
7
6
5
4
3
2
1
0
10
Figure 2.14
68
20
30
40
50
60
age (years)
70
80
90
100
(i)
(ii)
(iii)
(iv)
SOLUTION
(i)
(ii)
The bar with the greatest frequency density represents the modal age group.
So the modal age group is 20 a 30.
frequency
Frequency density =
class width
So,
Frequency = frequency density × class width
Age (years)
Frequency density
Class width
0 a 20
4
20
4 × 20 = 80
20 a 30
7
10
7 × 10 = 70
30 a 40
4.2
10
4.2 × 10 = 42
40 a 50
2.8
10
2.8 × 10 = 28
50 a 60
1
10
1 × 10 = 10
60 a 100
0.5
40
0.5 × 40 = 20
S1
2
Cumulative frequency curves
What is the modal age group?
How many people took part in the survey?
Find an estimate for the mean age that a person first met their partner.
Draw a cumulative frequency curve for the data and use the curve to provide
an estimate for the median.
Frequency
The total number of people is 80 + 70 + 42 + 28 + 10 + 20 = 250
!
(iii)
You will see that the class with the greatest frequency is 0 a 20, with 80
people. However, this is not the modal class because its frequency density
of 4 people per year is lower than the frequency density of 7 people per year
for the 20 a 30 class. The modal class is that with the highest frequency
density. The class width for 0 a 20 is twice that for 20 a 30 and
this is taken into account in working out the frequency density.
To find an estimate for the mean, work out the mid-point of each class
multiplied by its frequency; then sum the results and divide the answer by
the total frequency.
Estimated mean = 80 × 10 + 70 × 25 + 42 × 35 + 28 × 45 + 10 × 55 + 20 × 80
250
7430
=
250
= 29.7 years to 3 s.f.
69
(iv)
S1
2
250
Representing and interpreting data
225
200
cumulative frequency
175
150
125
100
75
50
25
0
10
20
30
40
50
60
age (years)
70
80
90
100
Figure 2.15
The median age is 26 years.
EXAMPLE 2.3
These are the times, in seconds, that 15 members of an athletics club took to run
800 metres.
139
182
145
(i)
(ii)
(iii)
70
148
154
178
151
171
132
140
157
148
162
142
166
Draw a stem-and-leaf diagram of the data.
Find the median, the upper and lower quartiles and the interquartile range.
Draw a box-and-whisker plot of the data.
SOLUTION
S1
2
n = 15
(i)
13
2 represents 132 seconds
13
14
15
16
17
18
2
0
1
2
1
2
Exercise 2B
9
2 5 8 8
4 7
6
8
There are 15 data values, so the median is the 8th data value.
So the median is 151 seconds.
(ii)
The upper quartile is the median of the upper half of the data set.
So the upper quartile is 166 seconds.
The lower quartile is the median of the lower half of the data set.
So the lower quartile is 142 seconds.
Interquartile range = upper quartile − lower quartile
= 166 − 142
= 24 seconds
(iii)
Draw a box that starts at the lower quartile and ends at the upper quartile.
Add a line inside the box to show the position of the median.
Extend the whiskers to the greatest and least values in the data set.
130
140
150
160
time (seconds)
170
180
190
Figure 2.16
EXERCISE 2B
1
For each of the following data sets, find
(a)
(b)
(c)
(d)
(e)
the range
the median
the lower and upper quartiles
the interquartile range
any outliers.
71
(i)
Representing and interpreting data
S1
2
(ii)
(iii)
(iv)
2 (i)
6 8 3
12 5 17
18 12 8
25 28 33
115 123
125 121
2 1
5 4 6 8 5
6 7 8 8
11 4
10 12 19 12 5
9 15 11 16
9 11
12 14 8 14 7
14 37
19 23 27 25 28
132 109 127
116 141 132 114 109
117 118 117
116 123 105 125
6
8
6
For the following data set, find the median and interquartile range.
2 8 4 6 3
5 1 8 2 5
8 0 3 7 8 5
Use your answers to part (i) to deduce the median and interquartile range for
each of the following data sets.
(ii)
(iii)
(iv)
3
32 38 34 36 33
20 80 40 60 30
50 110 70 90 60
35 31 38 32 35
50 10 80 20 50
80 40 110 50 80
Find
(i)
(ii)
(iii)
38 30 33 37 38 35
80 0 30 70 80 50
110 30 60 100 110 80
Score
the median
the upper and lower quartiles
the interquartile range
for the scores of golfers in the first round
of a competition.
Tally
70
71
72
73
74
75
76
77
78
79
80
81
82
(iv)
(v)
Illustrate the data with a box-and-whisker plot.
The scores for the second round are illustrated on the box-and-whisker
plot below. Compare the two and say why you think the differences might
have arisen.
67
72
68
70
74
77
4
The numbers of goals scored by a hockey team in its matches one season are
illustrated on the vertical line chart below.
5
Exercise 2B
4
matches
S1
2
3
2
1
0
1
2
3
4
goals
5
6
7
Draw a box-and-whisker plot to illustrate the same data.
State, with reasons, which you think is the better method of display in
this case.
(i)
(ii)
5
0
One year the yields, y, of a number of walnut trees were recorded to the
nearest kilogram as follows.
Yield, y (kg)
Frequency
40 y 50
1
50 y 60
5
60 y 70
7
70 y 80
4
80 y 90
2
90 y 100
1
(i)
(ii)
(iii)
(iv)
Construct the cumulative frequency table for these data.
Draw the cumulative frequency graph.
Use your graph to estimate the median and interquartile range of the yields.
Draw a box-and-whisker plot to illustrate the data.
The piece of paper where the actual figures had been recorded was then found,
and these were:
44 59 67 76 52
85 93 56 65 74
(v)
(vi)
62 68 78 53 63
69 82 53 65 70
Use these data to find the median and interquartile range and compare
your answers with those you obtained from the grouped data.
What are the advantages and disadvantages of grouping data?
73
The intervals of time, t seconds, between successive emissions from a weak
radioactive source were measured for 200 consecutive intervals, with the
following results.
Interval (t seconds)
Representing and interpreting data
S1
2
6
0t5
5 t 10
10 t 15
15 t 20
23
67
42
26
20 t 25
25 t 30
30 t 35
21
15
6
Frequency
Interval (t seconds)
Frequency
(i)
(ii)
(iii)
7
Draw a cumulative frequency graph for this distribution.
Use your graph to estimate
(a) the median
(b) the interquartile range.
Calculate an estimate of the mean of the distribution.
In a sample of 800 eggs from an egg farm each egg was weighed and classified
according to its mass, m grams. The frequency distribution was as follows.
Mass in grams
40 m 45
45 m 50
50 m 55
Number of eggs
36
142
286
Mass in grams
55 m 60
60 m 65
65 m 70
Number of eggs
238
76
22
Draw a cumulative frequency graph of the data, using a scale of 2 cm to
represent 5 grams on the horizontal axis (which should be labelled from
40 to 70 grams) and a scale of 2 cm to represent 100 eggs on the vertical axis.
Use your graph to estimate for this sample
the percentage of eggs which would be classified as large (over 62 grams)
the median mass of an egg
(iii) the interquartile range.
Indicate clearly on your diagram how you arrive at your results.
(i)
(ii)
8
The table summarises the observed lifetimes, x, in seconds, of 50 fruit flies
subjected to a new spray in a controlled experiment.
Interval
Mid-interval value
Frequency
0.5 x 5.5
3
3
5.5 x 10.5
8
22
10.5 x 15.5
13
12
15.5 x 20.5
18
9
20.5 x 25.5
23
2
25.5 x 30.5
28
1
30.5 x 35.5
33
1
74
(i)
9
During January the numbers of people entering a store during the first hour
after opening were as follows.
(i)
(ii)
(iii)
(iv)
(v)
Time after opening,
x minutes
Frequency
Cumulative
frequency
0 x 10
210
210
10 x 20
134
344
20 x 30
78
422
30 x 40
72
a
40 x 60
b
540
S1
2
Exercise 2B
(ii)
Making clear your methods and showing all your working, estimate
the mean and standard deviation of these lifetimes. Give your answers
correct to 3 significant figures and do not make any corrections for
grouping.
Draw the cumulative frequency graph and use it to estimate the
minimum lifetime below which 70% of all lifetimes lie.
Find the values of a and b.
Draw a cumulative frequency graph to represent this information. Take a
scale of 2 cm for 10 minutes on the horizontal axis and 2 cm for 50 people
on the vertical axis.
Use your graph to estimate the median time after opening that people
entered the store.
Calculate estimates of the mean, m minutes, and standard deviation,
s minutes, of the time after opening that people entered the store.
Use your graph to estimate the number of people entering the store
between (m – 12s) and (m + 12s) minutes after opening.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q6 June 2009]
10
The numbers of people travelling on a certain bus at different times of the
day are as follows.
17 5 2 23 16 31 8
22 14 25 35 17 27 12
6 23 19 21 23 8 26
(i)
(ii)
(iii)
Draw a stem-and-leaf diagram to illustrate the information given above.
Find the median, the lower quartile, the upper quartile and the
interquartile range.
State, in this case, which of the median and mode is preferable as a
measure of central tendency, and why.
[Cambridge International AS and A Level Mathematics 9709, Paper 61 Q2 June 2010]
75
Representing and interpreting data
S1
2
KEY POINTS
1
Histograms:
●
commonly used to illustrate continuous data
●
horizontal axis shows the variable being measured (cm, kg, etc.)
●
vertical axis labelled frequency density where
frequency density = frequency
class width
2
●
no gaps between columns
●
the frequency is proportional to the area of each column.
For a small data set with n items,
the median, Q2, is the value of the n + 1th item of data.
2
If n is even then
●
●
the lower quartile, Q1, is the median of the lower half of the data set
●
the upper quartile, Q3, is the median of the upper half of the data set.
If n is odd then exclude the median from either ‘half ’ and proceed as if n
were even.
3
Interquartile range (IQR) = Q3 − Q1.
4
When data are illustrated using a cumulative frequency curve the median
and the lower and upper quartiles are estimated by identifying the data
values with cumulative frequencies 12n, 14 n and 43n .
5
An item of data x may be identified as an outlier if it is more than
1.5 × IQR beyond the lower or upper quartile, i.e. if
x Q1 − 1.5 × (Q3 – Q1) or x Q3 + 1.5 × (Q3 − Q1).
6
76
A box-and-whisker plot is a useful way of summarising data and showing
the median, upper and lower quartiles and any outliers.
3
Probability
S1
3
the desert.
Stephen Jay Gould
; ;
Probability
If we knew Lady Luck better, Las Vegas would still be a roadstop in
thelibrarian.com ; ;
A library without books
If you plan to pop into your local library
and pick up the latest bestseller, then forget
it. All the best books ‘disappear’ practically
as soon as they are put on the shelves.
I talked about the problem with the local
senior librarian, Gina Clarke.
‘We have a real problem with unauthorised
loans at the moment,’ Gina told me. ‘Out
of our total stock of, say 80 000 books,
something like 44 000 are out on loan at any
one time. About 20 000 are on the shelves
and I’m afraid the rest are unaccounted for.’
Librarian Gina Clarke is worried about the
problem of ‘disappearing books’
That means that the probability of finding the particular book you want is exactly 1–4 . With
odds like that, don’t bet on being lucky next time you visit your library.
How do you think the figure of 14 at the end of the article was arrived at? Do you
agree that the probability is exactly 14 ?
The information about the different categories of book can be summarised as
follows.
Category of book
Typical numbers
On the shelves
20 000
Out on loan
44 000
Unauthorised loan
16 000
Total stock
80 000
On the basis of these figures it is possible to estimate the probability of finding
the book you want. Of the total stock of 80 000 books bought by the library, you
might expect to find about 20 000 on the shelves at any one time. As a fraction,
20
this is 80
or 14 of the total. So, as a rough estimate, the probability of your finding a
particular book is 0.25 or 25%.
77
Probability
S1
3
Similarly, 16 000 out of the total of 80 000 books are on unauthorised loan, a
euphemism for stolen, and this is 20%, or 51.
An important assumption underlying these calculations is that all the books are
equally likely to be unavailable, which is not very realistic since popular books are
more likely to be stolen. Also, the numbers given are only rough approximations,
so it is definitely incorrect to say that the probability is exactly 14 .
Measuring probability
Probability (or chance) is a way of describing the likelihood of different possible
outcomes occurring as a result of some experiment.
In the example of the library books, the experiment is looking in the library for
a particular book. Let us assume that you already know that the book you want
is on the library’s stocks. The three possible outcomes are that the book is on the
shelves, out on loan or missing.
It is important in probability to distinguish experiments from the outcomes
which they may generate. Here are a few examples.
Experiment
Guessing the answer to a four-option multiple choice
question
Possible outcomes
A
B
C
D
Predicting the next vehicle to go past the corner of
my road
car
bus
lorry
bicycle
van
other
Tossing a coin
heads
tails
Another word for experiment is trial. This is used in Chapter 6 of this book to
describe the binomial situation where there are just two possible outcomes.
Another word you should know is event. This often describes several outcomes
put together. For example, when rolling a die, an event could be ‘the die shows an
even number’. This event corresponds to three different outcomes from the trial,
the die showing 2, 4 or 6. However, the term event is also often used to describe a
single outcome.
78
Estimating probability
Probability is a number which measures likelihood. It may be estimated
experimentally or theoretically.
In many situations probabilities are estimated on the basis of data collected
experimentally, as in the following example.
Of 30 drawing pins tossed in the air, 21 of them were found to have landed with
their pins pointing up. From this you would estimate the probability that the
21
next pin tossed in the air will land with its pin pointing up to be 30
or 0.7.
Estimating probability
Experimental estimation of probability
S1
3
You can describe this in more formal notation.
Estimated P(U ) = n(U )
n(T )
Probability of the next
throw landing pin-up.
Number of times it
landed pin-up
Total number
of throws.
Theoretical estimation of probability
There are, however, some situations where you do not need to collect data to
make an estimate of probability.
For example, when tossing a coin, common sense tells you that there are only two
realistic outcomes and, given the symmetry of the coin, you would expect them
to be equally likely. So the probability, P(H), that the next coin will produce the
outcome heads can be written as follows:
Number of ways of getting
the outcome heads.
P(H) = 1
2
Total number of
possible outcomes.
Probability of the next
toss showing heads.
EXAMPLE 3.1
Using the notation described above, write down the probability that the correct
answer for the next four-option multiple choice question will be answer A.
What assumptions are you making?
SOLUTION
Assuming that the test-setter has used each letter equally often, the probability,
P(A), that the next question will have answer A can be written as follows:
P(A) = 1
4
Answer A.
Answers A, B, C and D.
79
Probability
S1
3
Notice that we have assumed that the four options are equally likely.
Equiprobability is an important assumption underlying most work on probability.
Expressed formally, the probability, P(A), of event A occurring is:
P(A) =
Number of ways that
event A can occur.
n(A)
n()
Probability of event
A occurring.
Total number of ways that the
possible events can occur. Notice the
use of the symbol , the universal
set of all the ways that the possible
events can occur.
Probabilities of 0 and 1
The two extremes of probability are certainty at one end of the scale and
impossibility at the other. Here are examples of certain and impossible events.
Experiment
Certain event
Impossible event
Rolling a single die
The result is in the range 1
to 6 inclusive
The result is a 7
Tossing a coin
Getting either heads or
tails
Getting neither heads nor
tails
Certainty
As you can see from the table above, for events that are certain, the number of
ways that the event can occur, n(A) in the formula, is equal to the total number
of possible events, n().
n(A) = 1
n( )
So the probability of an event which is certain is one.
Impossibility
For impossible events, the number of ways that the event can occur, n(A), is zero.
n(A) = 0 = 0
n() n()
So the probability of an event which is impossible is zero.
Typical values of probabilities might be something like 0.3 or 0.9. If you arrive
at probability values of, say, −0.4 or 1.7, you will know that you have made a
mistake since these are meaningless.
0 P(A) 1
80
Impossible event.
Certain event.
The complement of an event
The complement of an event A, denoted by A′, is the event not-A, that is the
event ‘A does not happen’.
It was found that, out of a box of 50 matches, 45 lit but the others did not.
What was the probability that a randomly selected match would not have lit?
Expectation
EXAMPLE 3.2
S1
3
SOLUTION
The probability that a randomly selected match lit was
P(A) = 45 = 0.9.
50
The probability that a randomly selected match did not light was
P(A′) = 50 – 45 = 5 = 0.1.
50
50
From this example you can see that
P(A′) = 1 − P(A)
The probability of
A not occurring.
The probability of
A occurring.
This is illustrated in figure 3.1.
�
A
A
Figure 3.1 Venn diagram showing events A and not-A (A’)
Expectation
Health services braced for flu epidemic
Local health services are poised for their biggest challenge in years. The virulent strain
of flu, named Trengganu B from its origins in Malaysia, currently sweeping across the
world is expected to hit any day.
With a chance of one in three of any individual contracting the disease, and 120 000
people within the Health Area, surgeries and hospitals are expecting to be swamped
with patients.
Local doctor Aloke Ghosh says ‘Immunisation seems to be ineffective against this strain’.
81
How many people can the health services expect to contract flu? The answer is
easily seen to be 120 000 × 13 = 40 000. This is called the expectation or expected
frequency and is given in this case by np, where n is the population size and p the
probability.
Probability
S1
3
Expectation is a technical term and need not be a whole number. Thus the
expectation of the number of heads when a coin is tossed 5 times is 5 × 12 = 2.5.
You would be wrong to go on to say ‘That means either 2 or 3’ or to qualify your
answer as ‘about 2 12’. The expectation is 2.5.
The idea of expectation of a discrete random variable is explored more thoroughly
in Chapter 4. Applications of the binomial distribution are covered in Chapter 6.
The probability of either one event or another
So far we have looked at just one event at a time. However, it is often useful
to bracket two or more of the events together and calculate their combined
probability.
EXAMPLE 3.3
The table below is based on the data at the beginning of this chapter and shows
the probability of the next book requested falling into each of the three categories
listed, assuming that each book is equally likely to be requested.
Category of book
Typical numbers
Probability
On the shelves (S)
20 000
0.25
Out on loan (L)
44 000
0.55
Unauthorised loan (U)
16 000
0.20
Total (S + L + U)
80 000
1.00
What is the probability that a randomly requested book is either out on loan or
on unauthorised loan (i.e. that it is not available)?
Number of books
on loan.
SOLUTION
44 000 + 16 000
80 000
Number of books on
unauthorised loan.
60
000
=
80 000
Number of books.
= 0.75
P(L or U ) =
This can be written in more formal notation as
P(L ∪ U) =
=
82
n(L ∪ U)
n()
n(L) n(U )
+
n() n()
P(L ∪ U ) = P(L) + P(U )
Notice the use of the union symbol, ∪, to mean or. This is illustrated in figure 3.2.
�
Key: L
U
U
out on loan
out on unauthorised loan
Figure 3.2 Venn diagram showing events L and U. It is not possible for both
to occur.
In this example you could add the probabilities of the two events to get the
combined probability of either one or the other event occurring. However, you
have to be very careful adding probabilities as you will see in the next example.
EXAMPLE 3.4
Below are further details of the categories of books in the library.
Category of book
The probability of either one event or another
L
S1
3
Number of books
On the shelves
20 000
Out on loan
44 000
Adult fiction
22 000
Adult non-fiction
40 000
Junior
18 000
Unauthorised loan
16 000
Total stock
80 000
Asaph is trying to find the probability that the next book requested will be either
out on loan or a book of adult non-fiction.
He writes
Assuming all the books in the library are equally
likely to be requested,
P(on loan) + P(adult non-fiction) = 44 000 + 40 000
80 000 80 000
= 0.55 + 0.5
= 1.05
Explain why Asaph’s answer must be wrong. What is his mistake?
83
Probability
S1
3
SOLUTION
This answer is clearly wrong as you cannot have a probability greater than 1.
The way this calculation was carried out involved some double counting. Some of
the books classed as adult non-fiction were counted twice because they were also
in the on-loan category, as you can see from figure 3.3.
�
L
Key: L
A
A
out on loan
adult non-fiction
Figure 3.3 Venn diagram showing events L and A. It is possible for both
to occur.
If you add all six of the book categories together, you find that they add up to
160 000, which represents twice the total number of books owned by the library.
A more useful representation of the data in the previous example is given in the
two-way table below.
On the shelves
Out on loan
Unauthorised loan
Totals
Adult fiction
Adult
non-fiction
Junior
Total
4 000
12 000
4 000
20 000
14 000
20 000
10 000
44 000
4 000
8 000
4 000
16 000
22 000
40 000
18 000
80 000
If you simply add 44 000 and 40 000, you double count the 20 000 books which
fall into both categories. So you need to subtract the 20 000 to ensure that it is
counted only once. Thus:
Number either out on loan or adult non-fiction
= 44 000 + 40 000 − 20 000
= 64 000 books.
So, the required probability = 64000 = 0.8.
80000
Mutually exclusive events
84
The problem of double counting does not occur when adding two rows in the
table. Two rows cannot overlap, or intersect, which means that those categories
are mutually exclusive (i.e. the one excludes the other). The same is true for two
columns within the table.
Where two events, A and B, are mutually exclusive, the probability that either A
or B occurs is equal to the sum of the separate probabilities of A and B occurring.
A
B
�
Figure 3.4 (a) Mutually exclusive events
A
B
�
(b) Not mutually exclusive events
P(A or B) = P(A) + P(B) − P(A and B)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A or B) = P(A) + P(B)
P(A ∪ B) = P(A) + P(B)
Notice the use of the intersection
sign, ∩, to mean both ... and ...
EXAMPLE 3.5
The probability of either one event or another
Where two events, A and B, are not mutually exclusive, the probability that
either A or B occurs is equal to the sum of the separate probabilities of A and B
occurring minus the probability of A and B occurring together.
S1
3
A fair die is thrown. What is the probability that it shows each of these?
(i)
(ii)
(iii)
Event A: an even number
Event B: a number greater than 4
Either A or B (or both): a number which is either even or greater than 4
SOLUTION
(i)
Event A:
Three out of the six numbers on a die are even, namely 2, 4 and 6.
So
(ii)
3
6
= 12 .
Event B:
Two out of the six numbers on a die are greater than 4, namely 5 and 6.
So
(iii)
P(A) =
P(B) =
2
6
= 13.
Either A or B (or both):
Four of the numbers on a die are either even or greater than 4, namely 2, 4, 5
and 6.
So
P(A ∪ B) = 46 = 23 .
This could also be found using
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A ∪ B) = 63 + 62 − 16
=
4
6
=
2
3
This is the number 6 which is
both even and greater than 4.
85
Probability
S1
3
EXERCISE 3A
1
Three separate electrical components, switch, bulb and contact point, are
used together in the construction of a pocket torch. Of 534 defective torches,
examined to identify the cause of failure, 468 are found to have a defective
bulb. For a given failure of the torch, what is the probability that either the
switch or the contact point is responsible for the failure? State clearly any
assumptions that you have made in making this calculation.
2
If a fair die is thrown, what is the probability that it shows
(i)
(ii)
(iii)
(iv)
3
4
4 or more
less than 4
an even number?
A bag containing Scrabble letters has the following letter distribution.
A
B
C
D
E
F
G
H
I
J
K
L
M
9
2
2
4
12
2
3
2
9
1
1
4
2
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
6
8
2
1
6
4
6
4
2
2
1
2
1
The first letter is chosen at random from the bag; find the probability that it is
(i)
(ii)
(iii)
(iv)
(v)
(vi)
4
86
an E
in the first half of the alphabet
in the second half of the alphabet
a vowel
a consonant
the only one of its kind.
A sporting chance
(i) Two players, A and B, play tennis. On the basis of their previous results,
the probability of A winning, P(A), is calculated to be 0.65. What is P(B),
the probability of B winning?
(ii) Two hockey teams, A and B, play a game. On the basis of their previous
results, the probability of team A winning, P(A), is calculated to be 0.65.
Why is it not possible to calculate directly P(B), the probability of team B
winning, without further information?
(iii) In a tennis tournament, player A, the favourite, is estimated to have a 0.3
chance of winning the competition. Player B is estimated to have a 0.15
chance. Find the probability that either A or B will win the competition.
(iv) In the Six Nations Rugby Championship, France and England are each
given a 25% chance of winning or sharing the championship cup. It is also
estimated that there is a 5% chance that they will share the cup. Estimate
the probability that either England or France will win or share the cup.
5
The diagram shows even (E), odd (O) and square (S) numbers.
E
�
Copy the diagram and place the numbers 1 to 20 on it.
The numbers 1 to 20 are written on separate cards.
(ii)
A card is chosen at random. Find the probability that the number showing
is:
(a) even, E
(b) square, S
(c) odd, O
(d) both even and square, E ∩ S
(e) either even or square, E ∪ S
(f) both even and odd, E ∩ O
(g) either even or odd, E ∪ O.
S1
3
Independent and dependent events
(i)
S
O
Write down equations connecting the probabilities of the following events.
(h)
(i)
E, S, E ∩ S, E ∪ S
E, O, E ∩ O, E ∪ O
Independent and dependent events
Veronica
My lucky day!
Won $100 when the number on my newspaper came up in the daily draw and $50 in
the weekly draw too. A chance in a million!
This story describes two pieces of good fortune on the same day. Veronica said
1
the probability was about 1000 000 . What was it really?
The two events resulted from two different experiments, the daily draw and the
weekly draw. Consequently this situation is different from those you met in
the previous section. There you were looking at two events from a single
experiment (like the number coming up when a die is thrown being even or
being greater than 4).
The total number of entrants in the daily draw was 1245 and in the weekly draw
324. The draws were conducted fairly, that is each number had an equal chance
of being selected. The following table sets out the two experiments and their
corresponding events with associated probabilities.
87
S1
3
Experiment
Events (and estimated probabilities)
Daily draw
Winning: 1
1245
Not winning: 1244
Probability
1245
Winning: 1
324
Weekly draw
Not winning: 323
324
The two events ‘win daily draw’ and ‘win weekly draw’ are independent events.
Two events are said to be independent when the outcome of the first event does
not affect the outcome of the second event. The fact that Veronica has won the
daily draw does not alter her chances of winning the weekly draw.
●
For two independent events, A and B, P(A ∩ B) = P(A) × P(B).
In situations like this the possible outcomes resulting from the different
experiments are often shown on a tree diagram.
EXAMPLE 3.6
Find, in advance of the results of the two draws, the probability that
(i) Veronica would win both draws
(ii) Veronica would fail to win either draw
(iii) Veronica would win one of the two draws.
SOLUTION
The possible results are shown on the tree diagram in figure 3.5.
daily
weekly
1
324
1
1245
win
1244
1245
not
win
323
324
1
324
323
324
combined
win
1
1245
1
324
(i)
not
win
1
1245
323
324
(iii)
win
1244
1245
1
324
not
win
1244
1245
323
324
(iii)
(ii)
Figure 3.5
88
(i)
The probability that Veronica wins both
1
= 1 × 1 =
1245 324 403380
This is not quite Veronica’s ‘one in a million’ but it is not very far off it.
(ii)
The probability that Veronica wins neither
401812
= 1244 × 323 =
1245 324 403380
This of course is much the most likely outcome.
(iii)
The probability that Veronica wins one but not the other
S1
3
= 1 × 323 + 1244 × 1 = 1567
1245
324
324
����
� 1245
����
� 403380
Wins weekly draw
but not daily draw.
Look again at the structure of the tree diagram in figure 3.5.
There are two experiments, the daily draw and the weekly draw. These are
considered as First, then experiments, and set out First on the left and Then on
the right. Once you understand this, the rest of the layout falls into place, with
the different outcomes or events appearing as branches. In this example there are
two branches at each stage; sometimes there may be three or more. Notice that
for a given situation the component probabilities sum to 1, as before.
1 + 323 + 1244 + 401812 = 403380 = 1
403380 403380 403380 403380 403380
EXAMPLE 3.7
Independent and dependent events
Wins daily draw but
not weekly draw.
Some friends buy a six-pack of potato crisps. Two of the bags are snake flavoured
(S), the rest are frog flavoured (F ). They decide to allocate the bags by lucky dip.
Find the probability that
(i) the first two bags chosen are the same as each other
(ii) the first two bags chosen are different from each other.
first dip
2
6
4
6
second dip
combined
1
5
S
P(S, S)
2
6
1
5
4
5
F
P(S, F )
2
6
4
5
(ii)
2
5
S
P(F, S)
4
6
2
5
(ii)
3
5
F
P(F, F)
4
6
3
5
S
F
(i)
(i)
Figure 3.6
SOLUTION
Note: P(F, S) means the probability of drawing a frog bag (F ) on the first dip and
a snake bag (S) on the second.
(i)
The probability that the first two bags chosen are the same as each other is
P(S, S) + P(F, F) = 62 × 51 + 46 × 53
1
6
= 15
+ 15
7
= 15
89
(ii)
S1
3
The probability that the first two bags chosen are different from each other is
P(S, F) + P(F, S) = 62 × 54 + 46 × 52
4
4
= 15
+ 15
Probability
8
= 15
Note
The answer to part (ii) above hinged on the fact that two orderings (S then F, and F
then S) are possible for the same combined event (that the two bags selected include
one snake and one frog bag).
The probabilities changed between the first dip and the second dip. This is
because the outcome of the second dip is dependent on the outcome of the first
one (with fewer bags remaining to choose from).
By contrast, the outcomes of the two experiments involved in tossing a coin
twice are independent, and so the probability of getting a head on the second toss
remains unchanged at 0.5, whatever the outcome of the first toss.
Although you may find it helpful to think about combined events in terms of how
they would be represented on a tree diagram, you may not always actually draw
them in this way. If there are several experiments and perhaps more than two
possible outcomes from each, drawing a tree diagram can be very time-consuming.
EXAMPLE 3.8
www.freeourdavid.com
Is this justice?
In 2012, David Starr was sentenced to 12 years’ imprisonment for armed robbery solely on
the basis of an identification parade. He was one of 12 people in the parade and was picked
out by one witness but not by three others.
Many people who knew David well believe he was incapable of such a crime.
Please add your voice to the clamour for a review of his case by clicking on
the ‘Free David’ button.
Free
David
How conclusive is this sort of evidence, or, to put it another way, how likely is it
that a mistake has been made?
Investigate the likelihood that David Starr really did commit the robbery.
SOLUTION
In this situation you need to assess the probability of an innocent individual
being picked out by chance alone. Assume that David Starr was innocent and the
1
witnesses were selecting in a purely random way (that is, with a probability of 12
11
of selecting each person and a probability of 12 of not selecting each person). If
90
each of the witnesses selected just one of the twelve people in the identity parade
in this random manner, how likely is it that David Starr would be picked out by
at least one witness?
P(at least one selection) = 1 − P(no selections)
= 1 − 0.706 = 0.294 (i.e. roughly 30%).
In other words, there is about a 30% chance of an innocent person being chosen
in this way by at least one of the witnesses.
Exercise 3B
11 11 11 11
×
×
×
= 1 − 12
12 12 12
S1
3
The website concluded:
Is 30% really the sort of figure we have in mind when judges use the phrase ‘beyond
reasonable doubt’? Because if it is, many innocent people will be condemned to a life
behind bars.
This raises an important statistical idea, which you will meet again if you study
Statistics 2 about how we make judgements and decisions.
Judgements are usually made under conditions of uncertainty and involve
us in having to weigh up the plausibility of one explanation against that of
another. Statistical judgements are usually made on such a basis. We choose one
explanation if we judge the alternative explanation to be sufficiently unlikely,
that is if the probability of its being true is sufficiently small. Exactly how small
this probability has to be will depend on the individual circumstances and is
called the significance level.
EXERCISE 3B
1
The probability of a pregnant woman giving birth to a girl is about 0.49.
Draw a tree diagram showing the possible outcomes if she has two babies
(not twins).
From the tree diagram, calculate the following probabilities:
(i)
(ii)
(iii)
2
that the babies are both girls
that the babies are the same sex
that the second baby is of different sex to the first.
In a certain district of a large city, the probability of a household suffering a
break-in in a particular year is 0.07 and the probability of its car being stolen
is 0.12.
Assuming these two trials are independent of each other, draw a tree diagram
showing the possible outcomes for a particular year.
Calculate, for a randomly selected household with one car, the following
probabilities:
(i)
(ii)
(iii)
that the household is a victim of both crimes during that year
that the household suffers only one of these misfortunes during that year
that the household suffers at least one of these misfortunes during that year.
91
3
There are 12 people at an identification parade. Three witnesses are called to
identify the accused person.
Assuming they make their choice purely by random selection, draw a tree
diagram showing the possible events.
Probability
S1
3
(i)
(ii)
4
From the tree diagram, calculate the following probabilities:
(a) that all three witnesses select the accused person
(b) that none of the witnesses selects the accused person
(c) that at least two of the witnesses select the accused person.
Suppose now that by changing the composition of people in the
identification parade, the first two witnesses increase their chances of
selecting the accused person to 0.25.
Draw a new tree diagram and calculate the following probabilities:
(a) that all three witnesses select the accused person
(b) that none of the witnesses selects the accused person
(c) that at least two of the witnesses select the accused person.
Ruth drives her car to work – provided she can get it to start! When she
remembers to put the car in the garage the night before, it starts next morning
with a probability of 0.95. When she forgets to put the car away, it starts next
morning with a probability of 0.75. She remembers to garage her car 90% of
the time.
What is the probability that Ruth drives her car to work on a randomly
chosen day?
5
Around 0.8% of men are red–green colour-blind (the figure is slightly
different for women) and roughly 1 in 5 men is left-handed.
Assuming these characteristics are inherited independently, calculate with the
aid of a tree diagram the probability that a man chosen at random will
(i)
(ii)
(iii)
(iv)
6
Three dice are thrown. Find the probability of obtaining
(i)
(ii)
(iii)
7
be both colour-blind and left-handed
be colour-blind and not left-handed
be colour-blind or left-handed
be neither colour-blind nor left-handed.
at least two 6s
no 6s
different scores on all the dice.
Explain the flaw in this argument and rewrite it as a valid statement.
The probability of throwing a 6 on a fair die = 16 . Therefore the probability
of throwing at least one 6 in six throws of the die is 16 + 16 + 16 + 16 + 16 + 16 = 1
92
so it is a certainty.
8
Two dice are thrown. The scores on the dice are added.
(i)
S1
3
Copy and complete this table showing all the possible outcomes.
First die
2
3
4
5
Exercise 3B
1
6
1
Second die
2
3
4
10
5
11
6
(ii)
(iii)
(iv)
7
8
9
10
11
12
What is the probability of a score of 4?
What is the most likely outcome?
Criticise this argument:
There are 11 possible outcomes, 2, 3, 4, up to 12. Therefore each of them has
1
a probability of 11.
9
The probability of someone catching flu in a particular winter when they
have been given the flu vaccine is 0.1. Without the vaccine, the probability of
catching flu is 0.4. If 30% of the population has been given the vaccine, what
is the probability that a person chosen at random from the population will
catch flu over that winter?
10
Kevin hosts the TV programme Thank Your Lucky Stars. During the show he
picks members of the large studio audience at random and asks them what
star sign they were born under.
(There are 12 star signs in all and you may assume that the probabilities that
a randomly chosen person will be born under each star sign are equal.)
(i)
(ii)
(iii)
(iv)
The first person Kevin picks says that he was born under the star sign
Aries. What is the probability that the next person he picks was not born
under Aries?
Show that the probability that the first three people picked were all born
under different star signs is approximately 0.764.
Calculate the probability that the first five people picked were all born
under different star signs.
What is the probability that at least two of the first five people picked
were born under the same star sign?
[MEI, part]
93
Probability
S1
3
11
One plastic toy aeroplane is given away free in each packet of cornflakes. Equal
numbers of red, yellow, green and blue aeroplanes are put into the packets.
Faye, a customer, has collected three colours of aeroplane but still wants a
yellow one.
Find the probability that
she gets a yellow aeroplane by opening just one packet
(ii) she fails to get a yellow aeroplane in opening four packets
(iii) she needs to open exactly five packets to get the yellow aeroplane she
wants.
Henry, a quality controller employed by the cornflakes manufacturer, opens
a number of packets chosen at random to check on the distribution of
colours. Find the probability that
(i)
(iv)
(v)
(vi)
the first two packets he opens both have red aeroplanes in
the first two packets he opens have aeroplanes of different colours in
he gets all four different colours by opening just four packets.
[MEI]
Conditional probability
Sad news
Myra
My best friend had a heart attack while out shopping. Sachit was rushed to
hospital but died on the way. He was only 47 – too young ...
What is the probability that somebody chosen at random will die of a heart attack
in the next 12 months?
One approach would be to say that, since there are about 300 000 deaths per year
from heart and circulatory diseases (H & CD) among the 57 000 000 population
of the country where Sachit lived,
probability =
number of deaths from H & CD per ye a r
total population
= 300 000 = 0.0053.
57 000 000
However, if you think about it, you will probably realise that this is rather a
meaningless figure. For a start, young people are much less at risk than those in
or beyond middle age.
94
So you might wish to give two answers:
deaths from H & CD among over-40s
population of over-40s
P2 =
deaths from H & CD among under-40s
population of under-40s
Typically only 1500 of the deaths would be among the under-40s, leaving (on the
basis of these figures) 298 500 among the over-40s. About 25 000 000 people in
the country are over 40, and 32 000 000 under 40 (40 years and 1 day counts as
over 40). This gives
P1 =
298500
deaths from H & CD among over-40s
=
25 000 000
population o f over-40s
S1
3
Conditional probability
P1 =
= 0.0119
and
P2 =
deaths from H & CD among under-40s
= 1500
32 000 000
population of under--40s
= 0.000 047.
So somebody in the older group is over 200 times more likely to die of a heart
attack than somebody in the younger group. Putting them both together as an
average figure resulted in a figure that was representative of neither group.
But why stop there? You could, if you had the figures, divide the population
up into 10-year, 5-year, or even 1-year intervals. That would certainly improve
the accuracy; but there are also more factors that you might wish to take into
account, such as the following.
●
Is the person overweight?
●
Does the person smoke?
●
Does the person take regular exercise?
The more conditions you build in, the more accurate will be the estimate of the
probability.
You can see how the conditions are brought in by looking at P1:
P1 =
deaths from H & CD among over-40s
298500
=
population o f over-40s
25 000 000
= 0.0119
You would write this in symbols as follows:
Event G: Somebody selected at random is over 40.
Event H: Somebody selected at random dies from H & CD.
The probability of someone dying from H & CD given that he or she is over 40 is
given by the conditional probability P(H G).
95
P(H G) = n(H ∩ G)
n(G)
S1
3
= n(H ∩ G)/ n()
n(G)/ n()
Probability
= P (H ∩ G).
P (G)
P(H G) means the probability
of event H occurring given that
event G has occurred.
This result may be written in general form for all cases of conditional probability
for events A and B.
The probability of
both B and A.
A
B
�
P(B A ) = P(B ∩ A)
P(A)
The probability of B
given A.
The probability of A.
B
A
Figure 3.7
Conditional probability is used when your estimate of the probability of an event
is altered by your knowledge of whether some other event has occurred. In this
case the estimate of the probability of somebody dying from heart and circulatory
diseases, P(H), is altered by a knowledge of whether the person is over 40 or not.
Thus conditional probability addresses the question of whether one event is
dependent on another one. If the probability of event B is not affected by the
occurrence of event A, we say that B is independent of A. If, on the other hand,
the probability of event B is affected by the occurrence (or not) of event A, we say
that B is dependent on A.
●
If A and B are independent, then P(B A) = P(B A′) and this is just P(B).
●
If A and B are dependent, then P(B A) ≠ P(B A′).
As you have already seen, the probability of a combined event is the product of
the separate probabilities of each event, provided the question of dependence
between the two events is properly dealt with. Specifically:
The probability of both A
and B occurring.
●
96
for dependent events P(A ∩ B) = P(A) × P(B A).
The probability of
A occurring.
The probability of B occurring,
given that A has occurred.
When A and B are independent events, then, because P(B A) = P(B), this can be
written as
●
A company is worried about the high turnover of its employees and decides to
investigate whether they are more likely to stay if they are given training. On
1 January one year the company employed 256 people (excluding those about to
retire). During that year a record was kept of who received training as well as who
left the company. The results are summarised in this table.
Still employed
Left company
Total
109
43
152
60
44
104
169
87
256
Given training
Not given training
Totals
(i)
(ii)
(iii)
Conditional probability
EXAMPLE 3.9
for independent events P(A ∩ B) = P(A) × P(B).
S1
3
Find the probability that a randomly selected employee
(a) received training
(b) received training and did not leave the company.
Are the events T and S independent?
Find the probability that a randomly selected employee
(a) did not leave the company, given that the person had received training
(b) did not leave the company, given that the person had not received training.
SOLUTION
Using the notation T: The employee received training
S: The employee stayed in the company
(i)
(a)
(b)
(ii)
P(T ) = n (T ) = 152 = 0.59
n ( ) 256
n(T ∩ S ) 109
=
= 0.43
P(T ∩ S ) =
n()
256
If T and S are independent events then P(T ∩ S) = P(T) × P(S).
P(S) = n (S) = 169 = 0.66
n () 256
P(T ) × P(S) = 152 × 169 = 0.392
256 2566
As P(T ∩ S) ≠ P(T) × P(S), the events T and S are not independent.
97
Probability
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3
(iii) (a)
P( S T ) = P( S ∩ T ) =
P( T )
(b)
P(S T ′) = P(S ∩ T ′) =
P(T ′)
109
256
152
256
60
256
104
256
= 109 = 0.72
152
= 60 = 0.58
104
Since P(S T ) is not the same as P(S T ′), the event S is not independent of
the event T. Each of S and T is dependent on the other, a conclusion which
matches common sense. It is almost certainly true that training increases
employees’ job satisfaction and so makes them more likely to stay, but it is
also probably true that the company is more likely to go to the expense of
training the employees who seem less inclined to move on to other jobs.
?
●
How would you show that the event T is not independent of the event S ?
In some situations you may find it helps to represent a problem such as this as a
Venn diagram.
T(152)
T (104) �(256)
S(169)
43
44
109
60
Figure 3.8
?
●
What do the various numbers and letters represent?
Where is the region S′?
How are the numbers on the diagram related to the answers to parts (i) to (v)?
In other situations it may be helpful to think of conditional probabilities in terms
of tree diagrams. Conditional probabilities are needed when events are dependent,
that is when the outcome of one trial affects the outcomes from a subsequent
trial, so, for dependent events, the probabilities of all but the first layer of a tree
diagram will be conditional.
98
EXAMPLE 3.10
Rebecca is buying two goldfish from a pet shop. The shop’s tank contains seven
male fish and eight female fish but they all look the same.
S1
3
Conditional probability
Figure 3.9
Find the probability that Rebecca’s fish are
(i)
(ii)
(iii)
both the same sex
both female
both female given that they are the same sex.
SOLUTION
The situation is shown on this tree diagram.
first fish
7
15
8
15
second fish
6
14
M
P(both male)
8
14
F
P(male, female)
7
15
8
14
56
210
7
14
M
P(female, male)
8
15
7
14
56
210
7
14
F
P(both female)
M
F
7
15
6
14
8
15
42
210
7
14
56
210
Figure 3.10
P(both the same sex) = P(both male) + P(both female)
= 42 + 56 = 98 = 7
210 210 210 15
4
56
P(both female) = 210 = 15
(i)
(ii)
(iii)
4
P(both female both the same sex)
15
= P(both female and the same sex) ÷ P(both the same sex) =
=4
7
7
15
This is the same as
P(both female).
99
The ideas in the last example can be expressed more generally for any two
dependent events, A and B. The tree diagram would be as shown in figure 3.11.
Probability
S1
3
P(A)
P(A )
P(B A)
B A
P(A
B)
P(B A)
B A
P(A
B)
P(A)
P(B A )
B A
P(A
B)
P(A )
P(B A )
B A
P(A
B)
P(A)
P(B A)
A
P(B A)
P(B A )
A
The probabilities in the second layer of
the tree diagram are conditional on the
outcome of the first experiment.
P(A )
P(B A )
These events are conditional
upon the outcome of the
first experiment.
Figure 3.11
The tree diagram shows you that
●
●
?
●
EXERCISE 3C
P(B) = P(A ∩ B) + P(A′ ∩ B)
= P(A) × P(B A) + P(A′) × P(B A′)
P(A ∩ B) = P(A) × P(B A)
⇒ P(B A) = P(A ∩ B)
P(A)
How were these results used in Example 3.10 about the goldfish?
1
In a school of 600 students, 360 are girls. There are 320 hockey players, of
whom 200 are girls. Among the hockey players there are 28 goalkeepers,
19 of them girls. Find the probability that
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
100
a student chosen at random is a girl
a girl chosen at random plays hockey
a hockey player chosen at random is a girl
a student chosen at random is a goalkeeper
a goalkeeper chosen at random is a boy
a male hockey player chosen at random is a goalkeeper
a hockey player chosen at random is a male goalkeeper
two students chosen at random are both goalkeepers
two students chosen at random are a male goalkeeper and a female
goalkeeper
two students chosen at random are one boy and one girl.
2
100 cars are entered for a road-worthiness test which is in two parts,
mechanical and electrical. A car passes only if it passes both parts. Half the cars
fail the electrical test and 62 pass the mechanical. 15 pass the electrical but fail
the mechanical test.
S1
3
Exercise 3C
Find the probability that a car chosen at random
(i) passes overall
(ii) fails on one test only
(iii) given that it has failed, failed the mechanical test only.
3
Two dice are thrown. What is the probability that the total is
(i) 7
(ii) a prime number
(iii) 7, given that it is a prime number?
4
A cage holds two litters of rats. One litter comprises three females and four
males, and the other comprises two females and six males. A random selection
of two rats is made. Find, as fractions, the probabilities that the two rats are
(i) from the same litter
(ii) of the same sex
(iii) from the same litter and of the same sex
(iv) from the same litter given that they are of the same sex.
[MEI]
5
A and B are two events with probabilities given by P(A) = 0.4, P(B) = 0.7 and
P(A ∩ B) = 0.35.
(i)
(ii)
6
Find P(A B) and P(B A).
Show that the events A and B are not independent.
Quark hunting is a dangerous occupation. On a quark hunt, there is a
1
probability of 4 that the hunter is killed. The quark is twice as likely to be
1
killed as the hunter. There is a probability of 3 that both survive.
(i)
Copy and complete this table of probabilities.
Hunter dies
Hunter lives
1
2
Quark dies
1
3
Quark lives
Totals
Total
1
4
1
2
1
Find the probability that
(ii) both the hunter and the quark die
(iii) the hunter lives and the quark dies
(iv) the hunter lives, given that the quark dies.
101
7
Probability
S1
3
ln a tea shop 70% of customers order tea with milk, 20% tea with lemon and
3
10% tea with neither. Of those taking tea with milk 5 take sugar, of those
1
taking tea with lemon 4 take sugar, and of those taking tea with neither milk
11
nor lemon 20 take sugar. A customer is chosen at random.
(i)
(ii)
(iii)
Represent the information given on a tree diagram and use it to find the
probability that the customer takes sugar.
Find the probability that the customer takes milk or sugar or both.
Find the probability that the customer takes sugar and milk. Hence find the
probability that the customer takes milk given that the customer takes sugar.
[MEI]
8
Every year two teams, the Ramblers and the Strollers, meet each other for a
quiz night. From past results it seems that in years when the Ramblers win,
the probability of them winning the next year is 0.7 and in years when the
Strollers win, the probability of them winning the next year is 0.5. It is not
possible for the quiz to result in the scores being tied.
The Ramblers won the quiz in 2009.
(i)
(ii)
(iii)
(iv)
Draw a probability tree diagram for the three years up to 2012.
Find the probability that the Strollers will win in 2012.
If the Strollers win in 2012, what is the probability that it will be their first
win for at least three years?
Assuming that the Strollers win in 2012, find the smallest value of n such
that the probability of the Ramblers winning the quiz for n consecutive
years after 2012 is less than 5%.
[MEI, adapted]
9
There are 90 players in a tennis club. Of these, 23 are juniors, the rest are
seniors. 34 of the seniors and 10 of the juniors are male. There are 8 juniors
who are left-handed, 5 of whom are male. There are 18 left-handed players in
total, 4 of whom are female seniors.
(i)
(ii)
10
102
Represent this information in a Venn diagram.
What is the probability that
(a) a male player selected at random is left-handed?
(b) a left-handed player selected at random is a female junior?
(c) a player selected at random is either a junior or a female?
(d) a player selected at random is right-handed?
(e) a right-handed player selected at random is not a junior?
(f) a right-handed female player selected at random is a junior?
Data about employment for males and females in a small rural area are
shown in the table.
Unemployed
Employed
Male
206
412
Female
358
305
A person from this area is chosen at random. Let M be the event that the
person is male and let E be the event that the person is employed.
Find P(M).
(ii)
Find P(M and E).
(iii)
Are M and E independent events? Justify your answer.
(iv)
Given that the person chosen is unemployed, find the probability that the
person is female.
Exercise 3C
(i)
S1
3
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q5 June 2005]
11
The probability that Henk goes swimming on any day is 0.2. On a day when
he goes swimming, the probability that Henk has burgers for supper is 0.75.
On a day when he does not go swimming, the probability that he has burgers
for supper is x. This information is shown on the following tree diagram.
0.75
0.2
burgers
goes
swimming
no burgers
x
burgers
does not go
swimming
no burgers
The probability that Henk has burgers for supper on any day is 0.5.
(i)
(ii)
Find x.
Given that Henk has burgers for supper, find the probability that he went
swimming that day.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q2 June 2006]
12
Boxes of sweets contain toffees and chocolate. Box A contains 6 toffees and
4 chocolates, box B contains 5 toffees and 3 chocolates, and box C contains
3 toffees and 7 chocolates. One of the boxes is chosen at random and two
sweets are taken out, one after the other, and eaten.
(i)
(ii)
Find the probability that they are both toffees.
Given that they are both toffees, find the probability that they both come
from box A.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q2 November 2005]
103
13
S1
3
There are three sets of traffic lights on Karinne’s journey to work. The
independent probabilities that Karinne has to stop at the first, second and
third set of lights are 0.4, 0.8 and 0.3 respectively.
Probability
(i)
(ii)
(iii)
(iv)
Draw a tree diagram to show this information.
Find the probability that Karinne has to stop at each of the first two sets
of lights but does not have to stop at the third set.
Find the probability that Karinne has to stop at exactly two of the three
sets of lights.
Find the probability that Karinne has to stop at the first set of lights,
given that she has to stop at exactly two sets of lights.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q6 November 2008]
KEY POINTS
1
The probability of an event A is
P(A) = n(A)
n( )
where n(A) is the number of ways that A can occur and n() is the total
number of ways that all possible events can occur, all of which are
equally likely.
2
For any two events, A and B, of the same experiment,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Where the events are mutually exclusive (i.e. where the events do not
overlap) the rule still holds but, since P(A ∩ B) is now equal to zero, the
equation simplifies to:
P(A ∪ B) = P(A) + P(B).
3
Where an experiment produces two or more mutually exclusive events, the
probabilities of the separate events sum to 1.
4
P(A) + P(A′) = 1
5
For two independent events, A and B,
P(A ∩ B) = P(A) × P(B).
6
P(B A) means the probability of event B occurring given that event A has
already occurred,
P(B A) = P(A ∩ B) .
P(A)
104
7
The probability that event A and then event B occur, in that order, is
P(A) × P(B A).
8
If event B is independent of event A,
P(B A) = P(B A′) = P(B).
Discrete random variables
An approximate answer to the right problem is worth a good deal
more than an exact answer to an approximate problem.
John Tukey
Carshare.com Q T T T T T
Discrete random variables
4
S1
4
Share life’s journey!
Towns and cities around the country are gridlocked with
traffic – many of these cars have just one occupant.
To solve this problem, Carshare.com is launching a new
scheme for people to car-share on journeys into all major
cities. Our comprehensive database can put interested
drivers into touch with each other and live updates via
your mobile will display the number of car-shares
available in any major city. Car-shares are available from
centralised locations for maximum convenience.
Carshare.com is running a small trial scheme in a busy town just south of the capital.
We will be conducting a survey to measure the success of the trial. Keep up to date
with the trial via the latest news on our website.
?
●
How would you collect information on the volume of traffic in the town?
A traffic survey, at critical points around the town centre, was conducted at peak
travelling times over a period of a working week. The survey involved 1000 cars.
The number of people in each car was noted, with the following results.
Number of people per car
Frequency
?
●
1
2
3
4
5
>5
560
240
150
40
10
0
How would you illustrate such a distribution?
What are the main features of this distribution?
105
The numbers of people per car are necessarily discrete. A discrete frequency
distribution is best illustrated by a vertical line chart, as in figure 4.1. This shows
you that the distribution has positive skew, with most of the data at the lower end
of the distribution.
600
500
frequency
Discrete random variables
S1
4
400
300
200
100
0
0
1
2
3
4
number of people
5
6
Figure 4.1
The survey involved 1000 cars. This is a large sample and so it is reasonable to
use the results to estimate the probabilities of the various possible outcomes:
1, 2, 3, 4, 5 people per car. You divide each frequency by 1000 to obtain the
relative frequency, or probability, of each outcome (number of people).
Outcome (Number of people)
Probability (Relative frequency)
1
2
3
4
5
>5
0.56
0.24
0.15
0.04
0.01
0
Discrete random variables
You now have a mathematical model to describe a particular situation. In
statistics you are often looking for models to describe and explain the data
you find in the real world. In this chapter you are introduced to some of the
techniques for working with models for discrete data. Such models use discrete
random variables.
The model is discrete since the number of passengers can be counted and takes
positive integer values only. The number of passengers is a random variable since
the actual value of the outcome is variable and can only be predicted with a given
probability, i.e. the outcomes occur at random.
Discrete random variables may have a finite or an infinite number of outcomes.
106
On the other hand, if you considered the number of hits on a website in a given
day, there may be no theoretical maximum, in which case the distribution may
be considered as infinite. A well known example of an infinite discrete random
variable is the Poisson distribution, which you will meet if you study Statistics 2.
The study of discrete random variables in this chapter will be limited to finite cases.
S1
4
Discrete random variables
The distribution we have outlined so far is finite – in the survey the maximum
number of people observed was five, but the maximum could be, say, eight,
depending on the size of car. In this case there would be eight possible outcomes.
A well known example of a finite discrete random variable is the binomial
distribution, which you will study in Chapter 6.
Notation and conditions for a discrete random variable
A discrete random variable is usually denoted by an upper case letter, such as X,
Y or Z. You may think of this as the name of the variable. The particular values
that the variable takes are denoted by lower case letters, such as r. Sometimes
these are given suffixes r1, r2, r3, ... . Thus P(X = r1) means the probability that the
discrete random variable X takes a particular value r1. The expression P(X = r) is
used to express a more general idea, as, for example, in a table heading.
Another, shorter way of writing probabilities is p1, p2, p3, … . If a finite
discrete random variable has n distinct outcomes r1, r2, …, rn, with associated
probabilities p1, p2, …, pn, then the sum of the probabilities must equal 1. Since
the various outcomes cover all possibilities, they are exhaustive.
Formally we have:
p1 + p2 + … + pn = 1
or
n
n
i =1
i =1
∑ pi =∑P(X = ri) = 1.
You should be
familiar with all
these notations.
n
If there is no ambiguity then ∑ P(X = ri) is often abbreviated to ∑P(X = r) or pr.
i =1
You will often see an alternative notation used, in which the values that the
variable takes are denoted by x rather than r. In this book x is used for a
continuous variable (see Statistics 2) and r for a discrete variable.
Diagrams of discrete random variables
Just as with frequency distributions for discrete data, the most appropriate
diagram to illustrate a discrete random variable is a vertical line chart. Figure 4.2
shows a diagram of the probability distribution of X, the number of people per
car. Note that it is identical in shape to the corresponding frequency diagram in
figure 4.1. The only real difference is the change of scale on the vertical axis.
107
P(X = r)
0.6
S1
4
0.5
Discrete random variables
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6 r
Figure 4.2
EXAMPLE 4.1
Two tetrahedral dice, each with faces labelled 1, 2, 3 and 4, are thrown and the
random variable X represents the sum of the numbers shown on the dice.
(i)
(ii)
(iii)
Find the probability distribution of X.
Illustrate the distribution and describe the shape of the distribution.
What is the probability that any throw of the dice results in a value of X
which is an odd number?
SOLUTION
(i)
The table shows all the possible totals when the two dice are thrown.
Second die
First die
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
You can use the table to write down the probability distribution for X.
108
r
2
3
4
5
6
7
8
P(X = r)
1
16
2
16
3
16
4
16
3
16
2
16
1
16
(ii)
The vertical line chart in figure 4.3 illustrates this distribution, which is
symmetrical.
Discrete random variables
P(X = r)
0.3
0.25
0.2
0.15
0.1
0.05
0
S1
4
0
1
2
3
4
5
6
7
8
9
10 r
Figure 4.3
(iii)
The probability that X is an odd number
= P(X = 3) + P(X = 5) + P(X = 7)
2
4
2
= 16
+ 16
+ 16
=
1
2
As well as defining a discrete random variable by tabulating the probability
distribution, another effective way is to use an algebraic definition of the form
P(X = r) = f(r) for given values of r.
The following example illustrates how this may be used.
EXAMPLE 4.2
The probability distribution of a random variable X is given by
P(X = r) = kr
P(X = r) = 0
(i)
(ii)
(iii)
for r = 1, 2, 3, 4
otherwise.
Find the value of the constant k.
Illustrate the distribution and describe the shape of the distribution.
Two successive values of X are generated independently of each other.
Find the probability that
(a) both values of X are the same
(b) the total of the two values of X is greater than 6.
109
SOLUTION
(i)
Discrete random variables
S1
4
Tabulating the probability distribution for X gives:
r
1
2
3
4
P(X = r)
k
2k
3k
4k
Since X is a random variable, ∑P(X = r) = 1
⇒
⇒
⇒
k + 2k + 3k + 4k = 1
10k = 1
k = 0.1
Hence P(X = r) = 0.1r, for r = 1, 2, 3, 4, which gives the following probability
distribution.
r
P(X = r)
(ii)
1
2
3
4
0.1
0.2
0.3
0.4
The vertical line chart in figure 4.4 illustrates this distribution. It has negative
skew.
P(X = r)
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5 r
Figure 4.4
(iii)
Let X1 represent the first value generated and X2 the second value generated.
(a)
P(both values of X are the same)
= P(X1 = X2 = 1 or X1 = X2 = 2 or X1 = X2 = 3 or X1 = X2 = 4)
= P(X1 = X2 = 1) + P(X1 = X2 = 2) + P(X1 = X2 = 3) + P(X1 = X2 = 4)
= P(X1 = 1) × P(X2 = 1) + P(X1 = 2) × P(X2 = 2)
+ P(X1 = 3) × P(X2 = 3) + P(X1 = 4) × P(X2 = 4)
= (0.1)2 + (0.2)2 + (0.3)2 + (0.4)2
= 0.01 + 0.04 + 0.09 + 0.16
110
= 0.3
(b)
P(total of the two values is greater than 6)
= P(X1 + X2 6)
= P(X1 + X2 = 7 or 8)
= P(X1 = 3) × P(X2 = 4) + P(X1 = 4) × P(X2 = 3)
+ P(X1 = 4) × P(X2 = 4)
= 0.3 × 0.4 + 0.4 × 0.3 + 0.4 × 0.4
Exercise 4A
= P(X1 + X2 = 7) + P(X1 + X2 = 8)
S1
4
= 0.12 + 0.12 + 0.16
= 0.4
EXERCISE 4A
1
The random variable X is given by the sum of the scores when two ordinary
dice are thrown.
(i)
(ii)
(iii)
2
The random variable Y is given by the absolute difference between the scores
when two ordinary dice are thrown.
(i)
(ii)
(iii)
3
Find the probability distribution of X.
Illustrate the distribution and describe the shape of the distribution.
Find the values of
(a) P(X 8)
(b) P(X is even)
(c) P( X − 7 3).
Find the probability distribution of Y.
Illustrate the distribution and describe the shape of the distribution.
Find the values of
(a) P(Y 3)
(b) P(Y is odd).
The probability distribution of a discrete random variable X is given by
P(X = r) = kr
8
P(X = r) = 0
(i)
(ii)
for r = 2, 4, 6, 8
otherwise.
Find the value of k and tabulate the probability distribution.
If two successive values of X are generated independently find the
probability that
(a) the two values are equal
(b) the first value is greater than the second value.
111
Discrete random variables
S1
4
4
An irregular die with six faces produces scores, X, for which the probability
distribution is given by
P(X = r) = k
r
P(X = r) = 0
(ii)
Three fair coins are tossed.
(i)
(ii)
(iii)
(iv)
6
By considering the set of possible outcomes, HHH, HHT, etc., tabulate the
probability distribution for X, the number of heads occurring.
Illustrate the distribution and describe the shape of the distribution.
Find the probability that there are more heads than tails.
Without further calculation, state whether your answer to part (iii) would
be the same if four fair coins were tossed. Give a reason for your answer.
Two fair tetrahedral dice, each with faces labelled 1, 2, 3 and 4, are thrown and
the random variable X is the product of the numbers shown on the dice.
Find the probability distribution of X.
What is the probability that any throw of the dice results in a value of X
which is an odd number?
(i)
(ii)
7
otherwise.
Find the value of k and illustrate the distribution.
Show that, when this die is thrown twice, the probability of obtaining two
equal scores is very nearly 14.
(i)
5
for r = 1, 2, 3, 4, 5, 6
An ornithologist carries out a study of the number of eggs laid per pair by a
species of rare bird in its annual breeding season. He concludes that it may
be considered as a discrete random variable X with probability distribution
given by
P(X = 0) = 0.2
P(X = r) = k(4r − r 2)
P(X = r) = 0
for r = 1, 2, 3, 4
otherwise.
Find the value of k and write the probability distribution as a table.
(i)
The ornithologist observes that the probability of survival (that is of an egg
hatching and of the chick living to the stage of leaving the nest) is dependent
on the number of eggs in the nest. He estimates the probabilities to be as
follows.
(ii)
112
r
Probability of survival
1
0.8
2
0.6
3
0.4
Find, in the form of a table, the probability distribution of the number of
chicks surviving per pair of adults.
8
A sociologist is investigating the changing pattern of the number of children
which women have in a country. She denotes the present number by
the random variable X which she finds to have the following probability
distribution.
P(X = r)
(i)
0
1
2
3
4
5+
0.09
0.22
a
0.19
0.08
negligible
Exercise 4A
r
S1
4
Find the value of a.
She is keen to find an algebraic expression for the probability distribution
and suggests the following model.
P(X = r) = k(r + l)(5 − r)
P(X = r) = 0
(ii)
(iii)
9
(ii)
(iii)
(iv)
Find the probability that X = 1, i.e. P(X = 1).
7
Find P(X 2) and deduce that P(X = 2) = 216
.
Find P(X r) and so deduce P(X = r), for r = 3, 4, 5, 6.
Illustrate and describe the probability distribution of X.
A box contains six black pens and four red pens. Three pens are taken at
random from the box.
(i)
(ii)
11
Find the value of k for this model.
Compare the algebraic model with the probabilities she found,
illustrating both distributions on one diagram.
Do you think it is a good model?
In a game, each player throws three ordinary six-sided dice. The random
variable X is the largest number showing on the dice, so for example, for
scores of 2, 5 and 4, X = 5.
(i)
10
for r = 0, 1, 2, 3, 4, 5
otherwise.
By considering the selection of pens as sampling without replacement,
illustrate the various outcomes on a probability tree diagram.
The random variable X represents the number of red pens obtained.
Find the probability distribution of X.
A vegetable basket contains 12 peppers, of which 3 are red, 4 are green and
5 are yellow. Three peppers are taken, at random and without replacement,
from the basket.
(i)
(ii)
(iii)
Find the probability that the three peppers are all different colours.
Show that the probability that exactly 2 of the peppers taken are green
is 12
55 .
The number of green peppers taken is denoted by the discrete random
variable X. Draw up a probability distribution table for X.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q7 June 2007]
113
Discrete random variables
S1
4
Expectation and variance
Carshare.com Q T T T T T
Share life’s journey!
Latest update …
Car-share trial a massive success. Traffic volume down and number of occupants per car up!
?
●
What statistical evidence do you think Carshare.com’s claim is based on?
A second traffic survey, at critical points around the town centre, was conducted
at peak travelling times over a period of a working week. This time the survey
involved 800 cars. The number of people in each car is shown in the table.
Number of people per car
Frequency
?
●
1
2
3
4
5
5
280
300
164
52
4
0
How would you compare the results in the two traffic surveys?
The survey involved 800 cars. This is a fairly large sample and so, once again, it is
reasonable to use the results to estimate the probabilities of the various possible
outcomes: 1, 2, 3, 4 and 5 people per car, as before.
Outcome (Number of people)
Probability (Relative frequency)
1
2
3
4
5
5
0.35
0.375
0.205
0.065
0.005
0
One way to compare the two probability distributions, before and after the carsharing campaign, is to calculate a measure of central tendency and a measure
of spread.
The most useful measure of central tendency is the mean or expectation of the
random variable and the most useful measure of spread is the variance. To a large
extent the calculation of these statistics mirrors the corresponding statistics for a
frequency distribution, x and sd2.
ACTIVITY 4.1
114
Find the mean and variance of the frequency distribution for the people-per-car
survey following the introduction of the car-sharing scheme.
Using relative frequencies generates an alternative approach which gives the
expectation E(X) = µ and variance Var(X) = σ2 for a discrete random variable.
We define the expectation, E(X) as
and Variance, Var(X) as
σ2 = E([X − µ]2) = ∑(r – µ)2 pr
or
σ2 = E(X 2) − µ2 =
σ2 is read as ‘sigma
squared’.
∑r 2pr – ∑rpr .
2
The second version of the variance is often written as E(X 2) − [E(X)]2, which
can be remembered as ‘the expectation of the squares minus the square of
the expectation’.
Expectation and variance
Notice the notation, µ for the
distribution’s mean and σ for its
standard deviation. Also notice the
shortened notation for P(X = r).
E(X) = µ = ∑r P(X = r) = ∑rpr
S1
4
These formulae can also be written as:
∑ xp
2
Var(X ) = ∑ x 2p − [ E(X )]
E(X ) =
Look at how expectation and variance are calculated using the probability
distribution developed from the second survey of number of people per car. You
can use these statistics to compare the distribution of number of people per car
before and after the introduction of the car-sharing scheme.
When calculating the expectation and variance of a discrete probability
distribution, you will find it helpful to set your work out systematically in a table.
(a)
(b)
r
pr
r pr
r 2pr
(r − µ)2pr
1
0.35
0.35
0.35
0.35
2
0.375
0.75
1.5
0
3
0.205
0.615
1.845
0.205
4
0.065
0.26
1.04
0.26
5
0.005
0.025
0.125
0.045
Totals
Σpr = 1
µ = E(X) = 2
4.86
Var(X) = 0.86
In this case:
E(X) = µ = ∑r pr
= 1 × 0.35 + 2 × 0.375 + 3 × 0.205 + 4 × 0.065 + 5 × 0.005
=2
115
S1
4
This is µ.
And either from (a)
Var(X) = σ2 = ∑r 2pr – ∑r pr
= 12 × 0.35 + 22 × 0.375 + 32 × 0.205 + 42 × 0.065 + 52 × 0.005 − 22
2
Discrete random variables
= 4.86 − 4
= 0.86
or from (b)
Var(X) = σ2 = ∑(r – µ)2pr
= (1 − 2)2 × 0.35 + (2 − 2)2 × 0.375 + (3 − 2)2 × 0.205
+ (4 − 2)2 × 0.065 + (5 − 2)2 × 0.005
= 0.86
The equivalence of the two methods is proved in Appendix 2 on the CD.
In practice, method (a) is to be preferred since the computation is usually easier,
especially when the expectation is other than a whole number.
ACTIVITY 4.2
EXAMPLE 4.3
Carry out similar calculations for the expectation and variance of the probability
distribution before the car-sharing experiment using the data on page 106.
Using these two statistics, judge the success or otherwise of the scheme.
The discrete random variable X has the following probability distribution:
r
0
1
2
3
pr
0.2
0.3
0.4
0.1
Find
(i)
(ii)
(iii)
E(X)
E(X 2)
Var(X) using
(a)
E(X 2) – µ2
(b)
E([X − µ]2).
SOLUTION
(b)
r
pr
r pr
r2 pr
(r − µ)2 pr
0
0.2
0
0
0.392
1
0.3
0.3
0.3
0.048
2
0.4
0.8
1.6
0.144
3
0.1
0.3
0.9
0.256
1
1.4
2.8
0.84
Totals
116
(a)
(i)
E(X) = µ = ∑r pr
S1
4
= 0 × 0.2 + 1 × 0.3 + 2 × 0.4 + 3 × 0.1 = 1.4
(ii)
E(X 2) = ∑r 2pr
Var(X) = E(X 2) − µ2
= 2.8 − 1.42 = 0.84
(iii) (a)
Var(X) = E([X − µ]2)
= ∑(r – µ)2pr
= (0 − 1.4)2 × 0.2 + (1 − 1.4)2 × 0.3 + (2 − 1.4)2 × 0.4 + (3 − 1.4)2 × 0.1
= 0.392 + 0.048 + 0.144 + 0.256 = 0.84
(b)
Expectation and variance
= 0 × 0.2 + 1 × 0.3 + 4 × 0.4 + 9 × 0.1 = 2.8
Notice that the two methods of calculating the variance in part (iii) give the same
result, since one formula is just an algebraic rearrangement of the other.
?
●
Look carefully at both methods for calculating the variance.
Are there any situations where one method might be preferred to the other?
As well as being able to carry out calculations for the expectation and variance
you are often required to solve problems in context. The following example
illustrates this idea.
EXAMPLE 4.4
Laura buys one litre of mango juice on three days out of every four and none
on the fourth day. A litre of mango juice costs 40c. Let X represent her weekly
juice bill.
(i)
(ii)
(iii)
Find the probability distribution of her weekly juice bill.
Find the mean ( µ) and standard deviation (σ) of her weekly juice bill.
Find
(a) P(X µ + σ)
(b) P(X µ − σ).
SOLUTION
(i)
The pattern repeats every four weeks.
M
Tu
W
Th
F
Sa
Su
Number of litres
Juice bill
✓
✓
✓
✗
✓
✓
✓
6
$2.40
✗
✓
✓
✓
✗
✓
✓
5
$2.00
✓
✗
✓
✓
✓
✗
✓
5
$2.00
✓
✓
✗
✓
✓
✓
✗
5
$2.00
117
Tabulating the probability distribution for X gives the following.
Discrete random variables
S1
4
(ii)
r ($)
2.00
2.40
P(X = r)
0.75
0.25
E(X) = µ = ∑rP(X = r)
= 2 × 0.75 + 2.4 × 0.25
= 2.1
Var(X) = σ2 = E(X 2) − µ2
= 4 × 0.75 + 5.76 × 0.25 − 2.12
= 0.03
⇒ σ = 0.03 = 0.17 (correct to 2 s.f.)
Hence her mean weekly juice bill is $2.10, with a standard deviation of about
17 cents.
(iii) (a)
(b)
EXERCISE 4B
1
P(X µ + σ) = P(X 2.27) = 0.25
P(X µ − σ) = P(X 1.93) = 0
Find by calculation the expectation of the outcome with the following
probability distribution.
Outcome
Probability
2
(ii)
3 (i)
(ii)
3
4
5
0.1
0.2
0.4
0.2
0.1
Find E(X) = µ.
Find P(X µ).
A discrete random variable X can take only the values 4 and 5, and has
expectation 4.2.
By letting P(X = 4) = p and P(X = 5) = 1 − p, solve an equation in p and so
find the probability distribution of X.
A discrete random variable Y can take only the values 50 and 100.
Given that E(Y) = 80, write out the probability distribution of Y.
The random variable X is given by the sum of the scores when two ordinary
dice are thrown.
(i)
(ii)
118
2
The probability distribution of the discrete random variable X is given by
P(X = r) = 2r – 1
for r = 1, 2, 3, 4
16
P(X = r) = 0
otherwise.
(i)
4
1
(iii)
Use the shape of the distribution to find E(X) = µ.
Confirm your answer by calculation.
Calculate Var(X) = σ2.
Find the values of the following.
(a) P(X µ)
(b) P(X µ + σ)
(c) P ( X − µ 2σ)
5
The random variable Y is given by the absolute difference between the scores
when two ordinary dice are thrown.
(i)
(ii)
Three fair coins are tossed. Let X represent the number of tails.
(i)
Find E(X). Show that this is equivalent to 3 × 12 .
(ii)
Find Var(X). Show that this is equivalent to 3 × 14 .
Exercise 4B
6
Find E(Y ) and Var(Y ).
Find the values of the following.
(a) P(Y µ)
(b) P(Y µ + 2σ)
S1
4
If instead ten fair coins are tossed, let Y represent the number of tails.
(iii)
7
Write down the values of E(Y ) and Var(Y ).
Birds of a particular species lay either 0, 1, 2 or 3 eggs in their nests with
probabilities as shown in the table.
Number of eggs
Probability
0
1
2
3
0.25
0.35
0.30
k
Find
(i) the value of k
(ii) the expected number of eggs laid in a nest
(iii) the standard deviation of the number of eggs laid in a nest.
8
An electronic device produces an output of 0, 1 or 3 volts, with probabilities
1 1
1
2 , 3 and 6 respectively. The random variable X denotes the result of adding the
outputs for two such devices, which act independently.
(i)
(ii)
(iii)
9
Show that P(X = 4) = 19.
Tabulate all the possible values of X with their corresponding probabilities.
Hence calculate E(X ) and Var(X ), giving your answers as fractions in their
lowest terms.
Bob earns $80 per day, Monday to Friday inclusive. He works every alternate
Saturday for which he earns ‘time and a half’ and every fourth Sunday, for
which he is paid ‘double time’.
(i)
(ii)
(iii)
By considering a typical four-week period, find the probability
distribution for his daily wage.
Calculate the expectation and variance of his daily wage.
Show that there are two possible patterns Bob could work over a typical
four-week period, depending on which Saturdays and Sunday he works.
Hence find the expectation and variance of his weekly wage under either
pattern.
119
10
A hotel caters for business clients who make short stays. Past records
suggest that the probability of a randomly chosen client staying X nights in
succession is as follows.
r
Discrete random variables
S1
4
P(X = r)
(i)
(ii)
(iii)
(iv)
1
2
3
4
5
6+
0.42
0.33
0.18
0.05
0.02
0
Draw a sketch of this distribution.
Find the mean and standard deviation of X.
Find the probability that a randomly chosen client who arrives on
Monday evening will still be in the hotel on Wednesday night.
Find the probability that a client who has already stayed two nights will
stay at least one more night.
[MEI]
11
The probability distribution of the discrete random variable X is shown in
the table below.
x
−3
−1
0
4
P(X = x)
a
b
0.15
0.4
Given that E(X ) = 0.75, find the values of a and b.
[Cambridge International AS and A Level Mathematics 9709, Paper 61 Q1 June 2010]
12
Every day Eduardo tries to phone his friend. Every time he phones there is a
50% chance that his friend will answer. If his friend answers, Eduardo does
not phone again on that day. If his friend does not answer, Eduardo tries
again in a few minutes’ time. If his friend has not answered after 4 attempts,
Eduardo does not try again on that day.
(i)
(ii)
Draw a tree diagram to illustrate this situation.
Let X be the number of unanswered phone calls made by Eduardo on a
day. Copy and complete the table showing the probability distribution
of X.
x
P(X = x)
(iii)
0
1
2
3
4
1
4
Calculate the expected number of unanswered phone calls on a day.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q6 June 2008]
120
13
Gohan throws a fair tetrahedral die with faces numbered 1, 2, 3, 4. If she
throws an even number then her score is the number thrown. If she throws
an odd number then she throws again and her score is the sum of both
numbers thrown. Let the random variable X denote Gohan’s score.
(ii)
5
Show that P(X = 2) = 16
.
The table below shows the probability distribution of X.
Exercise 4B
(i)
S1
4
x
2
3
4
5
6
7
P(X = x)
5
16
1
6 16
1 3
6 8
3 1
8 8
1 1
8 16
1
6 16
Calculate E(X ) and Var(X ).
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q2 June 2009]
14
The probability distribution of the random variable X is shown in the
following table.
x
P(X = x)
−2
−1
0
1
2
3
0.08
p
0.12
0.16
q
0.22
The mean of X is 1.05.
(i)
(ii)
Write down two equations involving p and q and hence find the values of
p and q.
Find the variance of X.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q2 November 2009]
15
The random variable X takes the values −2, 0 and 4 only. It is given that
P(X = −2) = 2p, P(X = 0) = p and P(X = 4) = 3p.
(i)
(ii)
Find p.
Find E(X ) and Var(X ).
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q2 November 2007]
121
16
S1
4
A fair die has one face numbered 1, one face numbered 3, two faces
numbered 5 and two faces numbered 6.
Discrete random variables
(i)
Find the probability of obtaining at least 7 odd numbers in 8 throws of
the die.
The die is thrown twice. Let X be the sum of the two scores. The following
table shows the possible values of X.
First throw
Second throw
(ii)
(iii)
(iv)
1
3
5
5
6
6
1
2
4
6
6
7
7
3
4
6
8
8
9
9
5
6
8
10
10
11
11
5
6
8
10
10
11
11
6
7
9
11
11
12
12
6
7
9
11
11
12
12
Draw up a table showing the probability distribution of X.
Calculate E(X ).
Find the probability that X is greater than E(X ).
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q7 November 2008]
KEY POINTS
1
For a discrete random variable, X, which can take only the values
r1, r2, … , rn, with probabilities p1, p2, … , pn respectively:
●
2
p1 + p2 + + pn =
n
n
i =1
i =1
∑ pi = ∑P(X = ri) = pr = 1; pi 0
A discrete probability distribution is best illustrated by a vertical line chart.
∑rP(x = r) = ∑rpr
●
The expectation = E(X) = µ =
●
The variance, where σ is the standard deviation, is
∑(r − µ)2pr
2
Var(X) = σ2 = E(X 2) − [E(X)]2 = ∑r 2pr − ∑rpr
Var(X) = σ2 = E(X − µ)2 =
or
3
Another common notation is to denote the values the variable may take by x.
∑ xp
2
2
● The variance = Var(X ) = ∑ x p − [ E(X )]
●
122
The expectation = E(X) =
S1
5
An estate had seven houses;
Each house had seven cats;
Each cat ate seven mice;
Each mouse ate seven grains of wheat.
Wheat grains, mice, cats and houses,
How many were there on the estate?
Ancient Egyptian problem
Permutations and combinations
5
Permutations and
combinations
ProudMum
My son is a genius!
I gave Oscar five bricks and straightaway he did this!
Is it too early to enrol him with MENSA?
What is the probability that Oscar chose the bricks at random and just happened
by chance to get them in the right order?
There are two ways of looking at the situation. You can think of Oscar selecting
the five bricks as five events, one after another. Alternatively, you can think of
1, 2, 3, 4, 5 as one outcome out of several possible outcomes and work out the
probability that way.
Five events
Look at the diagram.
1
2
3
4
5
Figure 5.1
If Oscar had actually chosen them at random:
the probability of first selecting 1 is 51
the probability of next selecting 2 is 14
1 correct choice from
4 remaining bricks.
123
the probability of next selecting 3 is 13
S1
5
the probability of next selecting 4 is 12
Permutations and combinations
then only 5 remains so the probability of selecting it is 1.
So the probability of getting the correct numerical sequence at random is
1
5
1
× 14 × 13 × 12 × 1 = 120
.
Outcomes
How many ways are there of putting five bricks in a line?
To start with there are five bricks to choose from, so there are five ways of
choosing brick 1. Then there are four bricks left and so there are four ways of
choosing brick 2. And so on.
The total number of ways is
5
×
4
×
3
×
2
×
1
= 120.
Brick 1
Brick 2
Brick 3
Brick 4
Brick 5
Only one of these is the order 1, 2, 3, 4, 5, so the probability of Oscar selecting it
1
at random is 120
.
Number of possible
outcomes.
?
●
Do you agree with Oscar’s mother that he is a child prodigy, or do you think it
was just by chance that he put the bricks down in the right order?
What further information would you want to be convinced that he is a budding
genius?
Factorials
In the last example you saw that the number of ways of placing five different
bricks in a line is 5 × 4 × 3 × 2 × 1. This number is called 5 factorial and is written
5!. You will often meet expressions of this form.
In general the number of ways of placing n different objects in a line is n!, where
n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1.
n must be a
positive integer.
124
EXAMPLE 5.1
Calculate 7!
SOLUTION
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
10! = 10 × 9!
or in general n! = n × [(n − 1)!]
10! = 10 × 9 × 8 × 7! or in general n! = n × (n − 1) × (n − 2) × [(n − 3)!]
Factorials
Some typical relationships between factorial numbers are illustrated below:
S1
5
These are useful when simplifying expressions involving factorials.
EXAMPLE 5.2
Calculate 5!
3!
SOLUTION
5! = 5 × 4 × 3! = 5 × 4 = 20
3!
3!
EXAMPLE 5.3
Calculate
7 ! × 5!
3! × 4!
SOLUTION
7! × 5! = 7 × 6 × 5 × 4 × 3! × 5 × 4 !
3! × 4 !
3! × 4 !
= 7 × 6 × 5 × 4 × 5 = 4200
EXAMPLE 5.4
Write 37 × 36 × 35 in terms of factorials only.
SOLUTION
37 × 36 × 35 = 37 × 36 × 35 × 34 !
34 !
!
37
=
34 !
EXAMPLE 5.5
(i)
(ii)
Find the number of ways in which all five letters in the word GREAT can be
arranged.
In how many of these arrangements are the letters A and E next to each
other?
SOLUTION
(i)
There are five choices for the first letter (G, R, E, A or T). Then there are four
choices for the next letter, then three for the third letter and so on. So the
number of arrangements of the letters is
5 × 4 × 3 × 2 × 1 = 5! = 120
125
(ii)
S1
5
The E and the A are to be together, so you can treat them as a single letter.
So there are four choices for the first letter (G, R, EA or T), three choices for
the next letter and so on.
Permutations and combinations
So the number of arrangements of these four ‘letters’ is
4 × 3 × 2 × 1 = 4! = 24
However
EA
G
R
T
is different from
AE
G
R
T
So each of the 24 arrangements can be arranged into two different orders.
The total number of arrangements with the E and A next to each other is
2 × 4! = 48
Note
The total number of ways of arranging the letters with the A and the E apart is
120 – 48 = 72
Sometimes a question will ask you to deal with repeated letters.
EXAMPLE 5.6
Find the number of ways in which all five letters in the word GREET can be
arranged.
SOLUTION
There are 5! = 120 arrangements of five letters.
However, GREET has two repeated letters and so some of these arrangements are
really the same.
For example,
E
E
G
R
T
is the same as
E
E
G
R
T
The two Es can be arranged in 2! = 2 ways, so the total number of arrangements is
5!
= 60.
2!
EXAMPLE 5.7
How many different arrangements of the letters in the word MATHEMATICAL
are there?
SOLUTION
There are 12 letters, so there are 12! = 479 001 600 arrangements.
126
However, there are repeated letters and so some of these arrangements are the same.
For example,
A
T
H
E
M
A
T
I
C
A
L
M
A
T
H
E
M
A
T
I
C
A
L
M
A
T
H
E
M
A
T
I
C
A
L
are the same.
In fact, there are 3! = 6 ways of arranging the As.
S1
5
Exercise 5A
and
M
So the total number of arrangements of
M
A
T
H
E
M
A
T
I
C
A
L is
12 letters
12!
––––––––– = 19 958 400
2! × 2! × 3!
Three As repeated
Two Ts repeated
Two Ms repeated
Example 5.7 illustrates how to deal with repeated objects. You can generalise
from this example to obtain the following:
●
The number of distinct arrangements of n objects in a line, of which p are
identical to each other, q others are identical to each other, r of a third type are
identical, and so on is n ! .
p !q !r !…
EXERCISE 5A
1
Calculate
(i)
8!
(ii)
8!
6!
2
Simplify
(i)
(n − 1)!
n!
(ii)
(n − 1)!
(n − 2)!
3
Simplify
(i)
(n + 3)!
(n + 1)!
(ii)
n!
(n − 2)!
4
Write in factorial notation.
8×7×6
(i)
5×4×3
(ii)
15 × 16
4×3×2
(ii)
n! + (n + 1)!
7! + 8!
(iii)
5! × 6!
7! × 4 !
(iii)
(n + 1)n(n − 1)
4×3×2
5
Factorise
6
How many different four letter words can be formed from the letters A, B, C and
D if letters cannot be repeated? (The words do not need to mean anything.)
7
How many different ways can eight books be arranged in a row on a shelf?
8
In a greyhound race there are six runners. How many different ways are there
for the six dogs to finish?
(i)
127
S1
5
9
In a 60-metre hurdles race there are five runners, one from each of the
nations Austria, Belgium, Canada, Denmark and England.
(i)
Permutations and combinations
(ii)
10
How many different finishing orders are there?
What is the probability of predicting the finishing order by choosing
first, second, third, fourth and fifth at random?
John has an MP3 player which can play tracks in ‘shuffle’ mode. If an album
is played in ‘shuffle’ mode the tracks are selected in a random order with a
different track selected each time until all the tracks have been played.
John plays a 14-track album in ‘shuffle’ mode.
(i)
(ii)
11
In how many different orders could the tracks be played?
What is the probability that ‘shuffle’ mode will play the tracks in the
normal set order listed on the album?
In a ‘Goal of the season’ competition, participants are asked to rank ten goals
in order of quality.
The organisers select their ‘correct’ order at random. Anybody who matches
their order will be invited to join the television commentary team for the
next international match.
(i)
(ii)
12
The letters O, P, S and T are placed in a line at random. What is the
probability that they form a word in the English language?
13
Find how many arrangements there are of the letters in each of these words.
(i)
(iv)
14
EXAM
PASS
(ii)
(v)
MATHS
SUCCESS
(iii)
(vi)
CAMBRIDGE
STATISTICS
How many arrangements of the word ACHIEVE are there if
(i)
(ii)
(iii)
(iv)
128
What is the probability of a participant’s order being the same as that of
the organisers?
Five million people enter the competition. How many people would be
expected to join the commentary team?
there are no restrictions on the order the letters are to be in
the first letter is an A
the letters A and I are to be together.
the letters C and H are to be apart.
INVESTIGATIONS
1
Solve the inequality n! 10m for each of the cases m = 3, 4, 5.
2
In how many ways can you write 42 using factorials only?
(ii)
There are 4! ways of placing the four letters S, T, A, R in a line, if each of
them must appear exactly once. How many ways are there if each letter
may appear any number of times (i.e. between 0 and 4)? Formulate a
general rule.
There are 4! ways of placing the letters S, T, A, R in line. How many ways
are there of placing in line the letters
(a) S, T, A, A
(b) S, T, T, T?
Formulate a general rule for dealing with repeated letters.
Permutations
3 (i)
S1
5
Permutations
I should be one of the judges! When I heard the 16 songs in the competition, I knew
which ones I thought were the best three. Last night they announced the results and I
had picked the same three songs in the same order as the judges!
Joyetta
What is the probability of Joyeeta’s result?
The winner can be chosen in 16 ways.
The second song can be chosen in 15 ways.
The third song can be chosen in 14 ways.
Thus the total number of ways of placing three songs in the first three positions is
1
.
16 × 15 × 14 = 3360. So the probability that Joyeeta’s selection is correct is 3360
In this example attention is given to the order in which the songs are placed. The
solution required a permutation of three objects from sixteen.
In general the number of permutations, nPr , of r objects from n is given by
nP
r
= n × (n − 1) × (n − 2) × ... × (n − r + 1).
This can be written more compactly as
●
nP
r
=
n!
(n − r)!
129
Six people go to the cinema. They sit in a row with ten seats. Find how many
ways can this be done if
(i) they can sit anywhere
(ii) all the empty seats are next to each other.
EXAMPLE 5.8
Permutations and combinations
S1
5
SOLUTION
(i)
The first person to sit down has a choice of ten seats.
The second person to sit down has a choice of nine seats.
The third person to sit down has a choice of eight seats.
...
The sixth person to sit down has a choice of five seats.
So the total number of arrangements is 10 × 9 × 8 × 7 × 6 × 5 = 151 200.
This is a permutation of six objects from ten, so a quicker way to work this
out is
number of arrangements = 10P6 = 151 200
(ii)
Since all four empty seats are to be together you can consider them to be a
single ‘empty seat’, albeit a large one!
So there are seven seats to seat six people.
So the number of arrangements is 7P6 = 5040
Combinations
It is often the case that you are not concerned with the order in which items are
chosen, only with which ones are picked.
To take part in the UK National Lottery you fill in a ticket by selecting six
numbers out of a possible 49 (numbers 1, 2, . . . , 49). When the draw is made a
machine selects six numbers at random. If they are the same as the six on your
ticket, you win the jackpot.
?
●
You have the six winning numbers. Does it matter in which order the machine
picked them?
The probability of a single ticket winning the jackpot is often said to be 1 in
14 million. How can you work out this figure?
The key question is, how many ways are there of choosing six numbers out of 49?
If the order mattered, the answer would be 49P6, or 49 × 48 × 47 × 46 × 45 × 44.
130
However, the order does not matter. The selection 1, 3, 15, 19, 31 and 48 is the
same as 15, 48, 31, 1, 19, 3 and as 3, 19, 48, 1, 15, 31, and lots more. For each set
of six numbers there are 6! arrangements that all count as being the same.
So, the number of ways of selecting six balls, given that the order does not
matter, is
49 × 48 × 47 × 46 × 45 × 44 .
6!
49P
6
This is –––
6!
?
●
Show that 49C6 can be written as
49!
.
6! × 43!
Combinations
This is called the number of combinations of 6 objects from 49 and is denoted
by 49C6.
S1
5
Returning to the UK National Lottery, it follows that the probability of your one
ticket winning the jackpot is 491 .
C6
?
●
Check that this is about 1 in 14 million.
This example shows a general result, that the number of ways of selecting r
objects from n, when the order does not matter, is given by
n
Cr =
?
●
n
P
n!
= r
r!(n − r)! r !
How can you prove this general result?
n
Another common notation for nCr is . Both notations are used in this book to
r
help you become familiar with both of them.
!
EXAMPLE 5.9
n
The notation looks exactly like a column vector and so there is the possibility
r
of confusing the two. However, the context should usually make the meaning clear.
A School Governors’ committee of five people is to be chosen from eight
applicants. How many different selections are possible?
SOLUTION
8
8! = 8 × 7 × 6 = 56
Number of selections = =
5 5! × 3! 3 × 2 × 1
131
Permutations and combinations
S1
5
In how many ways can a committee of four people be selected from four applicants?
EXAMPLE 5.10
SOLUTION
Common sense tells us that there is only one way to make the committee, that is
by appointing all applicants. So 4C4 = 1. However, if we work from the formula
4C
4
4! = 1
4! × 0! 0!
=
For this to equal 1 requires the convention that 0! is taken to be 1.
?
●
Use the convention 0! = 1 to show that nC0 = nCn = 1 for all values of n.
The binomial coefficients
n
In the last section you met numbers of the form nCr or . These are called the
r
binomial coefficients; the reason for this is explained in Appendix 3 (which you
can find on the CD) and in the next chapter.
ACTIVITY 5.1
n
n n
n!
Use the formula =
and the results = = 1 to check that
r r !(n − r)!
0 n
the entries in this table, for n = 6 and 7, are correct.
r
0
1
2
3
4
5
6
7
n=6
1
6
15
20
15
6
1
–
n=7
1
7
21
35
35
21
7
1
It is very common to present values of nCr in a table shaped like an isosceles
triangle, known as Pascal’s triangle.
This completes
the triangle.
1
1
1
This is 4C0.
1
1
…
4
5
6
15
…
1
3
10
6
…
2
3
1
1
1
4
10
20
…
1
5
15
…
This is 5C3.
1
1
6
…
1
…
…
The numbers in this row are the same as
those in the first row of the table above.
132
Pascal’s triangle makes it easy to see two important properties of binomial
coefficients.
1 Symmetry: nCr = nCn–r
This provides a short cut in calculations when r is large. For example
100C
96
= 100C4 = 100 × 99 × 98 × 97 = 3 921 225.
1× 2× 3× 4
It also shows that the list of values of nCr for any particular value of n is
unchanged by being reversed. For example, when n = 6 the list is the seven
numbers 1, 6, 15, 20, 15, 6, 1.
2 Addition:
n+1C
r+1
= nCr + nCr+1
Look at the entry 15 in the bottom row of Pascal’s triangle, towards the right.
The two entries above and either side of it are 10 and 5,
…
5C
3
10
…
…
5
15
…
5C
4
S1
5
Using binomial coefficients to calculate probabilities
If you are choosing 11 players from a pool of 15 possible players you can either
name the 11 you have selected or name the 4 you have rejected. Similarly, every
choice of r objects included in a selection from n distinct objects corresponds to a
choice of (n − r) objects which are excluded. Therefore nCr = nCn–r.
6C
4
and 15 = 10 + 5. In this case 6C4 = 5C3 + 5C4. This is an example of the general
result that n+1Cr+1 = n Cr + n Cr+1. Check that all the entries in Pascal’s triangle
(except the 1s) are found in this way.
This can be used to build up a table of values of n Cr without much calculation. If
you know all the values of n Cr for any particular value of n you can add pairs of
values to obtain all the values of n+1Cr , i.e. the next row, except the first and last,
which always equal 1.
Using binomial coefficients to calculate probabilities
EXAMPLE 5.11
A committee of 5 is to be chosen from a list of 14 people, 6 of whom are men
and 8 women. Their names are to be put in a hat and then 5 drawn out.
What is the probability that this procedure produces a committee with no
women?
SOLUTION
The probability of an all-male committee of 5 people is given by
There are 6 men.
6C
the number of ways of choosing 5 people out of 6
6
–––––––––––––––––––––––––––––––––––––––––––– = 14 5 =
≈ 0.003
2002
C
the number of ways of choosing 5 people out of 14
5
There are 14 people.
133
Permutations and combinations
S1
5
b GoByBus.com b
Help decide our new bus routes
The exact route for our new bus
service is to be announced in April.
Rest assured our service will run
from Amli to Chatra via Bawal and
will be extended to include Dhar
once our new fleet of buses arrives
in September. As local people know,
there are several roads connecting
these towns and we are keen to hear the views as to the most useful
routes from our future passengers. Please post your views below!
This consultation is a farce. The chance of
getting a route that suits me is less than one
RChowdhry in a hundred :(
Is RChowdhry right? How many routes are there from Amli to Dhar? Start by
looking at the first two legs, Amli to Bawal and Bawal to Chatra.
There are three roads from Amli to Bawal and two roads from Bawal to Chatra.
How many routes are there from Amli to Chatra passing through Bawal on
the way?
Look at figure 5.2.
Amli
x
Bawal
u
Chatra
y
z
v
Figure 5.2
The answer is 3 × 2 = 6 because there are three ways of doing the first leg,
followed by two for the second leg. The six routes are
x−u
x−v
134
y−u
y−v
z−u
z − v.
There are also four roads from Chatra to Dhar. So each of the six routes from
Amli to Chatral has four possible ways of going on to Dhar. There are now
6 × 4 = 24 routes. See figure 5.3.
Bawal
p
Chatra
u
y
q
v
z
Dhar
r
s
Figure 5.3
They can be listed systematically as follows:
x−u−p
x−u−q
x−u−r
x−u−s
y−u−p
................
................
................
z−u−p
................
................
................
x−v−p
................
................
................
y−v−p
................
................
................
z −v−p
................
................
z−v−s
In general, if there are a outcomes from experiment A, b outcomes from
experiment B and c outcomes from experiment C then there are a × b × c different
possible combined outcomes from the three experiments.
?
●
EXAMPLE 5.12
1
If GoByBus chooses its route at random, what is the probability that it will be
the one RChowdhry wants? Is the comment justified?
2
In this example the probability was worked out by finding the number of
possible routes. How else could it have been worked out?
Using binomial coefficients to calculate probabilities
x
Amli
S1
5
A cricket team consisting of 6 batsmen, 4 bowlers and 1 wicket-keeper is to be
selected from a group of 18 cricketers comprising 9 batsmen, 7 bowlers and 2
wicket-keepers. How many different teams can be selected?
SOLUTION
The batsmen can be selected in 9C6 ways.
The bowlers can be selected in 7C4 ways.
The wicket-keepers can be selected in 2C1 ways.
Therefore total number of teams = 9C6 × 7C4 × 2C1
9! × 7 ! × 2!
3! × 6! 3! × 4 ! 1! × 1!
=9×8×7×7×6×5×2
3× 2×1 3× 2×1
= 5880
=
135
Permutations and combinations
S1
5
EXAMPLE 5.13
In a dance competition, the panel of ten judges sit on the same side of a long
table. There are three female judges.
(i)
How many different arrangements are there for seating the ten judges?
(ii)
How many different arrangements are there if the three female judges all
decide to sit together?
(iii)
If the seating is at random, find the probability that the three female judges
will not all sit together.
(iv)
Four of the judges are selected at random to judge the final round of the
competition. Find the probability that this final judging panel consists of two
men and two women.
SOLUTION
(i)
(ii)
There are 10! = 3 628 800 ways of arranging the judges in a line.
If the three female judges sit together then you can treat them as a single
judge.
So there are eight judges and there are 8! = 40 320 ways of arranging the
judges in a line.
However, there are 3! = 6 ways of arranging the female judges.
So there are 3! × 8! = 241 920 ways of arranging the judges so that all the
female judges are together.
(iii)
There are 3 628 800 − 241 920 = 3 386 880 ways of arranging the judges so that
the female judges do not all sit together.
So the probability that the female judges do not all sit together is
3 386 880
= 0.933 (to 3 s.f.).
3 628 800
(iv)
The probability of selecting two men and two women on the panel of four is
3C
136
2×
10C
7C
4
2
3! × 7 ! ÷ 10!
1! × 2! 5! × 2! 6! × 4 !
= 3 × 21 ÷ 210
= 0.3
=
EXERCISE 5B
1 (i)
(ii)
(iii)
Find the values of
(a) 6P2
(b) 8P4
(c) 10P4.
Find the values of (a) 6C2
(b) 8C4
(c) 10C4.
Show that, for the values of n and r in parts (i) and (ii),
=
nP
r.
r!
2
There are 15 runners in a camel race. What is the probability of correctly
guessing the first three finishers in their finishing order?
3
To win the jackpot in a lottery a contestant must correctly select six numbers
from the numbers 1 to 30 inclusive. What is the probability that a contestant
wins the jackpot with one selection of six numbers?
4
A group of 5 computer programmers is to be chosen to form the night shift
from a set of 14 programmers. In how many ways can the programmers be
chosen if the 5 chosen must include the shift-leader who is one of the 14?
5
My brother Mark decides to put together a rock band from amongst his year at
school. He wants a lead singer, a guitarist, a keyboard player and a drummer.
He invites applications and gets 7 singers, 5 guitarists, 4 keyboard players and
2 drummers. Assuming each person applies only once, in how many ways can
Mark put the group together?
6
A touring party of cricket players is made up of 5 players from each of India,
Pakistan and Sri Lanka and 3 from Bangladesh.
(i)
(ii)
7
How many different selections of 11 players can be made for a team?
In one match, it is decided to have 3 players from each of India, Pakistan
and Sri Lanka and 2 from Bangladesh. How many different team selections
can now be made?
A committee of four is to be selected from ten candidates, six men and four
women.
(i)
(ii)
8
Exercise 5B
nC
r
S1
5
In how many distinct ways can the committee be chosen?
Assuming that each candidate is equally likely to be selected, determine the
probabilities that the chosen committee contains
(a) no women
(b) two men and two women.
A committee of four is to be selected from five boys and four girls. The
members are selected at random.
(i)
(ii)
How many different selections are possible?
What is the probability that the committee will be made up of
(a) all girls?
(b) more boys than girls?
137
Permutations and combinations
S1
5
9
Baby Imran has a set of alphabet blocks. His mother often uses the blocks
I, M, R, A and N to spell Imran’s name.
One day she leaves him playing with these five blocks. When she comes
back into the room Imran has placed them in the correct order to spell
his name. What is the probability of Imran placing the blocks in this
order? (He is only 18 months old so he certainly cannot spell!)
A couple of days later she leaves Imran playing with all 26 of the alphabet
blocks. When she comes back into the room she again sees that he has
placed the five blocks I, M, R, A and N in the correct order to spell his
name. What is the probability of him choosing the five correct blocks
and placing them in this order?
(i)
(ii)
A football team consists of 3 players who play in a defence position,
3 players who play in a midfield position and 5 players who play in a
forward position. Three players are chosen to collect a gold medal for the
team. Find in how many ways this can be done
(a) if the captain, who is a midfield player, must be included, together
with one defence and one forward player.
(b) if exactly one forward player must be included, together with any two
others.
10 (i)
Find how many different arrangements there are of the nine letters in the
words GOLD MEDAL
(a) if there are no restrictions on the order of the letters,
(b) if the two letters D come first and the two letters L come last.
(ii)
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q7 June 2005]
11
The diagram shows the seating plan for passengers in a minibus, which has
17 seats arranged in 4 rows. The back row has 5 seats and the other 3 rows
have 2 seats on each side. 11 passengers get on the minibus.
Back
(i)
(ii)
Front
How many possible seating arrangements are there for the 11 passengers?
How many possible seating arrangements are there if 5 particular people
sit in the back row?
Of the 11 passengers, 5 are unmarried and the other 6 consist of 3 married
couples.
(iii)
138
In how many ways can 5 of the 11 passengers on the bus be chosen if
there must be 2 married couples and 1 other person, who may or may
not be married?
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q4 June 2006]
12
Issam has 11 different CDs, of which 6 are pop music, 3 are jazz and 2 are
classical.
(i)
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q3 June 2008]
13
Exercise 5B
(ii)
How many different arrangements of all 11 CDs on a shelf are there if the
jazz CDs are all next to each other?
Issam makes a selection of 2 pop music CDs, 2 jazz CDs and 1 classical
CD. How many different possible selections can be made?
S1
5
A choir consists of 13 sopranos, 12 altos, 6 tenors and 7 basses. A group
consisting of 10 sopranos, 9 altos, 4 tenors and 4 basses is to be chosen from
the choir.
(i)
(ii)
(iii)
In how many different ways can the group be chosen?
In how many ways can the 10 chosen sopranos be arranged in a line if the
6 tallest stand next to each other?
The 4 tenors and the 4 basses in the group stand in a single line with all
the tenors next to each other and all the basses next to each other. How
many possible arrangements are there if three of the tenors refuse to
stand next to any of the basses?
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q4 June 2009]
14
A staff car park at a school has 13 parking spaces in a row. There are 9 cars to
be parked.
(i)
(ii)
(iii)
How many different arrangements are there for parking the 9 cars and
leaving 4 empty spaces?
How many different arrangements are there if the 4 empty spaces are
next to each other?
If the parking is random, find the probability that there will not be 4
empty spaces next to each other.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q3 November 2005]
15
A builder is planning to build 12 houses along one side of a road. He will
build 2 houses in style A, 2 houses in style B, 3 houses in style C, 4 houses in
style D and 1 house in style E.
(i)
Find the number of possible arrangements of these 12 houses.
(ii)
Road
First group
(iii)
Second group
The 12 houses will be in two groups of 6 (see diagram). Find the number
of possible arrangements if all the houses in styles A and D are in the first
group and all the houses in styles B, C and E are in the second group.
Four of the 12 houses will be selected for a survey. Exactly one house
must be in style B and exactly one house in style C. Find the number of
ways in which these four houses can be selected.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q4 November 2008]
139
16 (i)
Permutations and combinations
S1
5
(ii)
Find how many numbers between 5000 and 6000 can be formed from
the digits 1, 2, 3, 4, 5 and 6
(a) if no digits are repeated,
(b) if repeated digits are allowed.
Find the number of ways of choosing a school team of 5 pupils from
6 boys and 8 girls
(a) if there are more girls than boys in the team,
(b) if three of the boys are cousins and are either all in the team or all not
in the team.
[Cambridge International AS and A Level Mathematics 9709, Paper 61 Q5 November 2009]
KEY POINTS
1
The number of ways of arranging n unlike objects in a line is n!
2
n! = n × (n – 1) × (n – 2) × (n – 3) × ... × 3 × 2 × 1.
3
The number of distinct arrangements of n objects in a line, of which p are
identical to each other, q others are identical to each other, r of a third type
are identical, and so on is
n!
.
p !q !r !…
4
The number of permutations of r objects from n is
nP
r
5
n! .
(n − r)!
The number of combinations of r objects from n is
nC
r
140
=
=
n! .
(n − r)!r !
6
For permutations the order matters. For combinations it does not.
7
By convention 0! = 1.
To be or not to be, that is the question.
Shakespeare (Hamlet)
Innovate.com
Samantha’s great invention
Mother of three, Samantha Weeks, has
done more than her bit to protect the
environment. She has invented the first
full spectrum LED bulb to operate on
stored solar energy.
Now Samantha is out to prove that she
is not only a clever scientist but a smart
business women as well. For Samantha
is setting up her own factory to make
and sell her bulbs.
Samantha admits there are still some
technical problems ...
S1
6
The binomial distribution
6
The binomial distribution
Samantha Weeks hopes to make a big
success of her light industry
Samantha’s production process is indeed not very good and there is a probability
of 0.1 that any bulb will be substandard and so not last as long as it should.
She decides to sell her bulbs in packs of three. She believes that if one bulb in a
pack is substandard the customers will not complain but that if two or more are
substandard they will do so. She also believes that complaints should be kept
down to no more than 2.5% of customers. Does she meet her target?
Imagine a pack of Samantha’s bulbs. There are eight different ways that good (G)
and substandard (S) bulbs can be arranged in Samantha’s packs, each with its
associated probability.
Arrangement
Probability
Good
Substandard
G G G
0.9 × 0.9 × 0.9 = 0.729
3
0
G G S
0.9 × 0.9 × 0.1 = 0.081
2
1
G S G
0.9 × 0.1 × 0.9 = 0.081
2
1
S G G
0.1 × 0.9 × 0.9 = 0.081
2
1
G S S
0.9 × 0.1 × 0.1 = 0.009
1
2
S G S
0.1 × 0.9 × 0.1 = 0.009
1
2
S S G
0.1 × 0.1 × 0.9 = 0.009
1
2
S S S
0.1 × 0.1 × 0.1 = 0.001
0
3
141
Putting these results together gives this table.
The binomial distribution
S1
6
Good
Substandard
Probability
3
0
0.729
2
1
0.243
1
2
0.027
0
3
0.001
So the probability of more than one substandard bulb in a pack is
0.027 + 0.001 = 0.028 or 2.8%.
This is slightly more than the 2.5% that Samantha regards as acceptable.
?
●
What business advice would you give Samantha?
In this example we wrote down all the possible outcomes and found their
probabilities one at a time. Even with just three bulbs this was repetitive. If
Samantha had packed her bulbs in boxes of six it would have taken 64 lines to list
them all. Clearly you need a more efficient approach.
You will have noticed that in the case of two good bulbs and one substandard,
the probability is the same for each of the three arrangements in the box.
Arrangement
Probability
Good
Substandard
G G S
0.9 × 0.9 × 0.1 = 0.081
2
1
G S G
0.9 × 0.1 × 0.9 = 0.081
2
1
S G G
0.1 × 0.9 × 0.9 = 0.081
2
1
So the probability of this outcome is 3 × 0.081 = 0.243. The number 3 arises
because there are three ways of arranging two good and one substandard bulb in
the box. This is a result you have already met in the previous chapter but written
slightly differently.
EXAMPLE 6.1
How many different ways are there of arranging the letters GGS?
SOLUTION
Since all the letters are either G or S, all you need to do is to count the number of
ways of choosing the letter G two times out of three letters. This is
142
3C
2
=
3! = 6 = 3.
2! × 1! 2
So what does this tell you? There was no need to list all the possibilities for
Samantha’s boxes of bulbs. The information could have been written down
like this.
Substandard
Expression
3
0
3C
2
1
3C
1
2
3C
0
3
3C
Probability
3
0.729
2
1
2(0.9) (0.1)
0.243
3(0.9)
1
(0.9)1(0.1)2
3
0(0.1)
0.027
0.001
The binomial distribution
Good
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6
The binomial distribution
Samantha’s light bulbs are an example of a common type of situation which is
modelled by the binomial distribution. In describing such situations in this book,
we emphasise the fact by using the word trial rather than the more general term
experiment.
●
You are conducting trials on random samples of a certain size, denoted by n.
●
There are just two possible outcomes (in this case substandard and good).
These are often referred to as success and failure.
●
Both outcomes have fixed probabilities, the two adding to 1. The probability
of success is usually called p, that of failure q, so p + q = 1.
●
The probability of success in any trial is independent of the outcomes of
previous trials.
You can then list the probabilities of the different possible outcomes as in the
table above.
The method of the previous section can be applied more generally. You can call the
probability of a substandard bulb p (instead of 0.1), the probability of a good bulb
q (instead of 0.9) and the number of substandard bulbs in a packet of three, X.
Then the possible values of X and their probabilities are as shown in the table
below.
r
0
1
2
3
P(X = r)
q3
3pq 2
3p 2q
p3
This package of values of X with their associated probabilities is called a binomial
probability distribution, a special case of a discrete random variable.
If Samantha decided to put five bulbs in a packet the probability distribution
would be as shown in the following table.
143
The binomial distribution
S1
6
r
0
1
2
3
4
5
P(X = r)
q5
5pq 4
10p 2q 3
10p 3q 2
5p 4q
p5
10 is 5C2.
The entry for X = 2, for example, arises because there are two ‘successes’
(substandard bulbs), giving probability p 2, and three ‘failures’ (good bulbs),
giving probability q 3, and these can happen in 5C2 = 10 ways. This can be written
as P(X = 2) = 10p 2q 3.
If you are already familiar with the binomial theorem, you will notice that the
probabilities in the table are the terms of the binomial expansion of (q + p)5. This
is why this is called a binomial distribution. Notice also that the sum of these
probabilities is (q + p)5 = 15 = 1, since q + p = 1, which is to be expected since the
distribution covers all possible outcomes.
Note
The binomial theorem on the expansion of powers such as (q + p)n is covered in
Pure Mathematics 1. The essential points are given in Appendix 3 on the CD.
The general case
The general binomial distribution deals with the possible numbers of successes
when there are n trials, each of which may be a success (with probability p) or a
failure (with probability q); p and q are fixed positive numbers and p + q = 1. This
distribution is denoted by B(n, p). So, the original probability distribution for the
number of substandard bulbs in Samantha’s boxes of three is B(3, 0.1).
For B(n, p), the probability of r successes in n trials is found by the same argument
as before. Each success has probability p and each failure has probability q, so the
probability of r successes and (n − r) failures in a particular order is prqn−r. The
positions in the sequence of n trials which the successes occupy can be chosen in
nC ways. Therefore
r
P(X = r) = nCr prqn−r for 0 r n.
This can also be written as
n –r
n
pr = pr (1 – p ) .
r
The successive probabilities for X = 0, 1, 2, ..., n are the terms of the binomial
expansion of (q + p)n.
144
Notes
1 The number of successes, X, is a variable which takes a restricted set of values
(X = 0, 1, 2, ..., n) each of which has a known probability of occurring. This is an
S1
6
example of a random variable. Random variables are usually denoted by upper
case, such as r. To state that X has the binomial distribution B(n, p) you can use
the abbreviation X B(n, p), where the symbol means ‘has the distribution’.
Exercise 6A
case letters, such as X, but the particular values they may take are written in lower
2 It is often the case that you use a theoretical distribution, such as the binomial,
to describe a random variable that occurs in real life. This process is called
modelling and it enables you to carry out relevant calculations. If the theoretical
distribution matches the real life variable perfectly, then the model is perfect.
Usually, however, the match is quite good but not perfect. In this case the results
of any calculations will not necessarily give a completely accurate description of
the real life situation. They may, nonetheless, be very useful.
EXERCISE 6A
1
The recovery ward in a maternity hospital has six beds. What is the
probability that the mothers there have between them four girls and two
boys? (You may assume that there are no twins and that a baby is equally
likely to be a girl or a boy.)
2
A typist has a probability of 0.99 of typing a letter correctly. He makes his
mistakes at random. He types a sentence containing 200 letters. What is the
probability that he makes exactly one mistake?
3
In a well-known game you have to decide which your opponent is going to
choose: ‘Paper’, ‘Stone’ or ‘Scissors’. If you guess entirely at random, what is
the probability that you are right exactly 5 times out of 15?
4
There is a fault in a machine making microchips, with the result that only 80%
of those it produces work. A random sample of eight microchips made by this
machine is taken. What is the probability that exactly six of them work?
5
An airport is situated in a place where poor visibility (less than 800 m) can
be expected 25% of the time. A pilot flies into the airport on ten different
occasions.
(i) What is the probability that he encounters poor visibility exactly four
times?
(ii) What other factors could influence the probability?
6
Three coins are tossed.
(i) What is the probability of all three showing heads?
(ii) What is the probability of two heads and one tail?
(iii) What is the probability of one head and two tails?
(iv) What is the probability of all three showing tails?
(v) Show that the probabilities for the four possible outcomes add up to 1.
145
7
S1
6
A coin is tossed ten times.
What is the probability of it coming down heads five times and tails five
times?
(ii) Which is more likely: exactly seven heads or more than seven heads?
The binomial distribution
(i)
8
In an election 30% of people support the Progressive Party. A random
sample of eight voters is taken.
(i) What is the probability that it contains
(a) 0
(b) 1
(c) 2
(d) at least 3 supporters of the Progressive Party?
(ii) Which is the most likely number of Progressive Party supporters to find
in a sample of size eight?
9
There are 15 children in a class.
(i) What is the probability that
(a) 0
(b) 1
(c) 2
(d) at least 3 were born in January?
(ii) What assumption have you made in answering this question? How valid
is this assumption in your view?
10
Criticise this argument.
If you toss two coins they can come down three ways: two heads, one head and
one tail, or two tails. There are three outcomes and so each of them must have
probability one third.
The expectation and variance of B(n, p)
EXAMPLE 6.2
The number of substandard bulbs in a packet of three of Samantha’s bulbs is
modelled by the random variable X where X B(3, 0.1).
(i)
(ii)
Find the expected frequencies of obtaining 0, 1, 2 and 3 substandard bulbs in
2000 packets.
Find the mean number of substandard bulbs per packet.
SOLUTION
(i)
P(X = 0) = 0.729 (as on page 143), so the expected frequency of packets with
no substandard bulbs is 2000 × 0.729 = 1458.
Similarly, the other expected frequencies are
for 1 substandard bulb: 2000 × 0.243 = 486
for 2 substandard bulbs: 2000 × 0.027 = 54
for 3 substandard bulbs: 2000 × 0.001 = 2.
(ii)
The expected total of substandard bulbs in 2000 packets is
0 × 1458 + 1 × 486 + 2 × 54 + 3 × 2 = 600.
146
Check:
1458 + 486 + 54 + 2 = 2000
This is also called
the expectation.
600
Therefore the mean number of substandard bulbs per packet is 2000
= 0.3.
3
i.e. by finding ∑rP(X = r). This is the standard method for finding an
r =0
expectation, as you saw in Chapter 4.
Notice also that the mean or expectation of X is 0.3 = 3 × 0.1 = np. The result for
the general binomial distribution is the same:
●
if X B(n, p) then the expectation or mean of X = µ = np.
This seems obvious: if the probability of success in each single trial is p, then the
expected numbers of successes in n independent trials is np. However, since what
seems obvious is not always true, a proper proof is required.
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6
Using the binomial distribution
Notice in this example that to calculate the mean we have multiplied each
probability by 2000 to get the frequency, multiplied each frequency by the
number of faulty bulbs, added these numbers together and finally divided by
2000. Of course we could have obtained the mean with less calculation by just
multiplying each number of faulty bulbs by its probability and then summing,
Let us take the case when n = 5. The distribution table for B(5, p) is as on
page 144, and the expectation of X is
0 × q 5 + 1 × 5pq 4 + 2 × 10p 2q 3 + 3 × 10p 3q 2 + 4 × 5p 4q + 5 × p 5
= 5pq 4 + 20p 2q 3 + 30p 3q 2 + 20p 4 q + 5p 5
= 5p(q 4 + 4pq 3 + 6p 2q 2 + 4p 3q + p 4)
Take out the common
factor 5p.
= 5p(q + p) 4
= 5p
Since q + p = 1.
The proof in the general case follows the same pattern: the common factor is
now np, and the expectation simplifies to np(q + p)n−1 = np. The details are more
fiddly because of the manipulations of the binomial coefficients.
Similarly, you can show that in this case the variance of X is given by 5pq. This is
an example of the general results that for a binomial distribution
ACTIVITY 6.1
●
mean = µ = np
●
variance, Var(X) = σ 2 = npq = np(1 − p)
●
standard deviation = σ = npq = np (1 – p) .
If you want a challenge write out the details of the proof that if X B(n, p) then
the expectation of X is np.
Using the binomial distribution
EXAMPLE 6.3
Which is more likely: that you get at least one 6 when you throw a die six times,
or that you get at least two 6s when you throw it twelve times?
147
SOLUTION
S1
6
1
On a single throw of a die the probability of getting a 6 is 6 and that of not getting
a 6 is 56.
1
The binomial distribution
So the probability distributions for the two situations required are B(6, 6) and
1
B(12, 6) giving probabilities of:
6
1 − 6C0(56) = 1 − 0.335 = 0.665 (at least one 6 in six throws)
and
1−
[12C0(56)12 + 12C1(56)11(16)] = 1 − (0.112 + 0.269)
= 0.619 (at least two 6s in 12 throws)
So at least one 6 in six throws is somewhat more likely.
EXAMPLE 6.4
Extensive research has shown that 1 person out of every 4 is allergic to a particular
grass seed. A group of 20 university students volunteer to try out a new treatment.
(i)
(ii)
(iii)
(iv)
What is the expectation of the number of allergic people in the group?
What is the probability that
(a) exactly two
(b) no more than two of the group are allergic?
How large a sample would be needed for the probability of it containing at
least one allergic person to be greater than 99.9%?
What assumptions have you made in your answer?
SOLUTION
This situation is modelled by the binomial distribution with n = 20, p = 0.25 and
q = 0.75. The number of allergic people is denoted by X.
(i)
(ii)
(iii)
Expectation = np = 20 × 0.25 = 5 people.
X B(20, 0.25)
(a) P(X = 2) = 20C2(0.75)18(0.25)2 = 0.067
(b) P(X 2) = P(X = 0) + P(X = 1) + P(X = 2)
= (0.75)20 + 20C1(0.75)19(0.25) + 20C2(0.75)18(0.25)2
= 0.003 + 0.021 + 0.067
= 0.091
Let the sample size be n (people), so that X B(n, 0.25).
The probability that none of them is allergic is
P(X = 0) = (0.75)n
and so the probability that at least one is allergic is
P(X 1) = 1 − P(X = 0)
= 1 − (0.75)n
148
So we need
1 − (0.75)n 0.999
(0.75)n 0.001
(0.75)n = 0.001
Solving
n = log 0.001 ÷ log 0.75
You meet
logarithms in Pure
Mathematics 2.
= 24.01
So 25 people are required.
Notes
1 Although 24.01 is very close to 24 it would be incorrect to round down.
1 – (0.75)24 = 0.998 996 6 which is just less than 99.9%.
Using the binomial distribution
n log 0.75 = log 0.001
gives
S1
6
2 You can also use trial and improvement on a calculator to solve for n.
(iv)
The assumptions made are:
(a) That the sample is random. This is almost certainly untrue. University
students are nearly all in the 18–25 age range and so a sample of them
cannot be a random sample of the whole population. They may well also
be unrepresentative of the whole population in other ways. Volunteers
are seldom truly random.
(b) That the outcome for one person is independent of that for another. This
is probably true unless they are a group of friends from, say, an athletics
team, where those with allergies are less likely to be members.
EXPERIMENT
Does the binomial distribution really work?
In the first case in Example 6.3, you threw a die six times (or six dice once each,
which amounts to the same thing).
1
X B(6, 6) and this gives the probabilities in the following table.
Number of 6s
Probability
0
0.335
1
0.402
2
0.201
3
0.054
4
0.008
5
0.001
6
0.000
149
So if you carry out the experiment of throwing six dice 1000 times and record
the number of 6s each time, you should get no 6s about 335 times, one 6 about
402 times and so on. What does ‘about’ mean? How close an agreement can you
expect between experimental and theoretical results?
The binomial distribution
S1
6
You could carry out the experiment with dice, but it would be very tedious
even if several people shared the work. Alternatively you could simulate the
experiment on a spreadsheet using a random number generator.
EXERCISE 6B
1
In a game five dice are rolled together.
(i)
(ii)
2
A certain type of sweet comes in eight colours: red, orange, yellow, green,
blue, purple, pink and brown and these normally occur in equal proportions.
Veronica’s mother gives each of her children 16 of the sweets. Veronica says
that the blue ones are much nicer than the rest and is very upset when she
receives less than her fair share of them.
(i)
(ii)
(iii)
3
150
What is the probability that
(a) all five show 1
(b) exactly three show 1
(c) none of them shows 1?
What is the most likely number of times for 6 to show?
How many blue sweets did Veronica expect to get?
What was the probability that she would receive fewer blue ones than she
expected?
What was the probability that she would receive more blue ones than she
expected?
In a particular area 30% of men and 20% of women are overweight and there
are four men and three women working in an office there. Find the probability
that there are
(i) 0
(ii) 1
(iii) 2 overweight men;
(iv) 0
(v) 1
(vi) 2 overweight women;
(vii) 2 overweight people in the office.
What assumption have you made in answering this question?
4
On her drive to work Stella has to go through four sets of traffic lights.
2
She estimates that for each set the probability of her finding them red is 3 and
1
green 3. (She ignores the possibility of them being amber.) Stella also estimates
that when a set of lights is red she is delayed by one minute.
(ii)
5
Exercise 6B
(i)
S1
6
Find the probability of
(a) 0
(b) 1
(c) 2
(d) 3 sets of lights being against her.
Find the expected extra journey time due to waiting at lights.
Pepper moths are found in two varieties, light and dark. The proportion of
dark moths increases with certain types of atmospheric pollution. At the time
of the question 30% of the moths in a particular town are dark. A research
student sets a moth trap and catches nine moths, four light and five dark.
(i)
(ii)
What is the probability of that result for a sample of nine moths?
What is the expected number of dark moths in a sample of nine?
The next night the student’s trap catches ten pepper moths.
(iii)
6
What is the probability that the number of dark moths in this sample is
the same as the expected number?
Bella Cicciona, a fortune teller, claims to be able to predict the sex of unborn
children. In fact, on each occasion she is consulted, the probability that she
makes a correct prediction is 0.6, independent of any other prediction.
One afternoon, Bella is consulted by ten expectant mothers. Find, correct to 2
significant figures, the probabilities that
(i)
(ii)
(iii)
(iv)
her first eight predictions are correct and her last two are wrong
she makes exactly eight correct predictions
she makes at least eight correct predictions
she makes exactly eight correct predictions given that she makes at least
eight.
[MEI]
7
A general knowledge quiz has ten questions. Each question has three possible
‘answers’ of which one only is correct. A woman attempts the quiz by pure
guesswork.
(i)
(ii)
Find the probabilities that she obtains
(a) exactly two correct answers
(b) not more than two correct answers.
What is the most likely number of correct answers and the probability that
she just achieves this number?
[MEI]
151
The binomial distribution
S1
6
8
Five unbiased dice are thrown. Calculate the probabilities that
(i)
(ii)
(iii)
there will be exactly four 6s
there will be some one number appearing exactly four times
there will be some one number appearing exactly three times and a
second number appearing twice.
[MEI]
9
Six fair coins are tossed and those landing heads uppermost are eliminated.
The remainder are tossed again and the process of elimination is repeated.
Tossing and elimination continue in this way until no coins are left.
Find the probabilities of the following events.
(i)
(ii)
(iii)
(iv)
(v)
All six coins are eliminated in the first round.
Exactly two coins are eliminated in the first round.
Exactly two coins are eliminated in the first round and exactly two coins
are eliminated in the second round.
Exactly two coins are eliminated in each of the first three rounds.
Exactly two coins are eliminated in the first round and exactly two coins
are eliminated in the third round.
[MEI]
10
A supermarket gets eggs from a supplier in boxes of 12. The supermarket
manager is concerned at the number of eggs which are broken on arrival.
She has a random sample of 100 boxes checked and the numbers of broken
eggs per box are as follows.
Number of eggs broken
0
1
2
3
4
5+
Number of boxes
35
39
19
6
1
0
(i)
(ii)
(iii)
Calculate the mean and standard deviation of the number of broken eggs
in a box.
Show that a reasonable estimate for p, the probability of an egg being
broken on arrival, is 0.0825. Use this figure to calculate the probability
that a randomly chosen box will contain no broken eggs. How does this
probability relate to the observed number of boxes which contain no
broken eggs?
The manager tells the suppliers that they must reduce p to the value
which will ensure that, in the long run, 75% of boxes have no broken
eggs. To what value must the suppliers reduce p?
[MEI]
152
11
A box contains 300 discs of different colours. There are 100 pink discs, 100
blue discs and 100 orange discs. The discs of each colour are numbered from
0 to 99. Five discs are selected at random, one at a time, with replacement.
Find
(ii)
(iii)
(iv)
the probability that no orange discs are selected,
the probability that exactly 2 discs with numbers ending in a 6 are selected,
the probability that exactly 2 orange discs with numbers ending in a 6 are
selected,
the mean and variance of the number of pink discs selected.
Exercise 6B
(i)
S1
6
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q5 November 2005]
12
The mean number of defective batteries in packs of 20 is 1.6. Use a binomial
distribution to calculate the probability that a randomly chosen pack of 20
will have more than 2 defective batteries.
[Cambridge International AS and A Level Mathematics 9709, Paper 61 Q1 November 2009]
KEY POINTS
1
The binomial distribution may be used to model situations in which:
●
you are conducting trials on random samples of a certain size, n
●
in each trial there are two possible outcomes, often referred to as success
and failure
●
●
2
3
both outcomes have fixed probabilities, p and q, and p + q = 1
the probability of success in any trial is independent of the outcomes of
previous trials.
The probability that the number of successes, X, has the value r, is given by
n –r
n
n
P(X = r) = r prqn –r = r pr (1 – p )
n
An alternative notation for is nCr.
r
For B(n, p)
●
the expectation or mean of the number of successes, E(X) = µ = np.
●
the variance, Var(X) = σ 2 = npq = np(1 − p).
●
the standard deviation, σ =
npq = np (1 – p).
To be and not be to, that is the answer.
Piet Hein
153
The normal distribution
S1
7
7
The normal distribution
The
normal
law of error
stands out in the
experience of mankind
as one of the broadest
generalisations of natural
philosophy. It serves as the
guiding instrument in researches
in the physical and social sciences
and in medicine, agriculture and engineering.
It is an indispensable tool for the analysis and the
interpretation of the basic data obtained by observation and experiment.
W. J. Youden
UK Beanpole
Just had my height measured at the doctor’s − I’m 194.3 cm.
Can’t be many around as tall as me!
UK Beanpole is clearly exceptionally tall, but how much so? Is he one in a
hundred, or a thousand or even a million? To answer that question you need to
know the distribution of heights of adult British men.
The first point that needs to be made is that height is a continuous variable and
not a discrete one. If you measure accurately enough it can take any value.
This means that it does not really make sense to ask ‘What is the probability that
somebody chosen at random has height exactly 194.3 cm?’. The answer is zero.
However, you can ask questions like ‘What is the probability that somebody
chosen at random has height between 194.25 cm and 194.35 cm?’ and ‘What is
the probability that somebody chosen at random has height at least 194.3 cm?’.
When the variable is continuous, you are concerned with a range of values
rather than a single value.
154
Like many other naturally occurring variables, the heights of adult men may be
modelled by a normal distribution, shown in figure 7.1. You will see that this has
a distinctive bell-shaped curve and is symmetrical about its middle. The curve is
continuous as height is a continuous variable.
174
height (cm)
The normal distribution
On figure 7.1, area represents probability so the shaded area to the right of
194.3 cm represents the probability that a randomly selected adult male is over
194.3 cm tall.
S1
7
194.3
Figure 7.1
Before you can start to find this area, you must know the mean and standard
deviation of the distribution, in this case about 174 cm and 7 cm respectively.
So UK Beanpole’s height is 194.3 cm − 174 cm = 20.3 cm above the mean, and
that is
20.3
7
= 2.9 standard deviations.
The number of standard deviations beyond the mean, in this case 2.9, is denoted by
the letter z. Thus the shaded area gives the probability of obtaining a value of z 2.9.
You find this area by looking up the value of Φ(z) when z = 2.9 in a normal
distribution table of Φ(z) as shown in figure 7.2, and then calculating 1 − Φ(z).
(Φ is the Greek letter phi.)
z
0
1
2
3
4
5
6
7
8
9
1
2
3
4
5 6
ADD
7
8
9
0.0
0.5000
0.5040
0.5080
0.5120
0.5160
0.5199
0.5239
0.5279
0.5319
0.5359
4
8
12 16 20 24 28 32 36
0.1
0.5398
0.5438
0.5478
0.5517
0.5557
0.5596
0.5636
0.5675
0.5714
0.5753
4
8
12 16 20 24 28 32 36
0.2
0.5793
0.5832
0.5871
0.5910
0.5948
0.5987
0.6026
0.6064
0.6103
0.6141
4
8
12 15 19 23 27 31 35
0.3
0.6179
0.6217
0.6255
0.6293
0.6331
0.6368
0.6406
0.6443
06.480
0.6517
4
7
11 15 19 22 26 30 34
0.4
0.6554
0.6591
0.6628
0.6664
0.6700
0.6736
0.6772
0.6808
0.6844
0.6879
4
7
11 14 18 22 25 29 32
0.5
0.6915
0.6950
0.6985
0.7019
0.7054
0.7088
0.7123
0.7157
0.7190
0.7224
3
7
10 14 17 20 24 27 31
10 13 16 19 23 26 29
0.6
0.7257
0.7291
0.7324
0.7357
0.7389
0.7422
0.7454
0.7486
0.7517
0.7549
3
7
0.7
0.7580
0.7611
0.7642
0.7673
0.7704
0.7734
0.7764
0.7794
0.7823
0.7852
3
6
9
12 15 18 21 24 27
0.8
0.7881
0.7910
0.7939
0.7967
0.7995
0.8023
0.8051
0.8078
0.8106
0.8133
3
5
8
11 14 16 19 22 25
2.5
0.9938
0.9940
0.9941
0.9943
0.9945
0.9946
0.9948
0.9949
0.9951
0.9952
0
0
0
1
1
1
1
1
1
2.6
0.9953
0.9955
0.9956
0.9957
0.9959
0.9960
0.9961
0.9962
0.9963
0.9964
0
0
0
0
1
1
1
1
1
2.7
0.9965
0.9966
0.9967
0.9968
0.9969
0.9970
0.9971
0.9972
0.9973
0.9974
0
0
0
0
0
1
1
1
1
2.8
0.9974
0.9975
0.9976
0.9977
0.9977
0.9978
0.9979
0.9979
0.9980
0.9981
0
0
0
0
0
0
0
1
1
2.9
0.9981
0.9982
0.9982
0.9983
0.9984
0.9984
0.9985
0.9985
0.9986
0.9986
0
0
0
0
0
0
0
0
0
Φ(2.9) = 0.9981
Figure 7.2 Extract from tables of Φ(z)
155
The normal distribution
S1
7
This gives Φ(2.9) = 0.9981, and so 1 − Φ(2.9) = 0.0019.
The probability of a randomly selected adult male being 194.3 cm or over is
0.0019. One man in slightly more than 500 is at least as tall as UK Beanpole.
Using normal distribution tables
The function Φ(z) gives the area under the normal distribution curve to the
left of the value z, that is the shaded area in figure 7.3 (it is the cumulative
distribution function).The total area under the curve is 1, and the area given by
Φ(z) represents the probability of a value smaller than z.
Notice the scale for the
z values; it is in standard
deviations from the mean.
µ � 3σ
�3
µ � 2σ
�2
µ�σ
µ
µ�σ
�1
0
1
The shaded area is �(1)
µ � 2σ
2
µ � 3σ
3
x
z
Figure 7.3
If the variable X has mean µ and standard deviation σ then x, a particular value of
X, is transformed into z by the equation
z =
x − µ.
σ
z is a particular value of the variable Z which has mean 0 and standard deviation
1 and is the standardised form of the normal distribution.
Actual distribution, X
Standardised
distribution, Z
Mean
µ
0
Standard deviation
σ
1
Particular value
x
z =
x−µ
σ
Notice how lower case letters, x and z, are used to indicate particular values of
the random variables, whereas upper case letters, X and Z, are used to describe or
name those variables.
156
Normal distribution tables are easy to use but you should always make a point of
drawing a diagram and shading the region you are interested in.
It is often helpful to know that in a normal distribution, roughly
68% of the values lie within ±1 standard deviation of the mean
●
95% of the values lie within ±2 standard deviations of the mean
●
99.75% of the values lie within ±3 standard deviations of the mean.
Assuming the distribution of the heights of adult men is normal, with mean
174 cm and standard deviation 7 cm, find the probability that a randomly
selected adult man is
(i)
(ii)
(iii)
(iv)
(v)
under 185 cm
over 185 cm
over 180 cm
between 180 cm and 185 cm
under 170 cm
S1
7
Using normal distribution tables
EXAMPLE 7.1
●
giving your answers to 2 significant figures.
SOLUTION
The mean height, µ = 174 cm.
The standard deviation, σ = 7 cm.
(i)
The probability that an adult man selected at random is under 185 cm.
The area required is that shaded in figure 7.5.
µ � 174
z �0
x � 185
z � 1.571
Figure 7.5
x = 185 cm
and so
z = 185 − 174 = 1.571
7
157
The normal distribution
S1
7
Look up the value of Φ(z) in a normal distribution table.
z
0
1
2
3
4
5
6
7
8
9
1
2
3
4
5 6
ADD
7
8
9
0.0
0.5000
0.5040
0.5080
0.5120
0.5160
0.5199
0.5239
0.5279
0.5319
0.5359
4
8
12 16 20 24 28 32 36
0.1
0.5398
0.5438
0.5478
0.5517
0.5557
0.5596
0.5636
0.5675
0.5714
0.5753
4
8
12 16 20 24 28 32 36
1.4
0.9192
0.9207
0.9222
0.9236
0.9251
0.9265
0.9279
0.9292
0.9306
0.9319
1
3
4
6
7
8
1.5
0.9332
0.9345
0.9357
0.9370
0.9382
0.9394
0.9406
0.9418
0.9429
0.9441
1
2
4
5
6
7
10 11 13
8
1.6
0.9452
0.9463
0.9474
0.9484
0.9495
0.9505
0.9515
0.9525
0.9535
0.9545
1
2
3
4
5
6
7
8
9
1.7
0.9554
0.9564
0.9573
0.9582
0.9591
0.9599
0.9608
0.9616
0.9625
0.9633
1
2
3
4
4
5
6
7
8
1.8
0.9641
0.9649
0.9656
0.9664
0.9671
0.9678
0.9686
0.9693
0.9699
0.9706
1
1
2
3
4
4
5
6
6
10 11
Figure 7.4 Extract from tables of Φ(z)
Φ(1.571) = 0.9418 + 0.0001
= 0.9419
= 0.94
(2 s.f.)
Answer: The probability that an adult man selected at random is under
185 cm is 0.94.
(ii)
The probability that an adult man selected at random is over 185 cm.
The area required is the complement of that for part (i) (see figure 7.6).
Probability = 1 − Φ(1.571)
= 1 − 0.9419
= 0.0581
= 0.058
(2 s.f.)
µ � 174
z �0
x � 185
z � 1.571
Figure 7.6
Answer: The probability that an adult man selected at random is over 185 cm
is 0.058.
158
(iii)
The probability that an adult man selected at random is over 180 cm.
x = 180 cm
and so
z = 180 − 174 = 0.857
7
µ � 174
z �0
Using normal distribution tables
The area required = 1 − Φ(0.857)
= 1 − 0.8042
= 0.1958
= 0.20
(2 s.f.)
S1
7
x � 180
z � 0.857
Figure 7.7
Answer: The probability that an adult man selected at random is over 180 cm
is 0.20.
(iv)
The probability that an adult man selected at random is between 180 cm
and 185 cm.
The required area is shown in figure 7.8. It is
Φ(1.571) − Φ(0.857) = 0.9419 − 0.8042
= 0.1377
= 0.14
(2 s.f.)
µ � 174 x � 180, 185
z �0
z � 0.857, 1.571
Figure 7.8
Answer: The probability that an adult man selected at random is over 180 cm
but under 185 cm is 0.14.
159
(v)
The probability that an adult man selected at random is under 170 cm.
In this case
x = 170
and so
z = 170 − 174 = −0.571
7
The normal distribution
S1
7
x � 170
z � �0.571
µ � 174
z �0
Figure 7.9
However, when you come to look up Φ(−0.571), you will find that only
positive values of z are given in your tables. You overcome this problem by
using the symmetry of the normal curve. The area you want in this case is
that to the left of −0.571 and this is clearly just the same as that to the right of
+0.571 (see figure 7.10).
So
Φ(−0.571) = 1 − Φ(0.571)
= 1 − 0.716
= 0.284
= 0.28 (2 s.f.)
These graphs
illustrate that
Φ(–z) = 1 – Φ (z)
Φ(z)
Φ(�z)
�z
0
0
1 � Φ(z)
z
Figure 7.10
Answer: The probability that an adult man selected at random is under
170 cm is 0.28.
160
The normal curve
All normal curves have the same basic shape, so that by scaling the two axes
suitably you can always fit one normal curve exactly on top of another one.
−
φ(x) = 1 e 2
σ 2π
1
( x σ− µ )
2
The normal curve
The curve for the normal distribution with mean µ and standard deviation σ
(i.e. variance σ2) is given by the function φ(x) in
S1
7
The notation N(µ, σ2) is used to describe this distribution. The mean, µ, and
standard deviation, σ (or variance, σ2), are the two parameters used to define the
distribution. Once you know their values, you know everything there is to know
about the distribution. The standardised variable Z has mean 0 and variance 1,
so its distribution is N(0, 1).
x−µ
After the variable X has been transformed to Z using z =
the form of the
σ
curve (now standardised) becomes
− 1z 2
φ(z) = 1 e 2
2π
However, the exact shape of the normal curve is often less useful than the area
underneath it, which represents a probability. For example, the probability that
Z 2 is given by the shaded area in figure 7.11.
Easy though it looks, the function φ(z) cannot be integrated algebraically to
find the area under the curve; this can only be found by using a numerical
method. The values found by doing so are given as a table, and this area
function is called Φ(z).
φ(z)
z�0
2
z
Figure 7.11
161
The normal distribution
S1
7
EXAMPLE 7.2
Skilled operators make a particular component for an engine. The company
believes that the time taken to make this component may be modelled by the
normal distribution with mean 95 minutes and standard deviation 4 minutes.
Assuming the company’s belief to be true, find the probability that the time taken
to make one of these components, selected at random, was
(i)
(ii)
(iii)
over 97 minutes
under 90 minutes
between 90 and 97 minutes.
Sheila believes that the company is allowing too long for the job and invites them
to time her. They find that only 10% of the components take her over 90 minutes
to make, and that 20% take her less than 70 minutes.
(iv)
Estimate the mean and standard deviation of the time Sheila takes.
SOLUTION
According to the company µ = 95 and σ = 4 so the distribution is N(95, 42).
(i)
The probability that a component required over 97 minutes.
z = 97 − 95 = 0.5
4
The probability is represented by the shaded area in figure 7.12 and is given by
1 − Φ(0.5) = 1 − 0.6915
= 0.3085
= 0.309 (3 s.f.)
95
97
Figure 7.12
Answer: The probability it took the operator over 97 minutes to manufacture
a randomly selected component is 0.309.
162
(ii)
The probability that a component required under 90 minutes.
z = 90 − 95 = − 1.25
4
The probability is represented by the shaded area in figure 7.13 and given by
The normal curve
1 − Φ(1.25) = 1 − 0.8944
= 0.1056
= 0.106 (3 s.f.)
90
S1
7
95
Figure 7.13
Answer: The probability it took the operator under 90 minutes to
manufacture a randomly selected component is 0.106.
(iii)
The probability that a component required between 90 and 97 minutes.
The probability is represented by the shaded area in figure 7.14 and given by
1 − 0.1056 − 0.3085 = 0.5859
= 0.586 (3 s.f.)
0.1056
0.3085
90
95
97
Figure 7.14
Answer: The probability it took the operator between 90 and 97 minutes to
manufacture a randomly selected component is 0.586.
163
(iv)
S1
7
Estimate the mean and standard deviation of the time Sheila takes.
The normal distribution
The question has now been put the other way round. You have to infer the
mean, µ, and standard deviation, σ, from the areas under different parts of
the graph.
10% take her 90 minutes or more. This means that the shaded area in
figure 7.15 is 0.1.
90 − µ
σ
Φ(z) = 1 − 0.1 = 0.9
z=
0.1
µ
90
Figure 7.15
You now use the table of Φ(z) = p in reverse. z = 1.28 has a probability of
0.8997 which is as close to 0.9 as you can get using this middle part of the
table. However, you can achieve greater accuracy by looking at the righthand columns as well: z = 1.281 has a probability of 0.8999 and z = 1.282 has
a probability of 0.9001. So the best value for z is 1.2815.
z
0
1
2
3
4
5
6
7
8
9
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
0.5000
0.5398
0.5793
0.6179
0.6554
0.6915
0.7257
0.7580
0.7881
0.8159
0.8413
0.8643
0.8849
0.9032
0.9192
0.5040
0.5438
0.5832
0.6217
0.6591
0.6950
0.7291
0.7611
0.7910
0.8186
0.8438
0.8665
0.8869
0.9049
0.9207
0.5080
0.5478
0.5871
0.6255
0.6628
0.6985
0.7324
0.7642
0.7939
0.8212
0.8461
0.8686
0.8888
0.9066
0.9222
0.5120
0.5517
0.5910
0.6293
0.6664
0.7019
0.7357
0.7673
0.7967
0.8238
0.8485
0.8708
0.8907
0.9082
0.9236
0.5160
0.5557
0.5948
0.6331
0.6700
0.7054
0.7389
0.7704
0.7995
0.8264
0.8508
0.8729
0.8925
0.9099
0.9251
0.5199
0.5596
0.5987
0.6368
0.6736
0.7088
0.7422
0.7734
0.8023
0.8289
0.8531
0.8749
0.8944
0.9115
0.9265
0.5239
0.5636
0.6026
0.6406
0.6772
0.7123
0.7454
0.7764
0.8051
0.8315
0.8554
0.8770
0.8962
0.9131
0.9279
0.5279
0.5675
0.6064
0.6443
0.6808
0.7157
0.7486
0.7794
0.8078
0.8340
0.8577
0.8790
0.8980
0.9147
0.9292
0.5319
0.5714
0.6103
0.6480
0.6844
0.7190
0.7517
0.7823
0.8106
0.8365
0.8599
0.8810
0.8997
0.9162
0.9306
0.5359
0.5753
0.6141
0.6517
0.6879
0.7224
0.7549
0.7852
0.8133
0.8389
0.8621
0.8830
0.9015
0.9177
0.9319
Φ−1(0.9) = 1.2815
Figure 7.16 Extract from tables of Φ(z)
164
1
2
3
4
4
4
4
4
4
3
3
3
3
3
2
2
2
2
1
8
8
8
7
7
7
7
6
5
5
5
4
4
3
3
12
12
12
11
11
10
10
9
8
8
7
6
6
5
4
16
16
15
15
14
14
13
12
11
10
9
8
7
6
6
5 6
ADD
20
20
19
19
18
17
16
15
14
13
12
10
9
8
7
24
24
23
22
22
20
19
18
16
15
14
12
11
10
8
7
8
9
28
28
27
26
25
24
23
21
19
18
16
14
13
11
10
32
32
31
30
29
27
26
24
22
20
19
16
15
13
11
36
36
35
34
32
31
29
27
25
23
21
18
17
14
13
Returning to the problem, you now know that
90 − µ
= 1.2815
σ
⇒
90 − µ = 1.2815σ.
S1
7
The normal curve
0.2
70
µ
Figure 7.17
The second piece of information, that 20% of components took Sheila under
70 minutes, is illustrated in figure 7.17.
z=
70 − µ
σ
(z has a negative value in this case, the point being to the left of the mean.)
Φ(z) = 0.2
and so, by symmetry,
Φ(−z) = 1 − 0.2 = 0.8.
Using the table of the normal function gives
−z = 0.842
or
z = −0.842
This gives a second equation for µ and σ.
70 − µ = −0.842
⇒
70 − µ = −0.842σ.
σ
You now solve equations and simultaneously.
Subtract
and
90 − µ = 1.2815σ
70 − µ = −0.842σ
−−−−−−−−−−−−−−−−−
20
= 2.1235σ
σ = 9.418 = 9.42 (3 s.f.)
µ = 77.930 = 77.9 (3 s.f.)
Answer: Sheila’s mean time is 77.9 minutes with standard deviation
9.42 minutes.
165
S1
7
EXERCISE 7A
1
The distribution of the heights of some plants is normal and has a mean of
40 cm and a standard deviation of 2 cm. Find the probability that a randomly
selected plant is
The normal distribution
(i)
(ii)
(iii)
(iv)
2
The distribution of the masses of some baby parrots is normal and has a
mean of 60 g and a standard deviation of 5 g. Find the probability that a
randomly selected bird is
(i)
(ii)
(iii)
(iv)
3
(ii)
(iii)
(iv)
(ii)
(iii)
under 168.5 cm
over 174.5 cm
between 168.5 and 174.5 cm.
A pet shop has a tank of goldfish for sale. All the fish in the tank were hatched
at the same time and their weights may be taken to be normally distributed
with mean 100 g and standard deviation 10 g. Melanie is buying a goldfish and
is invited to catch the one she wants in a small net. In fact the fish are much
too quick for her to be able to catch any particular fish, and the one which she
eventually nets is selected at random. Find the probability that its weight is
(i)
(ii)
(iii)
166
under 98 g
over 98 g
under 102 g
between 98 and 102 g.
The distribution of the heights of 18-year-old girls may be modelled by the
normal distribution with mean 162.5 cm and standard deviation 6 cm. Find
the probability that the height of a randomly selected 18-year-old girl is
(i)
5
under 63 g
over 63 g
over 68 g
between 63 and 68 g.
The distribution of the mass of sweets in a bag is normal and has a mean
of 100 g and a standard deviation 2 g. Find the probability that a randomly
selected bag is
(i)
4
under 42 cm
over 42 cm
over 40 cm
between 40 and 42 cm.
over 115 g
under 105 g
between 105 and 115 g.
6
(i)
(ii)
(iii)
7
too strong
too weak
all right.
A biologist finds a nesting colony of a previously unknown sea bird on a remote
island. She is able to take measurements on 100 of the eggs before replacing
them in their nests. She records their weights, w g, in this frequency table.
Weight, w
Frequency
(i)
(ii)
(iii)
8
S1
7
Exercise 7A
When he makes instant coffee, Tony puts a spoonful of powder into a mug.
The weight of coffee in grams on the spoon may be modelled by the normal
distribution with mean 5 g and standard deviation 1 g. If he uses more than
6.5 g Julia complains that it is too strong and if he uses less than 4 g she tells
him it is too weak. Find the probability that he makes the coffee
25 w 27 27 w 29 29 w 31 31 w 33 33 w 35 35 w 37
2
13
35
33
17
0
Find the mean and standard deviation of these data.
Assuming the weights of the eggs for this type of bird are normally
distributed and that their mean and standard deviation are the same as
those of this sample, find how many eggs you would expect to be in each
of these categories.
Do you think the assumption that the weights of the eggs are normally
distributed is reasonable?
The length of life of a certain make of tyre is normally distributed about a
mean of 24 000 km with a standard deviation of 2500 km.
(i)
(ii)
What percentage of such tyres will need replacing before they have
travelled 20 000 km?
As a result of improvements in manufacture, the length of life is still
normally distributed, but the proportion of tyres failing before 20 000 km
is reduced to 1.5%.
(a) If the standard deviation has remained unchanged, calculate the new
mean length of life.
(b) If, instead, the mean length of life has remained unchanged, calculate
the new standard deviation.
[MEI]
9
A machine is set to produce nails of length 10 cm, with standard deviation
0.05 cm. The lengths of the nails are normally distributed.
Find the percentage of nails produced between 9.95 cm and 10.08 cm in
length.
The machine’s setting is moved by a careless apprentice with the consequence
that 16% of the nails are under 5.2 cm in length and 20% are over 5.3 cm.
(i)
(ii)
Find the new mean and standard deviation.
167
10
The normal distribution
S1
7
The concentration by volume of methane at a point on the centre line of
a jet of natural gas mixing with air is distributed approximately normally
with mean 20% and standard deviation 7%. Find the probabilities that the
concentration
(i)
(ii)
(iii)
exceeds 30%
is between 5% and 15%.
In another similar jet, the mean concentration is 18% and the standard
deviation is 5%. Find the probability that in at least one of the jets the
concentration is between 5% and 15%.
[MEI]
11
In a particular experiment, the length of a metal bar is measured many times.
The measured values are distributed approximately normally with mean
1.340 m and standard deviation 0.021 m. Find the probabilities that any one
measured value
(i)
(ii)
(iii)
(iv)
exceeds 1.370 m
lies between 1.310 m and 1.370 m
lies between 1.330 m and 1.390 m.
Find the length l for which the probability that any one measured value is
less than l is 0.1.
[MEI]
12
A factory produces a very large number of steel bars. The lengths of these
bars are normally distributed with 33% of them measuring 20.06 cm or more
and 12% of them measuring 20.02 cm or less.
Write down two simultaneous equations for the mean and standard deviation
of the distribution and solve to find values to 4 significant figures. Hence
estimate the proportion of steel bars which measure 20.03 cm or more.
The bars are acceptable if they measure between 20.02 cm and 20.08 cm.
What percentage are rejected as being outside the acceptable range?
13
[MEI]
The diameters D of screws made in a factory are normally distributed with
mean 1 mm. Given that 10% of the screws have diameters greater than
1.04 mm, find the standard deviation correct to 3 significant figures, and hence
show that about 2.7% of the screws have diameters greater than 1.06 mm.
Find, correct to 2 significant figures,
(i)
(ii)
the number d for which 99% of the screws have diameters that exceed d mm
the number e for which 99% of the screws have diameters that do not
differ from the mean by more than e mm.
[MEI]
168
14
A machine produces crankshafts whose diameters are normally distributed
with mean 5 cm and standard deviation 0.03 cm. Find the percentage of
crankshafts it will produce whose diameters lie between 4.95 cm and 4.97 cm.
Crankshafts with diameters outside the interval 5 ± 0.05 cm are rejected. If
the mean diameter of the machine’s production remains unchanged, to what
must the standard deviation be reduced if only 4% of the production is to
be rejected?
Exercise 7A
What is the probability that two successive crankshafts will both have a
diameter in this interval?
S1
7
[MEI]
15
In a reading test for eight-year-old children, it is found that a reading score X
is normally distributed with mean 5.0 and standard deviation 2.0.
(i)
(ii)
(iii)
What proportion of children would you expect to score between 4.5
and 6.0?
There are about 700 000 eight-year-olds in the country. How many
would you expect to have a reading score of more than twice the mean?
Why might educationalists refer to the reading score X as a ‘score out of 10’?
The reading score is often reported, after scaling, as a value Y which is
normally distributed, with mean 100 and standard deviation 15. Values of Y
are usually given to the nearest integer.
(iv)
(v)
Find the probability that a randomly chosen eight-year-old gets a score,
after scaling, of 103.
What range of Y scores would you expect to be attained by the best 20%
of readers?
[MEI]
16
Extralite are testing a new long-life bulb. The lifetimes, in hours, are
assumed to be normally distributed with mean µ and standard deviation σ.
After extensive tests, they find that 19% of bulbs have a lifetime exceeding
5000 hours, while 5% have a lifetime under 4000 hours.
(i) Illustrate this information on a sketch.
(ii) Show that σ = 396 and find the value of µ.
In the remainder of this question take µ to be 4650 and σ to be 400.
(iii)
(iv)
Find the probability that a bulb chosen at random has a lifetime between
4250 and 4750 hours.
Extralite wish to quote a lifetime which will be exceeded by 99% of bulbs.
What time, correct to the nearest 100 hours, should they quote?
A new school classroom has six light-fittings, each fitted with an Extralite
long-life bulb.
(v)
Find the probability that no more than one bulb needs to be replaced
within the first 4250 hours of use.
[MEI]
169
The normal distribution
S1
7
17
Tyre pressures on a certain type of car independently follow a normal
distribution with mean 1.9 bars and standard deviation 0.15 bars.
(i)
(ii)
Find the probability that all four tyres on a car of this type have pressures
between 1.82 bars and 1.92 bars.
Safety regulations state that the pressures must be between 1.9 − b bars
and 1.9 + b bars. It is known that 80% of tyres are within these safety
limits. Find the safety limits.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q6 June 2005]
18
The lengths of fish of a certain type have a normal distribution with mean
38 cm. It is found that 5% of the fish are longer than 50 cm.
(i)
Find the standard deviation.
(ii)
When fish are chosen for sale, those shorter than 30 cm are rejected. Find
the proportion of fish rejected.
(iii)
9 fish are chosen at random. Find the probability that at least one of
them is longer than 50 cm.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q3 June 2006]
19 (i)
(ii)
The random variable X is normally distributed. The mean is twice the
standard deviation. It is given that P(X 5.2) = 0.9. Find the standard
deviation.
A normal distribution has mean µ and standard deviation σ. If 800
observations are taken from this distribution, how many would you
expect to be between µ − σ and µ + σ?
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q3 June 2007]
20
In a certain country the time taken for a common infection to clear up is
normally distributed with mean µ days and standard deviation 2.6 days. 25%
of these infections clear up in less than 7 days.
(i)
Find the value of µ.
In another country the standard deviation of the time taken for the infection
to clear up is the same as in part (i) but the mean is 6.5 days. The time taken
is normally distributed.
(ii)
Find the probability that, in a randomly chosen case from this country,
the infection takes longer than 6.2 days to clear up.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q4 June 2008]
170
21
The random variable X has a normal distribution with mean 4.5. It is given
that P(X 5.5) = 0.0465 (see diagram).
Exercise 7A
0.0465
4.5
(i)
(ii)
5.5
S1
7
X
Find the standard deviation of X.
Find the probability that a random observation of X lies between 3.8
and 4.8.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q4 November 2007]
22 (i)
(ii)
The daily minimum temperature in degrees Celsius (°C) in January
in Ottawa is a random variable with distribution N(−15.1, 62.0). Find
the probability that a randomly chosen day in January in Ottawa has a
minimum temperature above 0 °C.
In another city the daily minimum temperature in °C in January is a
random variable with distribution N(µ, 40.0). In this city the probability
that a randomly chosen day in January has a minimum temperature
above 0 °C is 0.8888. Find the value of µ.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q3 November 2008]
23
The times for a certain car journey have a normal distribution with mean
100 minutes and standard deviation 7 minutes. Journey times are classified
as follows:
‘short’
(the shortest 33% of times),
‘long’
(the longest 33% of times),
‘standard’ (the remaining 34% of times).
(i)
(ii)
Find the probability that a randomly chosen car journey takes between
85 and 100 minutes.
Find the least and greatest times for ‘standard’ journeys.
[Cambridge International AS and A Level Mathematics 9709, Paper 61 Q3 November 2009]
171
Modelling discrete situations
Although the normal distribution applies strictly to a continuous variable, it is
also common to use it in situations where the variable is discrete providing that
The normal distribution
S1
7
●
the distribution is approximately normal; this requires that the steps in its
possible values are small compared with its standard deviation
●
continuity corrections are applied where appropriate.
The meaning of the term ‘continuity correction’ is explained in the following
example.
EXAMPLE 7.3
The result of an Intelligence Quotient (IQ) test is an integer score, X. Tests are
designed so that X has a mean value of 100 with standard deviation 15. A large
number of people have their IQs tested. What proportion of them would you
expect to have IQs measuring between 106 and 110 (inclusive)?
SOLUTION
Although the random variable X is an integer and hence discrete, the steps of 1 in
its possible values are small compared with the standard deviation of 15. So it is
reasonable to treat it as if it is continuous.
If you assume that an IQ test is measuring innate, natural intelligence (rather than
the results of learning), then it is reasonable to assume a normal distribution.
If you draw the probability distribution function for the discrete variable
X it looks like figure 7.18. The area you require is the total of the five bars
representing 106, 107, 108, 109 and 110.
section of
normal curve
z1
172
Figure 7.18
110.5
110
110
109.5
109
109
108.5
108
108
107.5
107
107
106.5
106
105.5
106
z2
The equivalent section of the normal curve would run not from 106 to 110
but from 105.5 to 110.5, as you can see in figure 7.18. When you change from
the discrete scale to the continuous scale, the numbers 106, 107, etc. no longer
represent the whole intervals, just their centre points.
where z1 = 105.5 − 100 and z2 = 110.5 − 100 .
15
15
This is
Φ(0.7000) − Φ(0.3667) = 0.7580 − 0.6431 = 0.1149
Answer: The proportion of IQs between 106 and 110 (inclusive) should be
approximately 11%.
In this calculation, both end values needed to be adjusted to allow for the
fact that a continuous distribution was being used to approximate a discrete
one. These adjustments, 106 → 105.5 and 110 → 110.5, are called continuity
corrections. Whenever a discrete distribution is approximated by a continuous
one a continuity correction may need to be used.
You must always think carefully when applying a continuity correction. Should
the corrections be added or subtracted? In this case 106 and 110 are inside the
required area and so any value (like 105.7 or 110.4) which would round to them
must be included. It is often helpful to draw a sketch to illustrate the region you
want, like the one in figure 7.18.
If the region of interest is given in terms of inequalities, you should look
carefully to see whether they are inclusive ( or ) or exclusive ( or ).
For example 20 X 30 becomes 19.5 X 30.5 whereas 20 X 30
becomes 20.5 X 29.5.
Using the normal distribution as an approximation for the binomial distribution
So the area you require under the normal curve is given by Φ(z2) − Φ(z1)
S1
7
Two particularly common situations are when the normal distribution is used to
approximate the binomial and the Poisson distributions. (You will learn about
the Poisson distribution if you study Statistics 2.)
Using the normal distribution as an approximation for the
binomial distribution
You may use the normal distribution as an approximation for the binomial,
B(n, p) (where n is the number of trials each having probability p of success) when
●
n is large
●
p is not too close to 0 or 1.
A rough way of judging whether n is large enough is to require that both np 5
and nq 5, where q = 1 – p.
173
These conditions ensure that the distribution is reasonably symmetrical and not
skewed away from either end, see figure 7.19.
The normal distribution
S1
7
positive skew
negative skew
symmetrical
Figure 7.19
The parameters for the normal distribution are then
Mean:
µ = np
Variance: σ2 = npq = np (1 − p)
so that it can be denoted by N(np, npq).
EXAMPLE 7.4
This is a true story. During voting at an election, an exit poll of 1700 voters
indicated that 50% of people had voted for a particular candidate. When the
votes were counted it was found that he had in fact received 57% support.
850 of the 1700 people interviewed said they had voted for the candidate but 57%
of 1700 is 969, a much higher number. What went wrong? Is it possible to be so
far out just by being unlucky and asking the wrong people?
SOLUTION
The situation of selecting a sample of 1700 people and asking them if they voted
for one candidate or not is one that is modelled by the binomial distribution, in
this case B(1700, 0.57).
In theory you could multiply out (0.43 + 0.57t)1700 and use that to find the
probabilities of getting 0, 1, 2, ..., 850 supporters of this candidate in your sample
of 1700. In practice such a method would be impractical because of the work
involved.
What you can do is to use a normal approximation. The required conditions are
fulfilled: at 1700, n is certainly not small; p = 0.57 is near neither 0 nor 1.
The parameters for the normal approximation are given by
µ = np = 1700 × 0.57 = 969
174
σ = npq = 1700 × 0.57 × 0.43 = 20.4
You will see that the standard deviation, 20.4, is large compared with the steps
of 1 in the number of supporters of this candidate.
The probability of getting no more than 850 supporters of this candidate,
P(X 850), is given by Φ(z), where
Exercise 7B
z =
850.5 − 969
= − 5.8
20.4
850.5
S1
7
969
Figure 7.20
(Notice the continuity correction making 850 into 850.5.)
This is beyond the range of most tables and corresponds to a probability of about
0.000 01. The probability of a result as extreme as this is thus 0.000 02 (allowing
for an equivalent result in the tail above the mean). It is clearly so unlikely that
this was a result of random sampling that another explanation must be found.
?
●
EXERCISE 7B
What do you think went wrong with the exit poll? Remember this really did
happen.
1
A certain examination has a mean mark of 100 and a standard deviation of 15.
The marks can be assumed to be normally distributed.
(i)
(ii)
(iii)
What is the least mark needed to be in the top 35% of students taking this
examination?
Between which two marks will the middle 90% of the students lie?
150 students take this examination. Calculate the number of students
likely to score 110 or over.
[MEI]
2
25% of Flapper Fish have red spots, the rest have blue spots. A fisherman nets
10 Flapper Fish. What are the probabilities that
(i)
(ii)
exactly 8 have blue spots
at least 8 have blue spots?
175
A large number of samples, each of 100 Flapper Fish, are taken.
S1
7
(iii)
The normal distribution
(iv)
3
What is the mean and the standard deviation of the number of red-spotted
fish per sample?
What is the probability of a sample of 100 Flapper Fish containing over
30 with red spots?
A fair coin is tossed 10 times. Evaluate the probability that exactly half of the
tosses result in heads.
The same coin is tossed 100 times. Use the normal approximation to the
binomial to estimate the probability that exactly half of the tosses result in
heads. Also estimate the probability that more than 60 of the tosses result
in heads.
Explain why a continuity correction is made when using the normal
approximation to the binomial and the reason for the adoption of this
correction.
[MEI]
4
During an advertising campaign, the manufacturers of Wolfitt (a dog food)
claimed that 60% of dog owners preferred to buy Wolfitt.
(i)
(ii)
Assuming that the manufacturer’s claim is correct for the population of
dog owners, calculate
(a) using the binomial distribution
(b) using a normal approximation to the binomial
the probability that at least 6 of a random sample of 8 dog owners prefer
to buy Wolfitt. Comment on the agreement, or disagreement, between
your two values. Would the agreement be better or worse if the proportion
had been 80% instead of 60%?
Continuing to assume that the manufacturer’s figure of 60% is correct,
use the normal approximation to the binomial to estimate the probability
that, of a random sample of 100 dog owners, the number preferring to buy
Wolfitt is between 60 and 70 inclusive.
[MEI]
5
176
A multiple-choice examination consists of 20 questions, for each of which the
candidate is required to tick as correct one of three possible answers. Exactly
one answer to each question is correct. A correct answer gets 1 mark and a
wrong answer gets 0 marks. Consider a candidate who has complete ignorance
about every question and therefore ticks at random. What is the probability
that he gets a particular answer correct? Calculate the mean and variance of
the number of questions he answers correctly.
The examiners wish to ensure that no more than 1% of completely ignorant
candidates pass the examination. Use the normal approximation to the
binomial, working throughout to 3 decimal places, to establish the pass mark
that meets this requirement.
[MEI]
6
Use the normal distribution to estimate the minimum number of lines that
would ensure that the probability that a call cannot be made because all the
lines are occupied is less than 0.01.
S1
7
Exercise 7B
A telephone exchange serves 2000 subscribers, and at any moment during
1
the busiest period there is a probability of 30 for each subscriber that he will
require a line. Assuming that the needs of subscribers are independent, write
down an expression for the probability that exactly N lines will be occupied
at any moment during the busiest period.
Investigate whether the total number of lines needed would be reduced if the
subscribers were split into two groups of 1000, each with its own set of lines.
[MEI]
7
It is known that, on average, 2 people in 5 in a certain country are overweight.
A random sample of 400 people is chosen. Using a suitable approximation,
find the probability that fewer than 165 people in the sample are overweight.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q1 June 2005]
8
A survey of adults in a certain large town found that 76% of people wore a
watch on their left wrist, 15% wore a watch on their right wrist and 9% did
not wear a watch.
(i)
(ii)
A random sample of 14 adults was taken. Find the probability that more
than 2 adults did not wear a watch.
A random sample of 200 adults was taken. Using a suitable approximation,
find the probability that more than 155 wore a watch on their left wrist.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q7 June 2006]
9
On a certain road 20% of the vehicles are trucks, 16% are buses and the
remainder are cars.
(i)
(ii)
A random sample of 11 vehicles is taken. Find the probability that fewer
than 3 are buses.
A random sample of 125 vehicles is now taken. Using a suitable
approximation, find the probability that more than 73 are cars.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q3 June 2009]
10
On any occasion when a particular gymnast performs a certain routine, the
probability that she will perform it correctly is 0.65, independently of all
other occasions.
(i)
(ii)
(iii)
Find the probability that she will perform the routine correctly on exactly
5 occasions out of 7.
On one day she performs the routine 50 times. Use a suitable
approximation to estimate the probability that she will perform the
routine correctly on fewer than 29 occasions.
On another day she performs the routine n times. Find the smallest value
of n for which the expected number of correct performances is at least 8.
[Cambridge International AS and A Level Mathematics 9709, Paper 6 Q6 November 2007]
177
11
S1
7
In the holidays Martin spends 25% of the day playing computer games.
Martin’s friend phones him once a day at a randomly chosen time.
The normal distribution
(i)
(ii)
(iii)
Find the probability that, in one holiday period of 8 days, there are
exactly 2 days on which Martin is playing computer games when his
friend phones.
Another holiday period lasts for 12 days. State with a reason whether it is
appropriate to use a normal approximation to find the probability that
there are fewer than 7 days on which Martin is playing computer games
when his friend phones.
Find the probability that there are at least 13 days of a 40-day holiday
period on which Martin is playing computer games when his friend
phones.
[Cambridge International AS and A Level Mathematics 9709, Paper 61 Q5 June 2010]
KEY POINTS
1
The normal distribution with mean µ and standard deviation σ is denoted
by N(µ, σ2).
2
This may be given in standardised form by using the transformation
z=
x−µ
σ
3
In the standardised form, N(0, 1), the mean is 0, and the standard deviation
and variance are both 1.
4
The standard normal curve is given by
− 1z 2
Φ(z) = 1 e 2
2π
5
The area to the left of the value z in the diagram below, representing the
probability of a value less than z, is denoted by Φ(z) and is read from tables.
φ(z)
z�0
178
z
6
The normal distribution may be used to approximate suitable discrete
distributions but continuity corrections are then required.
7
The binomial distribution B(n, p) may be approximated by N(np, npq),
provided n is large and p is not close to 0 or 1, so that np 5 and nq 5.
Statistics 2
S2
Hypothesis testing using the binomial distribution
S2
8
8
Hypothesis testing using
the binomial distribution
You may prove anything by figures.
An anonymous salesman
Machoman Dan
Just became a father again! 8 boys
in a row – how’s that for macho
chromosomes? Even at school I told
people I was a real man!
What do you think?
There are two quite different points here.
Maybe you think that Dan is prejudiced, preferring boys to girls. However,
you should not let your views on that influence your judgement on the second
point, his claim to be biologically different from other people, with special
chromosomes.
There are two ways this claim could be investigated, to look at his chromosomes
under a high magnification microscope or to consider the statistical evidence.
Since you have neither Dan nor a suitable microscope to hand, you must resort
to the latter.
If you have eight children you would expect them to be divided about evenly
between the sexes, 4 − 4, 5 − 3 or perhaps 6 − 2. When you realised that a baby
was on its way you would think it equally likely to be a boy or a girl until it was
born, or a scan was carried out, when you would know for certain one way or
the other.
In other words you would say that the probability of its being a boy was 0.5 and
that of its being a girl was 0.5. So you can model the number of boys among eight
children by the binomial distribution B(8, 0.5).
This gives the probabilities in the table, also shown in figure 8.1.
180
Girls
Probability
0
8
1
256
1
7
8
256
2
6
28
256
3
5
56
256
4
4
70
256
5
3
56
256
6
2
28
256
7
1
8
0
8
256
1
256
S2
8
Hypothesis testing using the binomial distribution
Boys
0.3
probability
0.2
0.1
0
1
2
3
4
5
6
number of boys in a family of 8 children
7
8
Figure 8.1
So you can say that, if a biologically normal man fathers eight children, the
1
probability that they will all be boys is 256 (dark green in figure 8.1).
This is unlikely but by no means impossible.
Note
The probability of a baby being a boy is not in fact 0.5 but about 0.503. Boys are less
tough than girls and so more likely to die in infancy and this seems to be nature’s
way of compensating. In most societies men have a markedly lower life expectancy
as well.
181
Hypothesis testing using the binomial distribution
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8
?
●
In some countries many people value boys more highly than girls. Medical
advances mean that it will soon be possible for parents to decide in advance the
sex of their next baby. What would be the effect of this on a country’s population
if, say, half the parents decided to have only boys and the other half to let nature
take its course?
(This is a real problem. The social consequences could be devastating.)
Defining terms
In the last example we investigated Dan’s claim by comparing it to the usual
situation, the unexceptional. If we use p for the probability that a child is a boy
then the normal state of affairs can be stated as
p = 0.5.
This is called the null hypothesis, denoted by H0.
Dan’s claim (made, he says, before he had any children) was that
p 0.5
and this is called the alternative hypothesis, H1.
The word hypothesis (plural hypotheses) means a theory which is put forward
either for the sake of argument or because it is believed or suspected to be true.
An investigation like this is usually conducted in the form of a test, called a
hypothesis test. There are many different sorts of hypothesis test used in statistics;
in this chapter you meet only one of them.
It is never possible to prove something statistically in the sense that, for example,
you can prove that the angle sum of a triangle is 180°. Even if you tossed a coin
a million times and it came down heads every single time, it is still possible that
the coin is unbiased and just happened to land that way. What you can say is that
it is very unlikely; the probability of it happening that way is (0.5)1 000 000 which is
a decimal that starts with over 300 000 zeros. This is so tiny that you would feel
quite confident in declaring the coin biased.
There comes a point when the probability is so small that you say ‘That’s good
enough for me. I am satisfied that it hasn’t happened that way by chance.’
The probability at which you make that decision is called the significance level of
the test. Significance levels are usually given as percentages; 0.05 is written as 5%,
0.01 as 1% and so on.
182
So in the case of Dan, the question could have been worded:
Test, at the 1% significance level, Dan’s claim that his children are more likely to be
boys than girls.
Null hypothesis, H0:
p = 0.5 (Boys and girls are equally likely)
Alternative hypothesis, H1: p 0.5 (Boys are more likely)
Significance level:
1%
Probability of 8 boys from 8 children =
1
256
= 0.0039 = 0.39%.
Since 0.39% < 1% we reject the null hypothesis and accept the alternative
hypothesis. We accept Dan’s claim.
Hypothesis testing checklist
The answer would then look like this:
S2
8
This example also illustrates some of the problems associated with hypothesis
testing. Here is a list of points you should be considering.
Hypothesis testing checklist
1 Was the test set up before or after the data were known?
The test consists of a null hypothesis, an alternative hypothesis and a significance
level.
In this case, the null hypothesis is the natural state of affairs and so does not
really need to be stated in advance. Dan’s claim ‘Even at school I told people I
was a real man’ could be interpreted as the alternative hypothesis, p 0.5.
The problem is that one suspects that whatever children Dan had he would find
an excuse to boast. If they had all been girls, he might have been talking about
‘my irresistible attraction for the opposite sex’ and if they had been a mixture of
girls and boys he would have been claiming ‘super-virility’ just because he had
eight children.
Any test carried out retrospectively must be treated with suspicion.
2 Was the sample involved chosen at random and are the data
independent?
The sample was not random and that may have been inevitable. If Dan had lots
of children around the country with different mothers, a random sample of eight
could have been selected. However, we have no knowledge that this is the case.
The data are the sexes of Dan’s children. If there are no multiple births (for
example, identical twins), then they are independent.
183
Hypothesis testing using the binomial distribution
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8
3 Is the statistical procedure actually testing the original claim?
Dan claims to have ‘macho chromosomes’ whereas the statistical test is of the
alternative hypothesis that p 0.5. The two are not necessarily the same. Even
if this alternative hypothesis is true, it does not necessarily follow that Dan has
macho chromosomes.
The ideal hypothesis test
In the ideal hypothesis test you take the following steps, in this order:
1
Establish the null and alternative hypotheses.
2
Decide on the significance level.
3
Collect suitable data using a random sampling procedure that ensures the
items are independent.
4
Conduct the test, doing the necessary calculations.
5
Interpret the result in terms of the original claim, theory or problem.
There are times, however, when you need to carry out a test but it is just not
possible to do so as rigorously as this.
If Dan been a laboratory rat you could have organised that he fathered further
babies but this is not possible with a human.
Choosing the significance level
If, instead of 1%, we had set the significance level at 0.1%, then we would have
rejected Dan’s claim, since 0.39% 0.1%. The lower the percentage in the
significance level, the more stringent is the test.
The significance level you choose for a test involves a balanced judgement.
Imagine that you are testing the rivets on an plane’s wing to see if they have lost
their strength. Setting a small significance level, say 0.1%, means that you will
only declare the rivets weak if you are very confident of your finding. The trouble
with requiring such a high level of evidence is that even when they are weak you
may well fail to register the fact, with the possible consequence that the plane
crashes. On the other hand if you set a high significance level, such as 10%, you
run the risk of declaring the rivets faulty when they are all right, involving the
company in expensive and unnecessary maintenance work.
The question of how you choose the best significance level is, however, beyond
the scope of this introductory chapter.
184
EXAMPLE 8.1
Leonora claims that a die is biased with a tendency to show the number 1. The
die was thrown 20 times and the results were as follows.
1
4
6
4
6
4
5
1
5
4
1
1
2
1
3
4
2
1
3
3
SOLUTION
Let p be the probability of getting 1 on any throw of the die.
p = 16
Null hypothesis, H0:
Alternative hypothesis, H1: p
Significance level:
(The die is unbiased)
1
6
(The die is biased towards 1)
5%
Choosing the significance level
Using a 5% significance level, test whether Leonora’s claim is correct.
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8
The results may be summarised as follows.
Score
1
2
3
4
5
6
Frequency
6
2
3
5
2
2
Under the null hypothesis, the number of 1s obtained is modelled by the
binomial distribution, B(20, 16) which gives these probabilities:
Number of 1s
Expression
Probability
0.0261
8
(56 )
19 1
20C 5
1( 6 ) ( 6 )
18 1 2
20C 5
2( 6 ) ( 6 )
17 1 3
20C 5
3( 6 ) ( 6 )
16 1 4
20C 5
4( 6 ) ( 6 )
15 1 5
20C 5
5( 6 ) ( 6 )
14 1 6
20C 5
6( 6 ) ( 6 )
13 1 7
20C 5
7( 6 ) ( 6 )
12 1 8
20C 5
8( 6 ) ( 6 )
20
(16 )20
0.0000
0
1
2
3
4
5
6
7
0.1043
0.1982
0.2379
0.2022
0.1294
0.0647
0.0259
0.0084
������������� �����������
20
The probability of 1 coming
up between 0 and 5 times
is found by adding these
probabilities. You get 0.8981
but working to more decimal
places and then rounding
gives 0.8982 which is correct
to 4 decimal places.
If you worked out all these
and added them you would
get the probability that the
number of 1s is 6 or more (up
to a possible 20). It is much
quicker, however, to find this
as 1 – 0.8982 (the answer
above) = 0.1018.
185
Calling X the number of 1s occurring when a die is rolled 20 times, the
probability of six or more 1s is given by
Hypothesis testing using the binomial distribution
S2
8
P(X 6) = 1 − P(X 5)
= 1 − 0.8982
= 0.1018,
about 10%.
Since 10% 5%, the null hypothesis (the die is unbiased) is accepted. So
Leonora’s claim is rejected at the 5% significance level.
The probability of a result at least as extreme as that observed is greater than
the 5% cut-off that was set in advance, that is, greater than the chosen
significance level.
The alternative hypothesis (the die is biased in favour of the number 1) is rejected,
even though the number 1 did come up more often than the other numbers.
?
●
Does the procedure in Example 8.1 follow the steps of the ideal hypothesis test?
Note
Notice that this is a test not of the particular result (six 1s) but of a result at least
as extreme as this (at least six 1s), the darker area in figure 8.2. A hypothesis test
deals with the probability of an event ‘as unusual as or more unusual than’ what
has occurred.
0.3
probability
0.2
0.1
0
186
Figure 8.2
1
2
3
4
5
6
number of 1s
7
8
9
10
20
EXERCISE 8A
In all these questions you should apply this checklist to the hypothesis test.
Was the test set up before or after the data were known?
b
Was the sample used for the test chosen at random and are the data
independent?
Is the statistical procedure actually testing the original claim?
c
You should also comment critically on whether these steps have been followed.
●
Establish the null and alternative hypotheses.
●
Decide on the significance level.
●
Collect suitable data using a random sampling procedure that ensures the
items are independent.
●
Conduct the test, doing the necessary calculations.
●
Interpret the result in terms of the original claim, theory or problem.
1
Mrs da Silva is running for President. She claims to have 60% of the
population supporting her.
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8
Exercise 8A
a
She is suspected of overestimating her support and a random sample of
12 people are asked whom they support. Only four say Mrs da Silva.
Test, at the 5% significance level, the hypothesis that she has overestimated her
support.
2
A company developed synthetic coffee and claim that coffee drinkers
could not distinguish it from the real product. A number of coffee drinkers
challenged the company’s claim, saying that the synthetic coffee tasted
synthetic. In a test, carried out by an independent consumer protection body,
20 people were given a mug of coffee. Ten had the synthetic brand and ten the
natural, but they were not told which they had been given.
Out of the ten given the synthetic brand, eight said it was synthetic and two said
it was natural. Use this information to test the coffee drinkers’ claim (as against
the null hypothesis of the company’s claim), at the 5% significance level.
3
A group of 18 students decides to investigate the truth of the saying that if
you drop a piece of toast it is more likely to land butter-side down. They each
take one piece of toast, butter it on one side and throw it in the air. Fourteen
land butter-side down, the rest butter-side up. Use their results to carry out
a hypothesis test at the 1% significance level, stating clearly your null and
alternative hypotheses.
4
On average 70% of people pass their driving test first time. There are
complaints that Mr McTaggart is too harsh and so, unknown to himself, his
work is monitored. It is found that he fails four out of ten candidates. Are the
complaints justified at the 5% significance level?
187
Hypothesis testing using the binomial distribution
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8
5
A machine makes bottles. In normal running 5% of the bottles are expected
to be cracked, but if the machine needs servicing this proportion will increase.
As part of a routine check, 50 bottles are inspected and 5 are found to be
unsatisfactory. Does this provide evidence, at the 5% significance level, that
the machine needs servicing?
6
A firm producing mugs has a quality control scheme in which a random
sample of 10 mugs from each batch is inspected. For 50 such samples, the
numbers of defective mugs are as follows.
Number of defective mugs
0
1
2
3
4
5
6+
Number of samples
5
13
15
12
4
1
0
(i)
(ii)
Find the mean and standard deviation of the number of defective mugs
per sample.
Show that a reasonable estimate for p, the probability that a mug is
defective, is 0.2. Use this figure to calculate the probability that a randomly
chosen sample will contain exactly two defective mugs. Comment on the
agreement between this value and the observed data.
The management is not satisfied with 20% of mugs being defective and
introduces a new process to reduce the proportion of defective mugs.
(iii)
(iv)
A random sample of 20 mugs, produced by the new process, contains just
one which is defective. Test, at the 5% level, whether it is reasonable to
suppose that the proportion of defective mugs has been reduced, stating
your null and alternative hypotheses clearly.
What would the conclusion have been if the management had chosen to
conduct the test at the 10% level?
[MEI]
7
An annual mathematics contest contains 15 questions, 5 short and 10 long.
The probability that I get a short question right is 0.9. The probability
that I get a long question right is 0.5. My performances on questions are
independent of each other. Find the probability of the following:
(i)
(ii)
(iii)
(iv)
I get all the 5 short questions right.
I get exactly 8 of the 10 long questions right.
I get exactly 3 of the short questions and all of the long questions right.
I get exactly 13 of the 15 questions right.
After some practice, I hope that my performance on the long questions will
improve this year. I intend to carry out an appropriate hypothesis test.
(v)
State suitable null and alternative hypotheses for the test.
In this year’s contest I get exactly 8 of the 10 long questions right.
188
(vi)
Is there sufficient evidence, at the 5% significance level, that my
performance on long questions has improved?
8
Isaac claims that 30% of cars in his town are red. His friend Hardip thinks that
the proportion is less than 30%. The boys decided to test Isaac’s claim at the
5% significance level and found that 2 cars out of the random sample of 18
were red. Carry out the hypothesis test and state your conclusion.
9
At the 2009 election, 13 of the voters in Chington voted for the Citizens
Party. One year later, a researcher questioned 20 randomly selected voters in
Chington. Exactly 3 of these 20 voters said that if there were an election next
week they would vote for the Citizens Party. Test at the 2.5% significance level
whether there is evidence of a decrease in support for the Citizens Party in
Chington, since the 2009 election.
[Cambridge International AS and A Level Mathematics 9709, Paper 73 Q1 June 2010]
Critical values and critical (rejection) regions
In Example 8.1 the number 1 came up six times and this was not enough for
Leonora to show that the die was biased. What was the least number of times 1
would have had to come up for the test to give the opposite result?
Critical values and critical (rejection) regions
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q1 November 2007]
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8
We again use X to denote the number of times 1 comes up in the 20 throws and
so X = 6 means that the number 1 comes up six times.
We know from our earlier work that the probability that X 5 is 0.8982 and
we can use the binomial distribution to work out the probabilities that X = 6,
X = 7, etc.
P(X = 6) = 20C2(56 )14( 16 )6 = 0.0647
P(X = 7) = 20C7(56 )13( 16 )7 = 0.0259
20C can
2
also be
written as 20 .
2
We know P(X 6) = 1 − P(X 5) = 1 − 0.8982 = 0.1018.
0.1018 is a little over 10% and so greater than the significance level of 5%. There
is no reason to reject H0.
What about the case when the number 1 comes up seven times, that is X = 7?
Since
P(X 6) = P(X 5) + P(X = 6)
P(X 6) = 0.8982 + 0.0647 = 0.9629
So
P(X 7) = 1 − P(X 6)
= 1 − 0.9629 = 0.0371 = 3.71%
Since 3.7% < 5%, H0 is now rejected in favour of H1.
You can see that Leonora needed the 1 to come up seven or more times if her
claim was to be upheld. She missed by just one. You might think Leonora’s ‘all
or nothing’ test was a bit harsh. Sometimes tests are designed so that if the result
falls within a certain region further trials are recommended.
189
In this example the number 7 is the critical value (at the 5% significance level),
the value at which you change from accepting the null hypothesis to rejecting it.
The range of values for which you reject the null hypothesis, in this case X 7, is
called the critical region or the rejection region.
Hypothesis testing using the binomial distribution
S2
8
It is sometimes easier in hypothesis testing to find the critical region and see if
your value lies in it, rather than working out the probability of a value at least as
extreme as the one you have, the procedure used so far.
The quality control department of a factory
tests a random sample of 20 items from
each batch produced. A batch is rejected
(or perhaps subject to further tests) if the
number of faulty items in the sample, X,
is more than 2.
This means that the rejection region is X 3.
It is much simpler for the operator carrying
out the test to be told the rejection region
(determined in advance by the person designing
the procedure) than to have to work out a
probability for each test result.
EXAMPLE 8.2
Test
procedure
Take 20 pistons
If 3 or more are
faulty REJECT
the batch
World-wide 25% of men are colour-blind but it is believed that the condition is
less widespread among a group of remote hill tribes. An anthropologist plans to
test this by sending field workers to visit villages in that area. In each village
30 men are to be tested for colour-blindness. Find the rejection region for the test
at the 5% level of significance.
SOLUTION
Let p be the probability that a man in that area is colour-blind.
p = 0.25
Null hypothesis, H0:
Alternative hypothesis, H1: p 0.25 (Less colour-blindness in this area)
Significance level:
5%
With the hypothesis H0, if the number of colour-blind men in a sample of 30 is
X, then X B(30, 0.25).
The rejection region is the region X k, where
P(X k) 0.05
190
and
P(X k + 1) 0.05.
P(X = 0) = (0.75)30 = 0.00018
P(X = 1) = 30(0.75)29(0.25) = 0.00179
30
P(X = 3) = (0.75)27(0.25)3 = 0.02685
3
30
P(X = 4) = (0.75)26(0.25)4 = 0.06042.
4
So
P(X 3) = 0.00018 + 0.00179 + 0.00863 + 0.02685 ≈ 0.0375 0.05
but
P(X 4) ≈ 0.0929 0.05.
Therefore the rejection region is X 3.
?
●
What is the rejection region at the 10% significance level?
Critical values and critical (rejection) regions
P(X = 2) = 30 (0.75)28(0.25)2 = 0.00863
2
S2
8
In many other hypothesis tests it is usual to find the critical values from tables.
EXPERIMENTS
Mind reading
Here is a simple experiment to see if you can read the mind of a friend whom
you know well. The two of you face each other across a table on which is placed a
coin. Your friend takes the coin and puts it in one or other hand under the table.
You have to guess which one.
Play this game at least 20 times and test at the 10% significance level whether you
can read your friend’s mind.
Left and right
It is said that if people are following a route which brings them to a T-junction
where they have a free choice between turning left and right the majority will
turn right.
Design and carry out an experiment to test this hypothesis.
Note
This is taken very seriously by companies choosing stands at exhibitions. It is
considered worth paying extra for a location immediately to the right of one of the
entrances.
191
Coloured sweets
Hypothesis testing using the binomial distribution
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8
Get a large box of coloured sweets, such as Smarties,
and taste the different colours. Choose the colour, C,
which you think has the most distinctive flavour.
Now close your eyes and get a friend to feed you
sweets. Taste each one and say if it is your chosen
colour or not. Do this for at least 20 sweets and test at
the 10% significance level whether you can pick out
those with colour C by taste.
EXERCISE 8B
1
In a certain country, 90% of letters are delivered the day after posting.
A resident posts eight letters on a certain day.
Find the probability that
(i) all eight letters are delivered the next day
(ii) at least six letters are delivered the next day
(iii) exactly half the letters are delivered the next day.
It is later suspected that the service has deteriorated as a result of mechanisation.
To test this, 17 letters are posted and it is found that only 13 of them arrive
the next day. Let p denote the probability, after mechanisation, that a letter is
delivered the next day.
Hint: You will find
it easier to work
out the probability
that the number not
arriving on time
is 3, 2, 1 or 0 than
to calculate the
probability that the
number arriving
on time is 0, 1, 2,
…, 13.
(iv)
(v)
(vi)
Write down suitable null and alternative hypotheses for the value of p.
Carry out the hypothesis test, at the 5% level of significance, stating your
results clearly.
Write down the critical region for the test, giving a reason for your choice.
[MEI]
2
For most small birds, the ratio of males to females may be expected to be
about 1:1. In one ornithological study birds are trapped by setting fine-mesh
nets. The trapped birds are counted and then released. The catch may be
regarded as a random sample of the birds in the area.
The ornithologists want to test whether there are more male blackbirds than
females.
(i)
(ii)
192
Assuming that the sex ratio of blackbirds is 1:1, find the probability that a
random sample of 16 blackbirds contains
(a) 12 males
(b) at least 12 males.
State the null and alternative hypotheses the ornithologists should use.
In one sample of 16 blackbirds there are 12 males and 4 females.
(iii)
(iv)
[MEI]
3
A seed supplier advertises that, on average, 80% of a certain type of seed will
germinate. Suppose that 18 of these seeds, chosen at random, are planted.
(i)
Find the probability that 17 or more seeds will germinate if
(a) the supplier’s claim is correct
(b) the supplier is incorrect and 82% of the seeds, on average, germinate.
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8
One-tail and two-tail tests
Carry out a suitable test using these data at the 5% significance level,
stating your conclusion clearly. Find the critical region for the test.
Another ornithologist points out that, because female birds spend much
time sitting on the nest, females are less likely to be caught than males.
Explain how this would affect your conclusions.
Mr Brewer is the advertising manager for the seed supplier. He thinks that
the germination rate may be higher than 80% and he decides to carry out a
hypothesis test at the 10% level of significance. He plants 18 seeds.
(ii)
(iii)
(iv)
Write down the null and alternative hypotheses for Mr Brewer’s test,
explaining why the alternative hypothesis takes the form it does.
Find the critical region for Mr Brewer’s test. Explain your reasoning.
Determine the probability that Mr Brewer will reach the wrong conclusion if
(a) the true germination rate is 80%
(b) the true germination rate is 82%.
[MEI]
One-tail and two-tail tests
Think back to the two examples in the first part of this chapter.
What would Dan have said if his eight children had all been girls? What would
Leonora have said if the number 1 had not come up at all?
In both our examples the claim was not only that something was unusual but
that it was so in a particular direction. So we looked only at one side of the
distributions when working out the probabilities, as you can see in figure 8.1 on
page 181 and figure 8.2 on page 186. In both cases we applied one-tail tests. (The
word ‘tail’ refers to the darker coloured part at the end of the distribution.)
If Dan had just claimed that there was something odd about his chromosomes,
then you would have had to work out the probability of a result as extreme on
either side of the distribution, in this case eight girls or eight boys, and you would
then apply a two-tail test.
193
Hypothesis testing using the binomial distribution
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8
Here is an example of a two-tail test.
EXAMPLE 8.3
The producer of a television programme claims that it is politically unbiased.
‘If you take somebody off the street it is 50 : 50 whether he or she will say the
programme favours the government or the opposition’, she says.
However, when ten people, selected at random, are asked the question ‘Does the
programme support the government or the opposition?’, nine say it supports the
government.
Does this constitute evidence, at the 5% significance level, that the producer’s
claim is inaccurate?
SOLUTION
Read the last sentence carefully and you will see that it does not say in which
direction the bias must be. It does not ask if the programme is favouring the
government or the opposition, only if the producer’s claim is inaccurate. So you
must consider both ends of the distribution, working out the probability of such
an extreme result either way: 9 or 10 saying it favours the government, or 9 or 10
the opposition. This is a two-tail test.
If p is the probability that somebody believes the programme supports the
government, you have
Claim accurate
Null hypothesis, H0:
p = 0.5
Alternative hypothesis, H1: p 0.5
Significance level:
5%
Two-tail test
Claim inaccurate
The situation is modelled by the binomial distribution B(10, 0.5) and is shown in
figure 8.3.
0.25
probability
0.2
0.15
0.1
0.05
0
0
Figure 8.3
194
1
2
3
4
5
6
Number of people
7
8
9
10
This gives
P(X = 0) = 1
1024
P(X = 10) = 1
1024
P(X = 1) = 10
1024
10
P(X = 9) =
1024
Since 2.15% 5% the null hypothesis is rejected in favour of the alternative, that
the producer’s claim is inaccurate.
Exercise 8C
where X is the number of people saying the programme favours the government.
22
Thus the total probability for the two tails is 1024
or 2.15%.
S2
8
Note
You have to look carefully at the way a test is worded to decide if it should be
one-tail or two-tail.
Dan claimed his chromosomes made him more likely to father boys than girls. That
requires a one-tail test.
Leonora claimed the die was biased in the direction of too many 1s. Again a
one-tail test.
The test of the television producer’s claim was for inaccuracy in either direction and
so a two-tail test was needed.
EXERCISE 8C
1
To test the claim that a coin is biased, it is tossed 12 times. It comes down
heads 3 times. Test at the 10% significance level whether this claim is
justified.
2
A biologist discovers a colony of a previously unknown type of bird nesting in
a cave. Out of the 16 chicks which hatch during his period of investigation, 13
are female. Test at the 5% significance level whether this supports the view that
the sex ratio for the chicks differs from 1 : 1.
3
People entering an exhibition have to choose whether to turn left or right.
Out of the first twelve people, nine turn left and three right. Test at the 5%
significance level whether people are more likely to turn one way than the other.
4
A multiple choice test has 15 questions, with the answer for each allowing five
options, A, B, C , D and E. All the students in a class tell their teacher that they
guessed all 15 answers. The teacher does not believe them. Devise a two-tail
test at the 10% significance level to apply to a student’s mark to test the
hypothesis that the answers were not selected at random.
5
When a certain language is written down, 15% of the letters are Z. Use this
information to devise a test at the 10% significance level which somebody who
does not know the language could apply to a short passage, 50 letters long, to
determine whether it is written in the same language.
195
6
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8
A seed firm states on a packet of rare seeds that the germination rate is 20%.
The packet contains 25 seeds.
(i)
Hypothesis testing using the binomial distribution
(ii)
How many seeds would you expect to germinate out of the packet?
What is the probability of exactly 5 seeds germinating?
A man buys a packet and only 1 seed germinates.
(iii)
7
Is he justified in complaining?
Given that X has a binomial distribution in which n = 15 and p = 0.5, find the
probability of each of the following events.
(i)
(ii)
(iii)
(iv)
X=4
X4
X = 4 or X = 11
X 4 or X 11
A large company is considering introducing a new selection procedure for
job applicants. The selection procedure is intended to result over a long
period in equal numbers of men and women being offered jobs. The new
procedure is tried with a random sample of applicants and 15 of them,
11 women and 4 men, are offered jobs.
(v)
(vi)
Carry out a suitable test at the 5% level of significance to determine
whether it is reasonable to suppose that the selection procedure is
performing as intended. You should state the null and alternative
hypotheses under test and explain carefully how you arrive at your
conclusions.
Suppose now that, of the 15 applicants offered jobs, w are women. Find
all the values of w for which the selection procedure should be judged
acceptable at the 5% level.
[MEI]
Type I and Type II errors
There are two types of error that can occur when a hypothesis test is carried out.
They are illustrated in the following example.
EXAMPLE 8.4
A gold coin is used for the toss at a country’s football matches but it is suspected
of being biased. It is suggested that it shows heads more often than it should.
A test is planned in which the coin is to be tossed 19 times and the results
recorded. It is decided to use a 5% significance level; so, if the coin shows heads
14 or more times, it will be declared biased.
What errors are possible in interpreting the test result?
196
SOLUTION
S2
8
Two types of error are possible.
A Type I error
The probabilities of possible outcomes from 19 tosses when p = 0.5 can be found
using the binomial distribution. Some of them are given, to 2 significant figures,
in the table below.
Number of heads
10
11
12
13
14
15
16
Probability
0.50
0.32
0.18
0.084
0.032
0.010
0.0022
Type I and Type II errors
In this case the coin is actually unbiased, so the probability, p, of it showing heads is
given by p = 0.5. However, it happens to come up heads 14 or more times and so is
incorrectly declared to be biased.
The table shows that the probability of getting 14 or more heads, and so
making the error of rejecting the true null hypothesis that p = 0.5, is 0.032
and so just less than the 5% significance level. This type of error, where a null
hypothesis is rejected despite being correct, is called a Type I error. The figures
in the table illustrate the fact that for a binomial test the probability of making
a Type I error is either equal to the significance level of the test or slightly less
than it. For most other hypothesis tests it is equal to the significance level;
indeed that is the meaning of the term significance level, the probability of
rejecting a true null hypothesis.
A Type II error
The other type of error occurs when the null hypothesis is in fact false but is
nonetheless accepted. Imagine that the gold coin is actually biased with p = 0.8
and that it shows heads 12 times. In this test the null hypothesis is rejected if the
number of heads is 14 or more, and so it is accepted if the number of heads is less
than 14.
Since 12 14, the null hypothesis is accepted, even though it is in fact false. This
is called a Type II error, where a false null hypothesis is accepted.
In this case, it is possible to use the binomial distribution to work out the
probability of a Type II error. When p = 0.8, the probability that when the coin is
tossed 19 times the number of heads is less than 14 can be found to be 0.163, and
so this is the probability of a Type II error in this example.
197
Notes
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8
1 Notice that it was only possible to find the probability of a Type II error in
Example 8.4 because the value of the population parameter under consideration
Hypothesis testing using the binomial distribution
was known: p = 0.8. Since finding out about this parameter is the object of the test,
it would be unusual for it to be known. So, in practice, it is often not possible to
calculate the probability of a Type II error. By contrast, no calculation at all is needed
to find the probability of a Type I error; it is the significance level of the test.
2 For a given sample size, the probabilities of the two types of errors are linked.
In Example 8.4, the probability of a Type II error could be reduced by making
the test more severe; instead of requiring 14 or more heads to declare the coin
biased, it could be reduced to 13 or perhaps 12. However, that would increase the
probability of a Type I error.
3 The circumstances under which these errors occur is shown below.
Reality
Decision
Accept H0
(decide the coin is unbiased)
Reject H0
(decide the coin is biased)
The null hypothesis,
H0, is true.
Correct decision
H0 wrongly rejected:
Type I error
The null hypothesis,
H0, is false.
H0 wrongly accepted:
Type II error
Correct decision
In summary
EXAMPLE 8.5
●
A type I error occurs when the sample leads you to wrongly reject H0 when it
is in fact true.
●
A type II error occurs when the sample leads you to wrongly accept H0 when it
is in fact false.
It is known that 60% of the moths of a certain species are red; the rest are yellow.
A biologist finds a new colony of these moths and observes that more of them
seem to be red than she would expect. She designs an experiment in which she
will catch 10 moths at random, observe their colour and then release them. She
will then carry out a hypothesis test using a 5% significance level.
(i)
(ii)
(iii)
(iv)
198
State the null and alternative hypotheses for this test.
Find the rejection region.
Find the probability of a Type I error.
If in fact the proportion of red moths is 80%, find the probability that the
test will result in a Type II error.
SOLUTION
(i)
Null hypothesis:
H0: p = 0.6
The proportion of red moths in
this colony is 60%.
Alternative hypothesis:
H1: p 0.6
The proportion of red moths is
greater than 60%.
Assuming H0 is true, you can calculate the following probabilities for the 10
moths in the sample.
All 10 moths are red:
(0.6)10 = 0.0060...
9 are red and 1 yellow:
10C
8 are red and 2 yellow:
10C
1
× (0.6)9 × 0.4 = 0.0403...
2
× (0.6)8 × (0.4)2 = 0.1209...
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8
Type I and Type II errors
(ii)
Let p be the probability that a randomly selected moth is red.
There is no need to go any further.
The probability that there are nine or ten red moths is
0.0403... + 0.0060... = 0.0463...
and this is less than the 5% significance level.
The probability that there are eight, nine or ten red moths is
0.1209... + 0.0403... + 0.0060... = 0.167...
and this is greater than 5%.
So the rejection region for this test is 9 or 10 red moths.
(iii)
A Type I error occurs when a true null hypothesis is rejected.
In this case if H0 is true, and so p = 0.6, the probability of it being rejected
because a particular sample has 9 or 10 red moths has already been worked
out to be 0.0463... in part (ii). When rounded to 3 significant figures, this
gives 0.0464.
So the probability of a Type 1 error is 0.0464 (to 3 s.f.).
(iv)
If the proportion of red moths is 80%, the correct result from the test
would be for the null hypothesis to be rejected in favour of the alternative
hypothesis. The probability of this happening is
10C
1
× (0.8)9 × 0.2 + (0.8)10 = 0.376 (to 3 s.f.)
A Type II error occurs when this result does not occur.
So in this situation the probability of a Type II error is 1 − 0.376 = 0.624.
199
Hypothesis testing using the binomial distribution
S2
8
EXERCISE 8D
1
At a certain airport 20% of people take longer than an hour to check in. A
new computer system is installed, and it is claimed that this will reduce the
time to check in. It is decided to accept the claim if, from a random sample
of 22 people, the number taking longer than an hour to check in is either
0 or 1.
(i)
(ii)
(iii)
Calculate the significance level of the test.
State the probability that a Type I error occurs.
Calculate the probability that a Type II error occurs if the probability that
a person takes longer than an hour to check in is now 0.09.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q4 June 2007]
2
A manufacturer claims that 20% of sugar-coated chocolate beans are red.
George suspects that this percentage is actually less than 20% and so he takes a
random sample of 15 chocolate beans and performs a hypothesis test with the
null hypothesis p = 0.2 against the alternative hypothesis p 0.2. He decides to
reject the null hypothesis in favour of the alternative hypothesis if there are 0
or 1 red beans in the sample.
(i)
(ii)
With reference to this situation, explain what is meant by a Type I error.
Find the probability of a Type I error in George’s test.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q2 November 2005]
3
In a certain city it is necessary to pass a driving test in order to be allowed to
drive a car. The probability of passing the driving test at the first attempt is
0.36 on average. A particular driving instructor claims that the probability of
his pupils passing at the first attempt is higher than 0.36. A random sample
of 8 of his pupils showed that 7 passed at the first attempt.
(i)
(ii)
Carry out an appropriate hypothesis test to test the driving instructor’s
claim, using a significance level of 5%.
In fact, most of this random sample happened to be careful and sensible
drivers. State which type of error in the hypothesis test (Type I or Type II)
could have been made in these circumstances and find the probability of
this type of error when a sample of size 8 is used for the test.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q4 June 2009]
4
It is claimed that a certain 6-sided die is biased so that it is more likely to show
a six than if it was fair. In order to test this claim at the 10% significance level,
the die is thrown 10 times and the number of sixes is noted.
(i)
Given that the die shows a six on 3 of the 10 throws, carry out the test.
On another occasion the same test is carried out again.
(ii)
(iii)
Find the probability of a Type I error.
Explain what is meant by a Type II error in this context.
[Cambridge International AS and A Level Mathematics 9709, Paper 71 Q6 November 2010]
200
KEY POINTS
1
●
Was the test set up before or after the data were known?
●
Was the sample involved chosen at random and are the data independent?
●
Is the statistical procedure actually testing the original claim?
Steps for conducting a hypothesis test
●
Establish the null and alternative hypotheses.
●
Decide on the significance level.
●
Collect suitable data using a random sampling procedure that ensures the
items are independent.
●
Conduct the test, doing the necessary calculations.
●
Interpret the result in terms of the original claim, theory or problem.
3
A Type I error occurs when a true null hypothesis is rejected. The
probability of a Type I error occurring is less than or equal to the
significance level of the test.
4
A Type II error occurs when a false null hypothesis is accepted. The
probability of a Type II error occurring depends on the (unknown) value of
the population parameter; in a binomial test the parameter is p.
S2
8
Key points
2
Hypothesis testing checklist
201
The Poisson distribution
S2
9
9
The Poisson distribution
If something can go wrong, sooner or later it will go wrong.
Murphy’s Law
ElectricsExpress.com
Since our ‘next day delivery guarantee’ went
live, the number of orders has increased
dramatically. We are now one of the most
popular websites for mail order electrical goods.
We would like to reassure our customers that
we have taken on more staff to cope with the
increased demand for our products.
It is impossible to predict the level of demand,
however, we do know that we are receiving
an average of 150 orders per hour!
The appearance of this update on their website prompted a statistician to contact
ElectricsExpress.com. She offered to analyse the data and see what suggestions
she could come up with.
For her detailed investigation, she considered the distribution of the number of
orders per minute. For a random sample of 1000 single-minute intervals during
the last month, she collected the following data.
Number of orders
per minute
0
1
2
3
4
5
6
7
7
Frequency
70
215
265
205
125
75
30
10
5
Summary statistics for this frequency distribution are as follows.
n = 1000,
Σxf = 2525 and Σ x2f = 8885
–
⇒
x = 2.525
and
sd = 1.58 (to 3 s.f.)
She also noted that
202
●
orders made on the website appear at random and independently of each
other
●
the average number of orders per minute is about 2.5 which is equivalent to
150 per hour.
She suggested that the appropriate probability distribution to model the number
of orders was the Poisson distribution.
The particular Poisson distribution, with an average number of 2.5 orders per
minute, is defined as an infinite discrete random variable given by
r
P(X = r) = e−2.5 × 2.5
r!
for
r = 0, 1, 2, 3, 4, ...
●
X represents the random variable ‘number of orders per minute’
●
e is the mathematical constant 2.718 281 828 459...
●
e–2.5 can be found from your calculator as 0.082 (to 3 d.p.)
●
r ! means r factorial, for example 5! = 5 × 4 × 3 × 2 × 1 = 120.
Values of the corresponding probability distribution may be tabulated using the
formula, together with the expected frequencies this would generate. For example
The Poisson distribution
where
S2
9
4
P(X = 4) = e−2.5 × 2.5
4!
= 0.133 60 ...
= 0.134 (to 3 s.f.)
Number of orders per minute (r)
0
1
2
3
4
5
6
7
>7
Observed frequency
70
215
265
205
125
75
30
10
5
0.082
0.205
0.257
0.214
0.134
0.067
0.028
0.010
0.003
82
205
257
214
134
67
28
10
3
P(X = r)
Expected frequency
The closeness of the observed and expected frequencies (see figure 9.1) implies
that the Poisson distribution is indeed a suitable model in this instance.
300
Observed frequencies
frequency
250
Expected frequencies
200
150
100
50
0
0
1
2
3
4
5
6
number of orders per minute
7
�7
Figure 9.1
Note also that the sample mean, x– = 2.525, is very close to the sample variance,
s 2 = 2.509 (to 4 s.f.). You will see later that, for a Poisson distribution, the
expectation and variance are the same. So the closeness of these two summary
statistics provides further evidence that the Poisson distribution is a suitable model.
203
The Poisson distribution
S2
9
The Poisson distribution
A discrete random variable may be modelled by a Poisson distribution provided
●
events occur at random and independently of each other, in a given interval of
time or space
●
the average number events in the given interval, λ, is uniform and finite.
Let X represent the number of occurrences in a given interval, then
P(X = r) = e−λ × λ
r!
r
for r = 0, 1, 2, 3, 4, ...
Like the discrete random variables you met in Chapter 4, the Poisson distribution
may be illustrated by a vertical line chart. The shape of the Poisson distribution
depends on the value of the parameter λ (pronounced ‘lambda’). The letter µ
(pronounced ‘mu’) is also commonly used to represent the Poisson parameter. If
λ is small the distribution has positive skew, but as λ increases the distribution
becomes progressively more symmetrical. Three typical Poisson distributions are
illustrated in figure 9.2.
(a)
p
1.0
(b)
0.8
p
0.4
λ�1
λ � 0.2
0.6
0.2
0.4
0.2
0
(c)
0
1
2
X
3
4
0
5
0
1
2
3
X
4
5
6
p
0.2
λ�5
0.1
0
0
1
2
3
4
5
X
6
7
8
9
10
11
12
Figure 9.2 The shape of the Poisson distribution for (a) λ = 0.2 (b) λ = 1 (c) λ = 5
204
EXAMPLE 9.1
The number of defects in a wire cable can be modelled by the Poisson
distribution with a uniform rate of 1.5 defects per kilometre.
S2
9
The Poisson distribution
There are many situations in which events happen singly and the average number
of occurrences per given interval of time or space is uniform and is known or can
be easily found. Such events might include: the number of goals scored by a team in
a football match, the number of telephone calls received per minute at an exchange,
the number of accidents in a factory per week, the number of particles emitted in
a minute by a radioactive substance whose half-life is relatively long, the number
of typing errors per page in a document, the number of flaws per metre in a roll of
cloth or the number of micro-organisms in 1 millilitre of pond water.
Find the probability that
(i)
(ii)
a single kilometre of wire will have exactly 3 defects
a single kilometre of wire will have at least 5 defects.
SOLUTION
Let X represent the number of defects per kilometre, then
r
P(X = r) = e−1.5 × 1.5
r!
for
r = 0, 1, 2, 3, 4, ....
3
(i)
(ii)
P(X = 3) = e−1.5 × 1.5
3!
= 0.125 510...
= 0.126 (to 3 s.f.)
P(X 5) = 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]
0
1
2
3
4
= 1 – e−1.5 × 1.5 + e−1.5 × 1.5 + e−1.5 × 1.5 + e−1.5 × 1.55 + e−1.5 × 1.5
0!
1!
2!
3!
4!
= 1 – [0.223 130... + 0.334 695... + 0.251 021... + 0.125 510...
+ 0.047 066...]
= 0.0186 (to 3 s.f.)
Calculating Poisson distribution probabilities
In Example 9.1, about the defects in a wire cable, you had to work out P(X 5).
To do this you used P(X 5) = 1 – P(X 4) which saved you having to work out
all the probabilities for five or more occurrences and adding them together. Such
calculations can take a long time even though the terms eventually get smaller and
smaller, so that after some time you will have gone far enough for the accuracy
you require and may stop.
However, Example 9.1 did involve working out and summing five probabilities
and so was quite time consuming. Here are two ways of cutting down on the
amount of work, and so on the time you take.
205
Recurrence relations
S2
9
Recurrence relations allow you to use the term you have obtained to work out the
next one. For the Poisson distribution with parameter λ,
The Poisson distribution
P(X = 0) = e–λ
You must use your calculator to find this term.
P(X = 1) = e–λ × λ = λP(X = 0)
Multiply the previous term by λ.
= λ P(X = 1)
2
λ
Multiply the previous term by .
2
3
P(X = 3) = e–λ × λ = λ P(X = 2)
3! 3
λ
Multiply the previous term by .
3
4
P(X = 4) = e–λ × λ = λ P(X = 3)
4! 4
λ
Multiply the previous term by .
4
P(X = 2) = e–λ ×
λ2
2!
In general, you can find P(X = r) by multiplying your previous probability,
λ
P(X = r − 1), by . You would expect to hold the latest value on your calculator
r
and keep a running total in the memory.
Setting this out on paper with λ = 1.5 (the figure from Example 9.1) gives these
figures.
No. of cases, r
Conversion
0
1
2
3
4
× 1.5
1.5
× –––
2
1.5
× –––
3
1.5
× –––
4
P(X = r)
Running total, P(X r)
0.223 130…
0.223 130…
0.334 695…
0.557 825…
0.251 021…
0.808 846…
0.125 510…
0.934 356…
0.047 066…
0.981 422…
Adapting the Poisson distribution for different time intervals
EXAMPLE 9.2
Jasmit is considering buying a telephone answering machine. He has one for
five days’ free trial and finds that 22 messages are left on it. Assuming that this is
typical of the use it will get if he buys it, find:
(i)
(ii)
(iii)
the mean number of messages per day
the probability that on one particular day there will be exactly six messages
the probability that there will be exactly six messages in two days.
SOLUTION
(i)
Converting the total for five days to the mean for a single day gives
daily mean = 22 = 4.4 messages per day
5
206
Calling X the number of messages per day,
(ii)
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9
6
P(X = 6) = e−4.4 × 4.4
6!
= 0.124
Exercise 9A
(iii)
The mean for two days is
22
2 × 5 = 8.8 messages
So the probability of exactly six messages is
6
e−8.8 × 8.8 = 0.0972
6!
Modelling with a Poisson distribution
In the example about ElectricsExpress.com, the mean and variance of the number
of orders placed per minute on the website were given by x– = 2.525 and s 2 = 2.51
(to 3 s.f.). The corresponding Poisson parameter, λ, was then taken to be 2.5.
It can be shown that for any Poisson distribution
Mean = E(X) = λ
and
Variance = Var(X) = λ.
The notation Po(λ) or Poisson(λ) is used to describe this distribution. Formal
derivations of the mean and variance of a Poisson distribution are given in
Appendix 4 on the CD.
When modelling data with a Poisson distribution, the closeness of the mean and
variance is one indication that the data fit the model well.
When you have collected the data, go through the following steps in order to
check whether the data may be modelled by a Poisson distribution.
EXERCISE 9A
●
Work out the mean and variance and check that they are roughly equal.
●
Use the sample mean to work out the Poisson probability distribution and a
suitable set of expected frequencies.
●
Compare these expected frequencies with your observations.
1
If X Po(1.75), use the Poisson formula to calculate
(i)
2
P(X = 2)
(ii)
P(X 0).
If X Po(3.1), use the Poisson formula to calculate
(i)
P(X = 3)
(ii)
P(X < 2)
(iii)
P(X 2).
207
The Poisson distribution
S2
9
3
The number of wombats that are killed on a particular stretch of road in
Australia in any one day can be modelled by a Po(0.42) random variable.
(i)
(ii)
Calculate the probability that exactly two wombats are killed on a given
day on this stretch of road.
Find the probability that exactly four wombats are killed over a 5-day
period on this stretch of road.
4
A typesetter makes 1500 mistakes in a book of 500 pages. On how many pages
would you expect to find (i) 0 (ii) 1 (iii) 2 (iv) 3 or more mistakes? State any
assumptions in your workings.
5
In a country the mean number of deaths per year from lightning strike is 2.2.
(i)
Find the probabilities of 0, 1, 2 and more than 2 deaths from lightning
strike in any particular year.
In a neighbouring country, it is found that one year in twenty nobody dies
from lightning strike.
(ii)
6
Estimate the mean number of deaths per year in that country from
lightning strike.
350 raisins are put into a mixture which is well stirred and made into 100
small buns. Estimate how many of these buns will
(i)
(ii)
be without raisins
contain five or more raisins.
In a second batch of 100 buns, exactly one has no raisins in it.
(iii)
7
Estimate the total number of raisins in the second mixture.
A ferry takes cars and small vans on a short journey from an island to the
mainland. On a representative sample of weekday mornings, the numbers of
vehicles, X, on the 8 am sailing were as follows.
20
21
(i)
24
21
24
22
22
21
23
23
21
22
20
20
22
22
23
20
22
24
Show that X does not have a Poisson distribution.
In fact 20 of the vehicles belong to commuters who use that sailing of the ferry
every weekday morning. The random variable Y is the number of vehicles
other than those 20 who are using the ferry.
(ii)
Investigate whether Y may reasonably be modelled by a Poisson
distribution.
The ferry can take 25 vehicles on any journey.
(iii)
208
On what proportion of days would you expect at least one vehicle to be
unable to travel on this particular sailing of the ferry because there was no
room left and so have to wait for the next one?
8
Small hard particles are found in the molten glass from which glass bottles
are made. On average, 15 particles are found per 100 kg of molten glass. If a
bottle contains one or more such particles it has to be discarded.
Exercise 9A
Suppose bottles of mass 1 kg are made. It is required to estimate the
percentage of bottles that have to be discarded. Criticise the following
‘answer’: Since the material for 100 bottles contains 15 particles, approximately
15% will have to be discarded.
S2
9
Making suitable assumptions, which should be stated, develop a correct
argument using a Poisson model, and find the percentage of faulty 1 kg
bottles to three significant figures.
Show that about 3.7% of bottles of mass 0.25 kg are faulty.
[MEI]
9
People arrive randomly and independently at the elevator in a block of flats
at an average rate of 4 people every 5 minutes.
(i)
(ii)
(iii)
Find the probability that exactly two people arrive in a 1-minute period.
Find the probability that nobody arrives in a 15-second period.
The probability that at least one person arrives in the next t minutes is
0.9. Find the value of t.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q6 June 2008]
10
A shopkeeper sells electric fans. The demand for fans follows a Poisson
distribution with mean 3.2 per week.
(i)
(ii)
(iii)
Find the probability that the demand is exactly 2 fans in any one week.
The shopkeeper has 4 fans in his shop at the beginning of a week. Find
the probability that this will not be enough to satisfy the demand for fans
in that week.
Given instead that he has n fans in his shop at the beginning of a week,
find, by trial and error, the least value of n for which the probability of
his not being able to satisfy the demand for fans in that week is less than
0.05.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q6 November 2005]
11
People arrive randomly and independently at a supermarket checkout at an
average rate of 2 people every 3 minutes.
(i)
Find the probability that exactly 4 people arrive in a 5-minute period.
At another checkout in the same supermarket, people arrive randomly and
independently at an average rate of 1 person each minute.
(ii)
Find the probability that a total of fewer than 3 people arrive at the two
checkouts in a 3-minute period.
[Cambridge International AS and A Level Mathematics 9709, Paper 71 Q2 November 2010]
209
The Poisson distribution
S2
9
12
A manufacturer of rifle ammunition tests a large consignment for accuracy
by firing 500 batches, each of 20 rounds, from a fixed rifle at a target. Those
rounds that fall outside a marked circle on the target are classified as misses.
For each batch of 20 rounds the number of misses is counted.
Misses, X
0
1
2
3
4
5
6–20
Frequency
230
189
65
15
0
1
0
(i)
(ii)
(iii)
(iv)
Estimate the mean number of misses per batch.
Use your mean to estimate the probability of a batch producing 0, 1, 2,
3, 4 and 5 misses using the Poisson distribution as a model.
Use your answers to part (ii) to estimate expected frequencies of 0, 1, 2,
3, 4 and 5 misses per batch in 500 batches and compare your answers
with those actually found.
Do you think the Poisson distribution is a good model for this situation?
The sum of two or more Poisson distributions
Safer crossing near our school?
A recent traffic survey has revealed that the
number of vehicles using the main road
outside the school has reached levels where
crossing has become a hazard to our students.
The survey, carried out by a group of our
students, show that the volume of traffic
has increased so much that our students
are almost taking their lives in their hands
when crossing the road.
At 3 pm, usually one of quieter periods of the day, the average number of vehicles passing our
school to go into the town is 3.5 per minute and the average number of vehicles heading out
of town is 5.7 per minute. A safe crossing is a must!
The town council has told our students that if they can show that there is a greater than 1 in 4
chance of more than 10 vehicles passing per minute, then we should be successful in getting a
safer crossing for outside the school.
Assuming that the flows of vehicles, into and out of town, can be modelled by
independent Poisson distributions, you can model the flow of vehicles in both
directions as follows.
210
Let X represent the number of vehicles travelling into town at 3 pm, then
X ∼ Po(3.5).
Let Y represent the number of vehicles travelling out of town at 3 pm, then
Y ∼ Po(5.7).
You can find the probability distribution for T as follows.
P(T = 0) = P(X = 0) × P(Y = 0)
= 0.0302 × 0.0033 = 0.0001
P(T = 1) = P(X = 0) × P(Y = 1) + P(X = 1) × P(Y = 0)
= 0.0302 × 0.0191 + 0.1057 × 0.0033 = 0.0009
P(T = 2) = P(X = 0) × P(Y = 2) + P(X = 1) × P(Y = 1) + P(X = 2) × P(Y = 0)
= 0.0302 × 0.0544 + 0.1057 × 0.0191 + 0.1850 × 0.0033 = 0.0043
and so on.
The sum of two or more Poisson distributions
Let T represent the number of vehicles travelling in either direction at 3 pm, then
T = X + Y.
S2
9
You can see that this process is very time consuming. Fortunately, you can make
life a lot easier by using the fact that if X and Y are two independent Poisson
random variables, with means λ and µ respectively, then if T = X + Y then T is a
Poisson random variable with mean λ + µ.
X Po(λ) and Y ∼ Po(µ)
⇒
X + Y ∼ Po(λ + µ)
Using T Po(9.2) gives the required probabilities straight away.
P(T = 0) = e− 9.2 = 0.0001
P(T = 1) = e− 9.2 × 9.2 = 0.0009
2
P(T = 2) = e− 9.2 × 9.2 = 0.0043
2!
and so on.
You can now use the distribution for T to find the probability that the total traffic
flow exceeds 10 vehicles per minute.
P(T 10) = 1 – P(T 10)
= 1 – 0.6820 = 0.318
Since there is a greater than 25% chance of more than 10 vehicles passing
per minute, the case for the crossing has been made, based on the Poisson
probability models.
211
A rare disease causes the death, on average, of 2.0 people per year in Sweden, 0.8
in Norway and 0.5 in Finland. As far as is known the disease strikes at random
and cases are independent of one another.
What is the probability of 4 or more deaths from the disease in these three
countries in any year?
The Poisson distribution
S2
9
EXAMPLE 9.3
SOLUTION
Notice first that:
●
P(4 or more deaths) = 1 – P(3 or fewer deaths)
●
each of the three distributions fulfils the conditions for it to be modelled by
the Poisson distribution.
You can therefore add the three distributions together and treat the result as a
single Poisson distribution.
The overall mean is given by
2.0
+
0.8
+ 0.5 = 3.3
Sweden
Norway
Finland Total
giving an overall distribution of Po(3.3).
The probability of 4 or fewer deaths is then
2
3
1 − e− 3.3 × 1 + 3.3 + 3.3 + 3.3
2!
3!
So the probability of 4 or more deaths is given by
1 – 0.580 = 0.420
Notes
1 You may only add Poisson distributions in this way if they are independent of
each other.
2 The proof of the validity of adding Poisson distributions in this way is given in
Appendix 5 on the CD.
EXAMPLE 9.4
On a lonely Highland road in Scotland, cars are observed passing at the rate of 6 per
day and lorries at the rate of 3 per day. On the road is an old cattle grid which will
soon need repair. The local works department decide that if the probability of more
than 2 vehicles per hour passing is less than 1% then the repairs to the cattle grid can
wait until next spring, otherwise it will have to be repaired before the winter.
When will the cattle grid have to be repaired?
SOLUTION
Let C be the number of cars per hour, L be the number of lorries per hour and V
be the number of vehicles per hour.
212
V=L+C
Assuming that a car or a lorry passing along the road is a random event and the
two are independent
6 cars a day is 6 = 0.25 cars in an hour.
24
Similarly, there are 3 = 0.125 lorries per hour.
24
The required probability is
P(V > 2) = 1 – P(V 2)
2
= 1 − e− 0.375 × 1 + 0.375 + 0.375
2!
Exercise 9B
and so
⇒
C Po(0.25), L Po(0.125)
V Po(0.25 + 0.125)
V Po(0.375)
S2
9
= 0.006 65
This is less than 1% and so the repairs are left until spring.
?
●
EXERCISE 9B
The modelling of this situation raises a number of questions.
1
Is it true that a car or lorry passing along the road is a random event, or are
some of these regular users, like the lorry collecting the milk from the farms
along the road? If, say, three of the cars and one lorry are regular daily users,
what effect does this have on the calculation?
2
Is it true that every car or lorry travels independently of every other one?
3
Are vehicles more likely in some hours than others?
4
There are no figures for bicycles or motorcycles or other vehicles. Why might
this be so?
1
The numbers of lorry drivers and car drivers visiting an all-night transport
cafe between 2 am and 3 am on a Sunday morning have independent Poisson
distributions with means 5.1 and 3.6 respectively.
(i)
(ii)
Find the probabilities that, between 2 am and 3 am on any Sunday,
(a) exactly five lorry drivers visit the cafe
(b) at least one car driver visits the cafe
(c) exactly five lorry drivers and exactly two car drivers visit the cafe.
By using the distribution of the total number of drivers visiting the cafe,
find the probability that exactly seven drivers visit the cafe between 2 am
and 3 am on any Sunday. Given that exactly seven drivers visit the cafe
between 2 am and 3 am on one Sunday, find the probability that exactly
five of them are driving lorries.
[MEI]
213
2
Telephone calls reach a departmental administrator independently and at
random, internal ones at a mean rate of two in any five-minute period, and
external ones at a mean rate of one in any five-minute period.
(i)
The Poisson distribution
S2
9
(ii)
(iii)
3
Find the probability that in a five-minute period, the administrator
receives
(a) exactly three internal calls
(b) at least two external calls
(c) at most five calls in total.
Given that the administrator receives a total of four calls in a five-minute
period, find the probability that exactly two were internal calls.
Find the probability that in any one-minute interval no calls are received.
Two random variables, X and Y, have independent Poisson distributions given
by X Po(1.4) and Y Po(3.6) respectively.
(i)
Using the distributions of X and Y only, calculate
(a) P(X + Y = 0)
(b) P(X + Y = 1)
(c) P(X + Y = 2).
The random variable T is defined by T = X + Y.
(ii)
(iii)
4
A boy is watching vehicles travelling along a motorway. All the vehicles he
sees are either cars or lorries; the numbers of each may be modelled by two
independent Poisson distributions. The mean number of cars per minute is
8.3 and the mean number of lorries per minute is 4.7.
(i)
(ii)
(iii)
5
214
Write down the distribution of T.
Use your distribution from part (ii) to check your results in part (i).
For a given period of one minute, find the probability that he sees
(a) exactly seven cars
(b) at least three lorries.
Calculate the probability that he sees a total of exactly ten vehicles in a
given one-minute period.
Find the probability that he observes fewer than eight vehicles in a given
period of 30 seconds.
The number of cats rescued by an animal shelter each day may be modelled by
a Poisson distribution with parameter 2.5, while the number of dogs rescued
each day may be modelled by an independent Poisson distribution with
parameter 3.2.
(i)
Calculate the probability that on a randomly chosen day the shelter rescues
(a) exactly two cats
(b) exactly three dogs
(c) exactly five cats and dogs in total.
(ii)
Given that one day exactly five cats and dogs were rescued, find the
conditional probability that exactly two of these animals were cats.
6
The numbers of emissions per minute from two radioactive substances, A and
B, are independent and have Poisson distributions with means 2.8 and 3.25
respectively.
Find the probabilities that in a period of one minute there will be
(ii)
(iii)
7
at least three emissions from substance A
one emission from one of the two substances and two emissions from the
other substance
a total of five emissions.
Exercise 9B
(i)
S2
9
The number of incoming telephone calls received per minute by a company’s
telephone exchange follows a Poisson distribution with mean 1.92.
(i)
Find the probabilities of the following events.
(a) Exactly two calls are received in a one-minute interval.
(b) Exactly two calls are received each minute in a five-minute interval.
(c) At least five calls are received in a five-minute interval.
The number of outgoing telephone calls made per minute at the same
exchange also follows a Poisson distribution, with mean λ. It is found that
the proportion of one-minute intervals containing no outgoing calls is 20%.
Incoming and outgoing calls occur independently.
(ii)
(iii)
(iv)
Find the value of λ.
Find the probability that a total of four calls, incoming and outgoing, pass
through the exchange in a one-minute interval.
Given that exactly four calls pass through the exchange in a one-minute
interval, find the probability that two are incoming and two are outgoing.
[MEI]
8
The numbers of goals per game scored by teams playing at home and away in
the Premier League are modelled by independent Poisson distributions with
means 1.63 and 1.17 respectively.
(i)
(ii)
Find the probability that, in a game chosen at random,
(a) the home team scores at least two goals
(b) the result is a 1–1 draw
(c) the teams score five goals between them.
Give two reasons why the proposed model might not be suitable.
[MEI, part]
215
The Poisson distribution
S2
9
9
Every day I check the number of emails on my computer at home. The
numbers of emails, x, received per day for a random sample of 100 days are
summarised by
Σx = 184,
(i)
(ii)
(iii)
Σx2 = 514.
Find the mean and variance of the data.
Give two reasons why the Poisson distribution might be thought to be a
suitable model for the number of emails received per day.
Using the mean as found in part (i), calculate the expected number of days,
in a period of 100 days, on which I will receive exactly two emails.
On a working day, I also receive emails at the office. The number of emails
received per day at the office follows a Poisson distribution with mean λ. On
1.5% of working days I receive no emails at the office.
(iv)
(v)
Show that λ = 4.2, correct to 2 significant figures. Hence find the probability
that on one working day I receive at least five emails at the office.
Find the probability that on one working day I receive a total of ten emails
(at home and at the office).
[MEI, part]
The Poisson approximation to the binomial distribution
Rare disease blights town
Chemical plant blamed
A rare disease is attacking residents of Avonford.
In the last year alone five people have been
diagnosed as suffering from it. This is over three
times the national average.
The disease (known as Palfrey’s condition)
causes nausea and fatigue. One sufferer, James
Louth (32), of Harpers Lane, has been unable to
work for the past six months. His wife Muriel (29)
said ‘I am worried sick, James has lost his job and
I am frightened that the children (Mark, 4, and
Samantha, 2) will catch it.’
Mrs Louth blames the chemical complex on
the industrial estate for the disease. ‘There were
never any cases before Avonford Chemicals
arrived.’
Local environmental campaigner Roy James
supports Mrs Louth. ‘I warned the local council
when planning permission was sought that this
would mean an increase in this sort of illness.
Normally we would expect 1 case in every 40 000
of the population in a year.’
Avonford Chemicals spokesperson, Julia
Millward said ‘We categorically deny that our
216
Muriel Louth believes that the local chemical plant
could destroy her family’s lives
plant is responsible for the disease. Our record on
safety is very good. None of our staff has had the
disease. In any case five cases in a population of
60 000 can hardly be called significant.’
1
The expected number of cases is 60 000 × 40000
or 1.5, so 5 does seem rather high.
Do you think that the chemical plant is to blame or do you think people are just
looking for an excuse to attack it? How do you decide between the two points of
view? Is 5 really that large a number of cases anyway?
1–
1
39999
40000 = 40000 .
The probability of 5 cases among 60 000 people (and so 59 995 people not getting
the disease) is given by
60000C 39999
5
40000
59995
5
1
40000 ≈ 0.0141.
What you really want to know, however, is not the probability of exactly 5 cases
but that of 5 or more cases. If that is very small, then perhaps something unusual
did happen in Avonford last year.
You can find the probability of 5 or more cases by finding the probability of up
to and including 4 cases, and subtracting it from 1.
The probability of up to and including 4 cases is given by:
39999
40000
The Poisson approximation to the binomial distribution
The situation could be modelled by the binomial distribution. The probability of
1
somebody getting the disease in any year is 40000 and so that of not getting it is
S2
9
60000
0 cases
+ 60000C1 39999
40000
59999
+ 60000C2 39999
40000
1
40000
59998
+ 60000C3 39999
40000
1 case
1
40 000
59997
+ 60000C4 39999
40000
2
1
40000
59996
2 cases
3
1
40000
3 cases
4
4 cases
It is messy but you can evaluate it on your calculator. It comes out to be
0.223 + 0.335 + 0.251 + 0.126 + 0.047 = 0.981.
(The figures are written to three decimal places but more places were used in the
calculation.)
217
The Poisson distribution
S2
9
So the probability of 5 or more cases in a year is 1 – 0.981 = 0.019. It is unlikely
but certainly could happen, see figure 9.3 below.
p
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
number of cases
(
5
6
7
)
1
Figure 9.3 Probability distribution B 60 000, –––––
40 000
Note
Two other points are worth making. First, the binomial model assumes the trials are
independent. If this disease is at all infectious, that certainly would not be the case.
Second, there is no evidence at all to link this disease with Avonford Chemicals.
There are many other possible explanations.
Approximating the binomial terms
Although it was possible to do the calculation using results derived from the
binomial distribution, it was distinctly cumbersome. In this section you will
see how the calculations can be simplified, a process which turns out to be
unexpectedly profitable. The work that follows depends upon the facts that the
event is rare but there are many opportunities for it to occur: that is, p is small
and n is large.
Start by looking at the first term, the probability of 0 cases of the disease. This is
39999
40000
218
60000
= k, a constant.
Now look at the next term, the probability of 1 case of the disease. This is
60000C 39999
1
40000
S2
9
( )
59999
1
40000
×
60000
=k×
60000
39 999
≈k×
60000
40000
The Poisson approximation to the binomial distribution
39999
40000
60000 ×
×
39999
40000
=
40000
= k × 1.5.
Now look at the next term, the probability of 2 cases of the disease. This is
60000C
2
39999
×
40000
59998
×
( )
1
40000
=
60000 × 59 999 39999
×
40000
2×1
=
k × 60000 × 59999
2 × 1 × 39999 × 39999
≈
k × 60000 × 60000
2 × 40000 × 40000
2
60000
2
40000
×
×
39999
( )
1
40000
2
2
= k × (1.5) .
2
Proceeding in this way leads to the following probability distribution for the
number of cases of the disease.
Number of cases
0
1
Probability
k
k × 1.5
2
k×
(1.5)2
2!
3
k×
(1.5)3
3!
k×
4
…
(1.5)4
4!
…
Since the sum of the probabilities = 1,
2
3
4
k + k × 1.5 + k × (1.5) + k × (1.5) + k × (1.5) + ... = 1
2!
3!
4!
2
3
(1.5) + (1.5) + (1.5)4 + ... = 1
k 1 + 1.5 +
2!
3!
4!
The terms in the square brackets form a well known series in pure mathematics,
the exponential series ex.
2
3
4
ex = 1 + x + x + x + x + ...
2! 3! 4 !
Since k × e1.5 = 1, k = e–1.5.
219
This gives the probability distribution for the number of cases of the disease as
S2
9
Number of cases
The Poisson distribution
Probability
0
1
2
e–1.5
e–1.5 × 1.5
e–1.5 ×
3
(1.5)2
2!
e–1.5 ×
(1.5)3
3!
4
…
(1.5)4
4!
…
e–1.5 ×
r
and in general for r cases the probability is e–1.5 × (1.5) .
r!
Accuracy
These expressions are clearly much simpler than those involving binomial
coefficients. How accurate are they? The following table compares the results
from the two methods, given to six decimal places.
Probability
Number
of cases
Exact binomial method
Approximate method
0
0.223 126
0.223 130
1
0.334 697
0.334 695
2
0.251 025
0.251 021
3
0.125 512
0.125 511
4
0.047 066
0.047 067
You will see that the agreement is very good; there are no differences until the
sixth decimal place.
The Poisson distribution may be used as an approximation to the binomial
distribution, B(n, p), when
EXAMPLE 9.5
●
n is large (typically n > 50)
●
p is small (and so the event is rare)
●
np is not too large (typically np < 5).
It is known that nationally one person in a thousand is allergic to a particular
chemical used in making a wood preservative. A firm that makes this wood
preservative employs 500 people in one of its factories.
(i)
(ii)
220
What is the probability that more than two people at the factory are allergic
to the chemical?
What assumption are you making?
SOLUTION
(i)
Let X be the number of people in a random sample of 500 who are allergic to
the chemical.
n = 500
p = 0.001
Since n is large and p is small, the Poisson approximation to the binomial is
appropriate.
Exercise 9C
X B(500, 0.001)
S2
9
λ = np
= 500 × 0.001
= 0.5
Consequently
r
P(X = r) = e–λ × λ
r!
= e–0.5 ×
0.5r
r!
P(X 2) = 1 – P(X 2)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]
2
= 1 – e− 0.5 + e– 0.5 × 0.5 + e− 0.5 × 0.5
2
= 1 – [0.6065 + 0.3033 + 0.0758]
= 1 – 0.9856
= 0.0144
(ii)
EXERCISE 9C
1
The assumption made is that people with the allergy are just as likely to
work in the factory as those without the allergy. In practice this seems rather
unlikely: you would not stay in a job that made you unhealthy.
For each of the following binomial distributions, use the binomial formula to
calculate P(X = 3). In each case use an appropriate Poisson approximation to
find P(X = 3) and calculate the percentage error in using this approximation.
Describe what you notice.
(i)
(ii)
(iii)
2
X ∼ B(25, 0.2)
X ∼ B(250, 0.02)
X ∼ B(2500, 0.002)
An automatic machine produces washers, 3% of which are defective according
to a severe set of specifications. A sample of 100 washers is drawn at random
from the production of this machine. Using a suitable approximating
distribution, calculate the probabilities of observing
(i)
(ii)
exactly 3 defectives
between 2 and 4 defectives inclusive.
221
The Poisson distribution
S2
9
3
The number of civil lawsuits filed in state and federal courts on a given day is
500. The probability that any such lawsuit is settled within one week is 0.01.
Use the Poisson approximation to find the probability that, of the original 500
lawsuits on a given day, the number that are settled within a week is
(i)
(ii)
(iii)
4
One per cent of the items produced by a certain process are defective. Using
the Poisson approximation, determine the probability that in a random
sample of 1000 articles
(i)
(ii)
5
exactly seven
at least five
at most six.
exactly five are defective
at most five are defective.
Betty drives along a 50-kilometre stretch of road 5 days a week 50 weeks a
year. She takes no notice of the 70 km h–1 speed limit and, when the traffic
allows, travels between 95 and 105 km h–1. From time to time she is caught
by the police and fined but she estimates the probability of this happening
1
on any day is 300
. If she gets caught three times within three years she will be
disqualified from driving. Use Betty’s estimates of probability to answer the
following questions.
(i)
(ii)
(iii)
What is the probability of her being caught exactly once in any year?
What is the probability of her being caught less than three times in three
years?
What is the probability of her being caught exactly three times in three
years?
Betty is in fact caught one day and decides to be somewhat cautious, reducing
her normal speed to between 85 and 95 km h–1. She believes this will reduce
1
the probability of her being caught to 500
.
What is the probability that she is caught less than twice in the next three
years?
Motorists in a particular part of Malaysia have a choice between a direct route
and a one-way scenic detour. It is known that on average one in forty of the
cars on the road will take the scenic detour. The road engineer wishes to do
some repairs on the scenic detour. He chooses a time when he expects 100 cars
an hour to pass along the road.
(iv)
6
Find the probability that, in any one hour,
(i)
(ii)
(iii)
222
no cars will turn on to the scenic detour
at most 4 cars will turn on to the scenic detour.
Between 10.30 am and 11.00 am it will be necessary to block the road
completely. What is the probability that no car will be delayed?
7
A sociologist claims that only 3% of all suitably qualified students from inner
city schools go on to university. Use his claim and the Poisson approximation
to the binomial distribution to estimate the probability that in a randomly
chosen group of 200 such students
(ii)
(iii)
exactly five go to university
more than five go to university.
If the probability that more than n of the 200 students go to university is
less than 0.2, find the lowest possible value of n.
Exercise 9C
(i)
S2
9
Another group of 100 students is also chosen. Find the probability that
(iv)
(v)
exactly five of each group go to university
exactly ten of all the chosen students go to university.
[MEI, adapted]
8
In one part of the country, one person in 80 has blood of Type P. A random
sample of 150 blood donors is chosen from that part of the country. Let X
represent the number of donors in the sample having blood of Type P.
(i)
(ii)
State the distribution of X. Find the parameter of the Poisson distribution
which can be used as an approximation. Give a reason why a Poisson
approximation is appropriate.
Using the Poisson distribution, calculate the probability that in the sample
of 150 donors at least two have blood of Type P.
[MEI, part]
9
An airline regularly sells more seats for its early morning flight from London
to Paris than are available. On average, 5% of customers who have purchased
tickets do not turn up. For this flight, the airline always sells 108 tickets. Let X
represent the number of customers who do not turn up for this flight.
(i)
State the distribution of X, giving one assumption you must make for it to
be appropriate.
There is room for 104 passengers on the flight. For the rest of the question
use a suitable Poisson approximation.
(ii)
Find the probability that
(a) there are exactly three empty seats on Monday’s flight
(b) Tuesday’s flight is full
(c) from Monday to Friday inclusive the flight is full on just one day.
[MEI, part]
223
The Poisson distribution
S2
9
10
The manufacturers of Jupiter Jellybabies have launched a promotion to boost
sales. One per cent of bags, chosen at random, contains a prize. A school
tuck-shop takes delivery of 500 bags of Jupiter Jellybabies. Let X represent the
number of bags in the delivery which contain a prize.
State clearly the distribution which X takes.
Using a Poisson approximating distribution, find P(3 X 7).
(i)
(ii)
The values of the prizes are in the following proportions.
Value of prize
$10
$100
$1000
Proportion
90%
9%
1%
(iii)
Suppose the tuck-shop receives five bags which contain prizes. Find the
probability that at least one of these prizes has value $1000.
[MEI]
Using the normal distribution as an approximation for the
Poisson distribution
You may use the normal distribution as an approximation for the Poisson
distribution, provided that its parameter (mean) λ is sufficiently large for the
distribution to be reasonably symmetrical and not positively skewed.
As a working rule λ should be at least 15.
If λ = 15, mean = 15
The letter µ is also commonly
used in place of λ for the
Poisson parameter.
and standard deviation = 15 = 3.87 (to 3 s.f.).
A normal distribution is almost entirely contained within 3 standard deviations
of its mean and in this case the value 0 is between 3 and 4 standard deviations
away from the mean value of 15.
The parameters for the normal distribution are then
Mean:
µ=λ
Variance: σ2 = λ
so that it can be denoted by N(λ, λ).
(Remember that, for a Poisson distribution, mean = variance.)
For values of λ larger than 15 the Poisson probability graph becomes less
positively skewed and more bell-shaped in appearance thus making the normal
approximation appropriate. Figure 9.4 shows the Poisson probability graph for
the two cases λ = 3 and λ = 25. You will see that for λ = 3 the graph is positively
skewed but for λ = 25 it is approximately bell-shaped.
224
S2
9
0.08
probability
0.15
0.1
0.05
0.06
0.04
0.02
0 1
2 3
4 5
λ�3
6
7 8
10
15
20
25 30
λ � 25
35
40
Figure 9.4
EXAMPLE 9.6
The annual number of deaths nationally from a rare disease, X, may be modelled
by the Poisson distribution with mean 25. One year there are 31 deaths and it is
suggested that the disease is on the increase.
What is the probability of 31 or more deaths in a year, assuming the mean has
remained at 25?
SOLUTION
The Poisson distribution with mean 25 may be approximated by the normal
distribution with parameters
Mean: 25
Standard deviation: 25 = 5
25
Using the normal distribution as an approximation for the Poisson distribution
probability
0.2
30.5
Figure 9.5
The probability of there being 31 or more deaths in a year, P(X 31), is given by
1 – Φ(z), where
z = 30.5 − 25 = 1.1.
5
(Note the continuity correction, replacing 31 by 30.5.)
The required area is
1 – Φ(1.1) = 1 – 0.8643
= 0.1357
This is not a particularly low probability; it is quite likely that there would be that
many deaths in any one year.
225
Hypothesis test for the mean of a Poisson distribution
The Poisson distribution
S2
9
The next example shows you how to carry out a hypothesis test for the mean of a
Poisson distribution.
EXAMPLE 9.7
An old university has a high tower that is quite often struck by lightning. Records
going back over hundreds of years show that on average the tower is struck on
3.2 days per year.
It is suggested that a likely effect of global warming would be an increase in the
number of days on which the tower is struck. The following year the tower is
struck by lightning on 7 days.
Carry out a suitable hypothesis test at the 5% significance level and state your
conclusion.
What is the probability of a Type I error in this test?
SOLUTION
The number of days per year that the tower is struck by lightning is modelled by
X where
X Po(3.2).
So the null and alternative hypotheses may be stated as follows.
H0: µ = 3.2
H1: µ 3.2
The population mean, µ, is unchanged.
The population mean, µ, has increased.
One-tail test
Significance level: 0.05
The test is one-tailed because
it is for an increase in lightning
strikes. A test for a change
would be two-tailed.
Number of strikes
0
e–3.2 = 0.040 76...
1
3.2 × e–3.2 = 0.130 43...
2
3
The probability of 7 or more days
with lightning strikes is (1 – the
probability of 6 or fewer such days).
The calculation is shown in the
table.
Probability
4
5
6
Total
3.22
2!
3.23
3!
3.24
4!
3.25
5!
3.26
6!
× e–3.2 = 0.208 70...
× e–3.2 = 0.222 61...
× e–3.2 = 0.178 09...
× e–3.2 = 0.113 97...
× e–3.2 = 0.060 78...
P(X 6) = 0.955 38...
So the probability that X 7 is
1 – 0.955 38... = 0.044 61... = 0.045 (3 d.p.)
226
Since 0.045 0.05 (the significance level), the null hypothesis is rejected in
favour of the alternative hypothesis at the 5% significance level.
The evidence does not support the hypothesis that there has been no change in
the incidence of lightning strikes.
Exercise 9D
A Type I error occurs when a true null hypothesis is rejected. In this test, the
rejection region is X 6 and the probability of such a result is 0.044 61... if
µ = 3.2. So the probability of a Type I error is 0.0446 (to 3 s.f.).
S2
9
Note
Notice that the result in Example 9.7 does not prove that there has been an increase
in the incidence of lightning strikes; it does, however, suggest that this may well
be the case. The test was about lightning strikes and so in itself says nothing about
global warming. Whether global warming is connected to the incidence of lightning
strikes is not what was being tested; it formed no part of either the null or the
alternative hypothesis.
EXERCISE 9D
1
The number of cars per minute entering a multi-storey car park can be
modelled by a Poisson distribution with mean 2. What is the probability that
three cars enter during a period of one minute?
What are the mean and the standard deviation of the number of cars entering
the car park during a period of 30 minutes? Use the normal approximation to
the Poisson distribution to estimate the probability that at least 50 cars enter
in any one 30-minute period.
[MEI]
2
A large computer system which is in constant operation requires an average
of 30 service calls per year.
(i)
(ii)
(iii)
State the average number of service calls per month, taking a month
1
to be 12
of a year. What assumptions need to be made for the Poisson
distribution to be used to model the number of calls in a given month?
Use the Poisson distribution to find the probability that at least one service
call is required in January. Obtain the probability that there is at least one
service call in each month of the year.
The service contract offers a discount if the number of service calls in the
year is 24 or fewer. Use a suitable approximating distribution to find the
probability of obtaining the discount in any particular year.
[MEI]
3
The number of night calls to a fire station in a small town can be modelled by
a Poisson distribution with a mean of 4.2 per night.
Use the normal approximation to the Poisson distribution to estimate the
probability that in any particular week (Sunday to Saturday inclusive) the
number of night calls to the fire station will be
(i)
(ii)
(iii)
at least 30
exactly 30
between 25 and 35 inclusive.
227
The Poisson distribution
S2
9
4
At a busy intersection of roads, accidents requiring the summoning of
an ambulance occur with a frequency, on average, of 1.8 per week. These
accidents occur randomly, so that it may be assumed that they follow a
Poisson distribution.
Use a suitable approximating distribution to find the probability that in any
particular year (of 52 weeks) the number of accidents at the intersection will be
(i)
(ii)
(iii)
5
Tina is a traffic warden. The number of parking tickets she issues per day, from
Monday to Saturday inclusive, may be modelled by a Poisson distribution with
mean 11.5. By using suitable approximating distributions, find
(i)
(ii)
(iii)
6
the probability that on a particular Tuesday she issues at least 15 parking
tickets
the probability that during any week (excluding Sunday) she issues at least
50 parking tickets
the probability that during four consecutive weeks she issues
(a) at least 50 parking tickets each week
(b) at least 200 parking tickets altogether.
Account for the difference in the two answers.
The number of emails I receive per day on my computer may be modelled by a
Poisson distribution with mean 8.5.
(i)
(ii)
(iii)
7
at most 100
exactly 100
between 95 and 105 inclusive.
Use the most appropriate method to calculate the probability that I receive
(a) at least 8 emails tomorrow
(b) at least 240 emails next June.
What assumption do you have to make to find the probability in part
(i) (b)?
Compare your answers to parts (i) (a) and (b) and account for the variation.
At a petrol station cars arrive independently and at random times at constant
average rates of 8 cars per hour travelling east and 5 cars per hour travelling
west.
(i)
Find the probability that, in a quarter-hour period
(a) one or more cars travelling east and one or more cars travelling west
will arrive,
(b) a total of 2 or more cars will arrive.
(ii)
Find the approximate probability that, in a 12-hour period, a total of more
than 175 cars will arrive.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q6 June 2005]
228
8
(i)
(ii)
(iii)
State the null and alternative hypotheses.
Find the rejection region for the test.
Find the probability of a Type I error.
S2
9
Exercise 9D
Some ancient documents from the pharaoh’s astronomer are discovered
in one of the pyramids. They include records, covering many years, of
shooting stars during a certain part of one particular night of the year. The
data are well modelled by a Poisson distribution with mean 5.6. A modern
astronomer has a theory that there are now fewer shooting stars and so, on
the right day and time, repeats the observation and carries out a suitable
hypothesis test, using a 10% significance level.
The astronomer observes three shooting stars.
(iv)
9
Carry out the hypothesis test.
A dressmaker makes dresses for Easifit Fashions. Each dress requires 2.5 m2
of material. Faults occur randomly in the material at an average rate of
4.8 per 20 m2.
(i)
Find the probability that a randomly chosen dress contains at least 2 faults.
Each dress has a belt attached to it to make an outfit. Independently of faults
in the material, the probability that a belt is faulty is 0.03. Find the probability
that, in an outfit,
(ii)
(iii)
neither the dress nor its belt is faulty,
the dress has at least one fault and its belt is faulty.
The dressmaker attaches 300 randomly chosen belts to 300 randomly chosen
dresses. An outfit in which the dress has at least one fault and its belt is faulty
is rejected.
(iv)
Use a suitable approximation to find the probability that fewer than 3
outfits are rejected.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q6 June 2006]
10
It is proposed to model the number of people per hour calling a car breakdown
service between the times 0900 and 2100 by a Poisson distribution.
(i)
Explain why a Poisson distribution may be appropriate for this situation.
People call the car breakdown service at an average rate of 20 per hour, and a
Poisson distribution may be assumed to be a suitable model.
(ii)
(iii)
Find the probability that exactly 8 people call in any half hour.
By using a suitable approximation, find the probability that exactly 250
people call in the 12 hours between 0900 and 2100.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q5 June 2007]
229
The Poisson distribution
S2
9
11
Major avalanches can be regarded as randomly occurring events. They occur at
a uniform average rate of 8 per year.
(i)
(ii)
(iii)
Find the probability that more than 3 major avalanches occur in a
3-month period.
Find the probability that any two separate 4-month periods have a total
of 7 major avalanches.
Find the probability that a total of fewer than 137 major avalanches occur
in a 20-year period.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q3 June 2009]
12
When a guitar is played regularly, a string breaks on average once every
15 months. Broken strings occur at random times and independently of
each other.
(i)
Show that the mean number of broken strings in a 5-year period is 4.
A guitar is fitted with a new type of string which, it is claimed, breaks less
frequently. The number of broken strings of the new type was noted after a
period of 5 years.
(ii)
(iii)
The mean number of broken springs of the new type in a 5-year period is
denoted by λ. Find the rejection region for a test at the 10% significance
level when the null hypothesis λ = 4 is tested against the alternative
hypothesis λ 4.
Hence calculate the probability of making a Type I error.
The number of broken guitar strings of the new type, in a 5-year period, was
in fact 1.
(iv)
State, with a reason, whether there is evidence at the 10% significance level
that guitar strings of the new type break less frequently.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q5 June 2008]
13
Every month Susan enters a particular lottery. The lottery company states
that the probability, p, of winning a prize is 0.0017 each month. Susan thinks
that the probability of winning is higher than this, and carries out a test based
on her 12 lottery results in a one-year period. She accepts the null hypothesis
p = 0.0017 if she has no wins in the year and accepts the alternative
hypothesis p 0.0017 if she wins a prize in at least one of the 12 months.
(i)
(ii)
(iii)
Find the probability of the test resulting in a Type I error.
If in fact the probability of winning a prize each month is 0.0024, find
the probability of the test resulting in a Type II error.
Use a suitable approximation, with p = 0.0024, to find the probability
that in a period of 10 years Susan wins a prize exactly twice.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q5 November 2008]
230
14
Carry out the test at the 10% level of significance.
(ii) What would your conclusion have been if you had tested at the 5% level
of significance?
Jack decides that he will reject the null hypothesis that the average number
is 0.8 pieces per hectare if he finds 4 or more pieces of metal in another
ploughed field of area 3 hectares.
(i)
(iii)
S2
9
Exercise 9D
Pieces of metal discovered by people using metal detectors are found
randomly in fields in a certain area at an average rate of 0.8 pieces per
hectare. People using metal detectors in this area have a theory that
ploughing the fields increases the average number of pieces of metal found
per hectare. After ploughing, they tested this theory and found that a
randomly chosen field of area 3 hectares yielded 5 pieces of metal.
If the true mean after ploughing is 1.4 pieces per hectare, calculate the
probability that Jack makes a Type II error.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q6 November 2006]
15
A hospital patient’s white blood cell count has a Poisson distribution. Before
undergoing treatment the patient had a mean white blood cell count of 5.2.
After the treatment a random measurement of the patient’s white blood cell
count is made, and is used to test at the 10% significance level whether the
mean white blood cell count has decreased.
(i)
(ii)
(iii)
State what is meant by a Type I error in the context of the question, and
find the probability that the test results in a Type I error.
Given that the measured value of the white blood cell count after the
treatment is 2, carry out the test.
Find the probability of a Type II error if the mean white blood cell count
after the treatment is actually 4.1.
[Cambridge International AS and A Level Mathematics 9709, Paper 71 Q7 June 2010]
Historical note
Simeon Poisson was born in Pithiviers in France in 1781. Under family pressure he began to study
medicine but after some time gave it up for his real interest, mathematics. For the rest of his life Poisson
lived and worked as a mathematician in Paris. His contribution to the subject spanned a broad range
of topics in both pure and applied mathematics, including integration, electricity and magnetism and
planetary orbits as well as statistics. He was the author of between 300 and 400 publications and
originally derived the Poisson distribution as an approximation to the binomial distribution.
When he was a small boy, Poisson had his hands tied by his nanny who then hung him from a hook on
the wall so that he could not get into trouble while she went out. In later life he devoted a lot of time to
studying the motion of a pendulum and claimed that this interest derived from his childhood experience
of swinging against the wall.
231
The Poisson distribution
S2
9
KEY POINTS
1
The Poisson probability distribution
If X Po(λ), the parameter λ 0.
r
P(X = r) = e–λ × λ
r!
E(X) = λ
Var(X) = λ
2
3
r 0, r is an integer
Conditions under which the Poisson distribution may be used
●
The Poisson distribution is generally thought of as the probability
distribution for the number of occurrences of a rare event.
●
Situations in which the mean number of occurrences is known (or can
easily be found) but in which it is not possible, or even meaningful, to
give values to n or p may be modelled using the Poisson distribution
provided that the occurrences are
– random
– independent.
The sum of two Poisson distributions
If X ∼ Po(λ), Y Po(µ) and X and Y are independent
X + Y Po(λ + µ)
4
Approximating to the binomial distribution
The Poisson distribution may be used as an approximation to the binomial
distribution, B(n, p), when
– n is large (typically n > 50)
– p is small (and so the event is rare)
– np is not too large (typically np < 5).
It would be unusual to use the Poisson distribution with parameter, λ,
greater than about 20.
5
232
The Poisson distribution Po(λ) may be approximated by N(λ, λ), provided λ
is about 15 or more.
S2
10
A theory is a good theory if it satisfies two requirements: It must
accurately describe a large class of observations on the basis of a
model that contains only a few arbitrary elements, and it must make
definite predictions about the results of future observations.
Stephen Hawking
A Brief History of Time
Continuous random variables
10
Continuous random
variables
Lucky escape for local fisherman
Local fisherman Zhang Wei stared death in the face
yesterday as he was plucked from his boat by a freak wave.
Only the quick thinking of his brother Xiuying who grabbed
hold of his legs, saved Wei from a watery grave.
‘It was a bad day and suddenly this lump of water came
down on us,’ said Wei. ‘It was a wave in a million. It must
have been higher than our house, which is about 11 m high,
and it caught me off guard’.
Hero Xiuying is a man of few words. ‘All in the day’s work’
was his only comment.
Freak waves do occur and they can have serious consequences in terms of
damage to shipping, oil rigs and coastal defences, sometimes resulting in loss of
life. It is important to estimate how often they will occur, and how high they will
be. Was Zhang Wei’s one in a million estimate for a wave higher than 11 metres
at all accurate?
Before you can answer this question, you need to know the probability density of
the heights of waves at that time of the year in the area where the Zhang brothers
were fishing. The graph in figure 10.1 shows this sort of information; it was
collected in the same season of the year as the Zhang accident.
To obtain figure 10.1 a very large amount of wave data had to be collected. This
allowed the class interval widths of the wave heights to be sufficiently small for
the outline of the curve to acquire this shape. It also ensured that the sample data
were truly representative of the population of waves at that time of the year.
In a graph such as figure 10.1 the vertical scale is a measure of probability density.
Probabilities are found by estimating the area under the curve. The total area is
1.0, meaning that effectively all waves at this place have heights between 0.6 and
12.0 m, see figure 10.2.
233
S2
10
0.20
0.18
probability density
Continuous random variables
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0
0
1
2
3
4
5
6
7
wave height (m)
8
9
10
11
12
Figure 10.1
If this had been the place where the Zhang brothers were fishing, the probability
of encountering a wave at least 11 m high would have been 0.003, about 1 in 300.
Clearly Wei’s description of it as ‘a wave in a million’ was not justified purely by
its height. The fact that he called it a ‘lump of water’ suggests that perhaps it may
have been more remarkable for its steep sides than its height.
Area
� 12 (0.16 � 0.12) � 1
� 0.14
0.14
0.12
0.10
0.08
0.02
0
0.05
0.08
0.12
0.16
0.19
0.15
0.13
0.04
0.02
0
0.063
0.11 0.14 0.17 0.175 0.14 0.10 0.065 0.035
1
2
3
4
5
6
7
8
9
10
wave height (m)
This area is
approximately a
trapezium.
0.02
1
Area
� 12 (0.02 � 0.006) � 1
� 0.013
234
1
0.006
0.06
0.09
probability density
0.16
0.12
0.18
0.16
This area is
approximately a
trapezium.
0.20
Figure 10.2
11
12
This area is
approximately a
triangle.
0.006
0.006
Area
� 12 � 1.0 � 0.006
� 0.003
1.0
Probability density function
?
●
Is it reasonable to describe the height of a wave as random?
A function represented by a curve of this type is called a probability density
function, often abbreviated to p.d.f.. The probability density function of a
continuous random variable, X, is usually denoted by f(x). If f(x) is a p.d.f. it
follows that:
●
●
f(x) 0 for all x
∫∫ f(x) dx = 1
S2
10
Probability density function
In the wave height example the curve was determined experimentally. The curve
is continuous because the random variable, the wave height, is continuous and
not discrete. The possible heights of waves are not restricted to particular steps
(say every 12 metre), but may take any value within a range.
You cannot have negative probabilities.
The total area under the curve is 1.
All
values
of x
For a continuous random variable with probability density function f(x), the
probability that X lies in the interval [a, b] is given by
P(a X b) =
∫∫a f(x) dx
b
Looking at figure 10.1, you will see that in this case the probability density
function has quite a complicated curve and so it is not possible to find a simple
algebraic expression with which to model it.
Most of the techniques in this chapter assume that you do in fact have a
convenient algebraic expression with which to work. However, the methods
are still valid if this is not the case, but you would need to use numerical, rather
than algebraic, techniques for integration and differentiation. In the high-wave
incident mentioned on pages 233–234, the areas corresponding to wave heights
of less than 2 m and of at least 11 m were estimated by treating the shape as a
triangle: other areas were approximated by trapezia.
Note: Class boundaries
If you were to ask the question ‘What is the probability of a randomly selected
wave being exactly 2 m high?’ the answer would be zero. If you measured a likely
looking wave to enough decimal places (assuming you could do so), you would
eventually come to a figure which was not zero. The wave height might be 2.01... m
or 2.000 003... m but the probability of it being exactly 2 m is infinitesimally small.
Consequently in theory it makes no difference whether you describe the class
interval from 2 to 2.5 m as 2 h 2.5 or as 2 h 2.5.
235
However, in practice, measurements are always rounded to some extent. The reality
S2
10
of measuring a wave’s height means that you would probably be quite happy
to record it to the nearest 0.1 m and get on with the next wave. So, in practice,
measurements of 2.0 m and 2.5 m probably will be recorded, and intervals have to be
Continuous random variables
defined so that it is clear which class they belong to. You would normally expect at
one end of the interval and at the other: either 2 h 2.5 or 2 h 2.5. In either
case the probability of the wave being within the interval would be given by
2.5
∫2
f(x) dx
Rufus foils council office break-in
Somewhere an empty-pocketed thief is nursing a
sore leg and regretting the loss of a pair of trousers.
Council porter Fred Lamming, and Rufus, a Jack
Russell, were doing a late-night check round the
council head office when they came upon the intruder
on the ground floor.
‘I didn’t need to say anything,’ Fred told me; ‘Rufus
went straight for him and grabbed him by the leg.’
After a tussle the man’s trousers tore, leaving Rufus
with a mouthful of material while the man made good
his escape out of the window.
Following the incident, the town council are looking at
an electronic security system. ‘Rufus won’t live for ever,’
explained Council leader Sandra Martin.
EXAMPLE 10.1
The town council are thinking of fitting an electronic security system inside head
office. They have been told by manufacturers that the lifetime, X years, of the
system they have in mind has the p.d.f.
3x(20 − x)
4000
f(x) = 0
f(x) =
and
(i)
(ii)
236
for 0 x 20,
otherwise
Show that the manufacturers’ statement is consistent with f(x) being a
probability density function.
Find the probability that:
(a)
it fails in the first year
(a)
it lasts 10 years but then fails in the next year.
SOLUTION
(i)
S2
10
The condition f(x) 0 for all values of x between 0 and 20 is satisfied, as
shown by the graph of f(x), figure 10.3.
Probability density function
f(x)
0.1
0.075
0.05
0.025
0
1
10 11
20
x
This area gives the
probability that it lasts
10 years but then fails in the
next year, part (ii)(b).
This area gives the
probability it fails in the
first year, part (ii)(a).
Figure 10.3
The other condition is that the area under the curve is 1.
Area =
∞
20
∫∫−∞ f(x) dx = ∫∫0
3x(20 − x) dx
4000
∫∫ 0 (20x − x 2) dx
20
=
3
4000
=
3 2 x3
x
4000 10 − 3 0
=
3
203
2
4000 10 × 20 − 3
20
= 1, as required.
(ii) (a)
It fails in the first year.
This is given by P(X 1) =
∫0
1
3x(20 − x) dx
4000
∫ (20x − x ) dx
1
=
3
4000
=
3 2 x3
4000 10x − 3 0
2
0
1
(
3
3
2 1
4000 10 × 1 − 3
= 0.00725
=
)
237
(b)
Continuous random variables
S2
10
It fails in the 11th year.
This is given by P(10 X 11)
=
∫∫10
11
3x(20 − x) dx
4000
3 2 1 3 11
10x − 3 x
10
4000
= 3 10 × 112 − 13 × 113 − 3 10 × 102 − 13 × 103
4000
4000
= 0.07475
=
(
EXAMPLE 10.2
)
(
)
The continuous random variable X represents the amount of sunshine in hours
between noon and 4 pm at a skiing resort in the high season. The probability
density function, f(x), of X is modelled by
kx 2 for 0 x 4
f(x) =
otherwise.
0
(i)
(ii)
Find the value of k.
Find the probability that on a particular day in the high season there is more
than two hours of sunshine between noon and 4 pm.
SOLUTION
(i)
To find the value of k you must use the fact that the area under the graph of
f(x) is equal to 1.
∫∫
∞
−∞
∫
4
f(x) dx = ∫ kx 2 dx = 1
0
4
kx 3 = 1
3 0
Therefore
64k = 1
3
3
k=
64
So
(ii)
f(x)
1
0.5
238
Figure 10.4
0
1
2
3
4
x
The probability of more than 2 hours of sunshine is given by
P(X 2) =
∞
∫∫2
f(x) dx =
∫∫
4
3x 2 dx
2 64
4
= 64 − 8
64
= 56
64
= 0.875
In the next example, the probability density function is in two parts.
EXAMPLE 10.3
Probability density function
3
= x
64 2
S2
10
The number of hours Darren spends each day working in his garden is modelled
by the continuous random variable X, with p.d.f. f(x) defined by
kx
f(x) = k(6 − x)
0
(i)
(ii)
(iii)
for 0 x 3
for 3 x 6
otherwise.
Find the value of k.
Sketch the graph of f(x).
Find the probability that Darren will work between 2 and 5 hours in his
garden on a randomly selected day.
SOLUTION
(i)
To find the value of k you must use the fact that the area under the graph of
f(x) is equal to 1. You may find the area by integration, as shown below.
∫∫
∞
−∞
f(x) dx =
∫∫ kx dx + ∫ k(6 − x) dx = 1
3
6
0
3
6
3
kx 2 + 6kx − kx 2 = 1
2 0
2 3
Therefore
(
)
9k + (36k − 18k) −
18k − 9k = 1
2
2
9k = 1
k=
So
1
9
Note
In this case you could have found k without integration because the graph of the
p.d.f. is a triangle, with area given by 21 × base × height, resulting in the equation
1
2
hence
and
× 6 × k(6 – 3) = 1
9k = 1
k = 91
239
Sketch the graph of f(x).
(ii)
S2
10
f(x)
Continuous random variables
1
3
0
1
2
3
4
5
6 x
Figure 10.5
(iii)
To find P(2 X 5), you need to find both P(2 X 3) and P(3 X 5)
because there is a different expression for each part.
P(2 X 5) = P(2 X 3) + P(3 X 5)
=
∫2 9x dx + ∫3 9(6 − x) dx
3
5
1
3
1
5
2
2
= x + 2x − x
18 2 3 18 3
(
) ( )
= 9 − 4 + 10 − 25 − 2 − 1
18 18
3 18
2
= 0.72 to two decimaal places.
The probability that Darren works between 2 and 5 hours in his garden on a
randomly selected day is 0.72.
EXERCISE 10A
1
The continuous random variable X has probability density function f(x)
where
f(x) = kx
=0
(i)
(ii)
(iii)
(iv)
2
for 1 x 6
otherwise.
Find the value of the constant k.
Sketch y = f(x).
Find P(X 5).
Find P(2 X 3).
The continuous random variable X has p.d.f. f(x) where
f(x) = k(5 – x)
=0
(i)
(ii)
(iii)
240
for 0 x 4
otherwise.
Find the value of the constant k.
Sketch y = f(x).
Find P(1.5 X 2.3).
3
The continuous random variable X has p.d.f. f(x) where
f(x) = ax3
=0
(ii)
(iii)
4
Find the value of the constant a.
Sketch y = f(x).
Find P(X 2).
The continuous random variable X has p.d.f. f(x) where
f(x) = c
=0
(i)
(ii)
(iii)
(iv)
5
for –3 x 5
otherwise.
Find c.
Sketch y = f(x).
Find P( X 1).
Find P( X 2.5)
A continuous random variable X has p.d.f
f(x) = k(x – 1)(6 – x)
=0
(i)
(ii)
(iii)
6
for 1 x 6
otherwise.
Find the value of k.
Sketch y = f(x).
Find P(2 X 3).
A random variable X has p.d.f
f(x) = kx(3 − x)
=0
(i)
(ii)
7
Exercise 10A
(i)
S2
10
for 0 x 3
otherwise.
for 0 x 3
otherwise.
Find the value of k.
The lifetime (in years) of an electronic component is modelled by this
distribution. Two such components are fitted in a radio which will only
function if both devices are working. Find the probability that the radio will
still function after two years, assuming that their failures are independent.
The planning officer in a council needs information about how long cars stay
in the car park, and asks the attendant to do a check on the times of arrival and
departure of 100 cars. The attendant provides the following data.
Length of stay
Number of cars
Under 1
hour
1–2 hours
2–4 hours
4–10 hours
More than
10 hours
20
14
32
34
0
241
The planning officer suggests that the length of stay in hours may be
modelled by the continuous random variable X with probability density
function of the form
S2
10
Continuous random variables
f(x) = k (20 − 2x)
=0
(i)
(ii)
(iii)
(iv)
(v)
8
for 0 x 10
otherwise.
Find the value of k.
Sketch the graph of f(x).
According to this model, how many of the 100 cars would be expected to
fall into each of the four categories?
Do you think the model fits the data well?
Are there any obvious weaknesses in the model? If you were the planning
officer, would you be prepared to accept the model as it is, or would you
want any further information?
A fish farmer has a very large number of trout in a lake. Before deciding
whether to net the lake and sell the fish, she collects a sample of 100 fish and
weighs them. The results (in kg) are as follows.
Weight, W
Frequency
Weight, W
Frequency
0 W 0.5
2
2.0 W 2.5
27
0.5 W 1.0
10
2.5 W 3.0
12
1.0 W 1.5
23
3.0 W
1.5 W 2.0
26
(i)
0
Illustrate these data on a histogram, with the number of fish on the
vertical scale and W on the horizontal scale. Is the distribution of the data
symmetrical, positively skewed or negatively skewed?
A friend of the farmer suggests that W can be modelled as a continuous
random variable and proposes four possible probability density functions.
f1(w) = 92 w (3 − w)
f2(w) =
10 2
w (3 − w)2
81
f3(w) =
f4(w) =
4
w (3 − w)2
27
4 2
w (3 − w)
27
in each case for 0 W 3.
(ii)
(iii)
(iv)
242
Sketch the curves of the four p.d.f.s and state which one matches the data
most closely in general shape.
Use this p.d.f. to calculate the number of fish which that model predicts
should fall within each group.
Do you think it is a good model?
9
A random variable X has a probability density function f given by
f(x) = cx (5 − x )
=0
(ii)
6
Show that c = 125
.
The lifetime X (in years) of an electric light bulb has this distribution.
Given that a standard lamp is fitted with two such new bulbs and that their
failures are independent, find the probability that neither bulb fails in the
first year and the probability that exactly one bulb fails within two years.
Exercise 10A
(i)
S2
10
0x5
otherwise.
[MEI]
This graph shows the probability density function, f(x), for the heights, X, of
waves at the point with Latitude 44 °N Longitude 41 °W.
0.20
probability density
10
0.15
0.10
0.05
0
(i)
(ii)
2
4
6
wave height (m)
8
10
12
Write down the values of f(x) when x = 0, 2, 4, ..., 12.
Hence estimate the probability that the height of a randomly selected
wave is in the interval
(a) 0–2 m
(b) 2–4 m
(c) 4–6 m
(d) 6–8 m
(e) 8–10 m
(f) 10–12 m.
A model is proposed in which
f(x) = kx(12 – x)2
=0
(iii)
(iv)
(v)
for 0 x 12
otherwise.
Find the value of k.
Find, according to this model, the probability that a randomly selected
wave is in the interval
(a) 0–2 m
(b) 2–4 m
(c) 4–6 m
(d) 6–8 m
(e) 8–10 m
(f) 10–12 m.
By comparing the figures from the model with the real data, state
whether you think it is a good model or not.
243
Continuous random variables
S2
10
11
The continuous random variable X has p.d.f. f(x) where
f(x) = kx
= 4k – kx
=0
(i)
(ii)
(iii)
12
for 0 x 2
for 2 x 4
otherwise.
Find the value of the constant k.
Sketch y = f(x).
Find P(1 X 3.5).
A random variable X has p.d.f.
(x − 1)(2 − x)
f(x) = a
0
(i)
(ii)
(iii)
(iv)
for 1 x 2
for 2 x 4
otherwise.
Find the value of the constant a.
Sketch y = f(x).
Find P(1.5 X 2.5).
Find P( X − 2 1).
Mean and variance
You will recall from Chapter 4 that, for a discrete random variable, mean and
variance are given by: µ
∑ xi pi
µ = E(X ) =
i
Var(X) = ∑(xi − µ)2pi =
i
∑ xi2pi − [E(X )]2
i
where µ is the mean and pi is the probability of the outcome xi for i = 1, 2, 3, ...,
with the various outcomes covering all possibilities.
The expressions for the mean and variance of a continuous random variable are
equivalent, but with summation replaced by integration.
∫
µ= E(X) = ∫ x f(x) dx
All
values
of x
∫
∫
Var(X) = ∫ (x − µ)2 f(x) dx = ∫ x 2 f(x) dx − [E(X)]2
All
values
of x
All
values
of x
E(X) is the same as the population mean, µ, and is often called the mean of X.
244
EXAMPLE 10.4
The response time, in seconds, for a contestant in a general knowledge quiz is
modelled by a continuous random variable X whose p.d.f. is
(i)
(ii)
the mean time in seconds for a contestant to respond to a particular question
the standard deviation of the time taken.
Mean and variance
f(x) = x for 0 x 10.
50
The rules state that a contestant who makes no answer is disqualified from the
whole competition. This has the consequence that everybody gives an answer, if
only a guess, to every question. Find
S2
10
The organiser estimates the proportion of contestants who are guessing by
assuming that they are those whose time is at least one standard deviation greater
than the mean.
(iii)
Using this assumption, estimate the probability that a randomly selected
response is a guess.
SOLUTION
(i)
Mean time: E(X) =
=
∫0 x f (x) dx
10
∫0 50 dx
10 2
x
10
3
= x = 1000 = 20
150
3
150 0
= 6 23
The mean time is 6 23 seconds.
(ii)
Variance: Var(X ) =
=
∫0
10
x 2 f(x) dx − [E(X )]2
∫0 50 dx − (6 23 )
10 3
x
2
10
( )2
4
= x − 6 23
200 0
= 5 59
Standard deviation = variance = 5.5
The standard deviation of the times is 2.357 seconds (to 3 d.p.).
(iii)
All those with response times greater than 6.667 + 2.357 = 9.024 seconds are
taken to be guessing. The longest possible time is 10 seconds.
The probability that a randomly selected response is a guess is given by
∫
10
x dx = x 2
100 9.024
50
9.024
= 0.186
10
245
So just under 1 in 5 answers are deemed to be guesses.
S2
10
f(x)
GUESSES
Continuous random variables
0.2
0.1
2
0
4
6
8
10
x
Figure 10.6
Note
Although the intermediate answers have been given rounded to three decimal
places, more figures have been carried forward into subsequent calculations.
The median
The median value of a continuous random variable X with p.d.f. f(x) is the value
m for which
P(X m) = P(X m) = 0.5.
∫
m
Consequently
−∞
f(x) dx = 0.5
∞
∫m f(x) dx = 0.5.
and
The median is the value m such that the line x = m divides the area under the
curve f(x) into two equal parts. In figure 10.7 a is the smallest possible value of
X, b the largest. The line x = m divides the shaded region into two regions A and
B, both with area 0.5.
f(x)
∞
m
∫m f(x) dx = 0.5
∫−∞ f(x) dx = 0.5
A
0
246
Figure 10.7
a
B
m
b
x
! In general the mean does not divide the area into two equal parts but it will do so
if the curve is symmetrical about it because, in that case, it is equal to the median.
S2
10
The mode
The mode
The mode of a continuous random variable X whose p.d.f. is f(x) is the value of
x for which f(x) has the greatest value. Thus the mode is the value of X where the
curve is at its highest.
If the mode is at a local maximum of f(x), then it may often be found by
differentiating f(x) and solving the equation
f ′(x) = 0.
?
●
For which of the distributions in figure 10.8 could the mode be found by
differentiating the p.d.f.?
f(x)
f(x)
0
x
(a) The exponential
�λx
distribution f(x) � λe
(d) A bimodal distribution
0
(b) A distribution with
negative skew
0
x
(c) A triangular distribution
x
f(x)
f(x)
0
f(x)
x
0
(e) Pascal’s distribution
f(x) � 12 e��x�
f(x)
x
0
x
(f ) A uniform (rectangular)
distribution
Figure 10.8
247
Continuous random variables
S2
10
EXAMPLE 10.5
The continuous random variable X has p.d.f. f(x) where
f(x) = 4x (1 – x2)
=0
for 0 x 1
otherwise.
Find
(i) the mode
(ii) the median.
x = –0.577 is also a root
of f ′(x) = 0 but is outside
the range 0 x 1.
SOLUTION
(i)
The mode is found by differentiating f(x) = 4x – 4x3
f ′(x) = 4 – 12x2
1
Solving f ′(x) = 0
x=
= 0.577 to 3 decimal places.
3
It is easy to see from the shape of the graph (see figure 10.9) that this must be
a maximum, and so the mode is 0.577.
f(x)
2
1
Figure 10.9
0
0.2
0.4
0.6
0.8
1.0
∫∫−∞ f(x) dx = 0.5
m
(ii)
The median, m, is given by
⇒
∫0 (4x − 4x 3) dx = 0.5
m
Since x 0
[2x 2 − x 4 ]m0 = 0.5
2m2 − m4 = 0.5
Rearranging gives
2m4 – 4m2 + 1 = 0.
This is a quadratic equation in m2. The formula gives
m2 = 4 ± 16 − 8
4
m = 0.541 or 1.307 to 3 decimal places.
248
Since 1.307 is outside the domain of X, the median is 0.541.
x
The uniform (rectangular) distribution
It is common to describe distributions by the shapes of the graphs of their p.d.f.s:
U-shaped, J-shaped, etc
In figure 10.10, X may take values between a and b, and is zero elsewhere. Since
1
the area under the graph must be 1, the height is
. The term ‘uniform
b−a
distribution’ can be applied to both discrete and continuous variables so in the
continuous case it is often written as ‘uniform (rectangular)’.
f(x)
1
b�a
0
a
b
The uniform (rectangular) distribution
The uniform (rectangular) distribution is particularly simple since its p.d.f. is
constant over a range of values and zero elsewhere.
S2
10
x
Figure 10.10
EXAMPLE 10.6
A junior gymnastics league is open to children who are at least five years old but
have not yet had their ninth birthday. The age, X years, of a member is modelled
by the uniform (rectangular) distribution over the range of possible values
between five and nine. Age is measured in years and decimal parts of a year,
rather than just completed years. Find
(i)
(ii)
(iii)
(iv)
(v)
the p.d.f. f(x) of X
P(6 X 7)
E(X)
Var(X)
the percentage of the children whose ages are within one standard deviation
of the mean.
SOLUTION
(i)
The p.d.f. f(x) = 1 = 1
9−5 4
=0
for 5 x 9
otherwise.
249
(ii)
S2
10
P(6 X 7) = 14 by inspection of the rectangle in figure 10.11.
area
–14
f(x)
Continuous random variables
1
4
0
5
6
7
9
x
Figure 10.11
Alternatively, using integration
P(6 X 7) =
∫6 f(x) dx = ∫6 4 dx
7
7
1
7
= x
4 6
=7−6
4 4
= 1.
4
(iii)
By the symmetry of the graph, E(X ) = 7. Alternatively, using integration
∞
E(X ) =
∫∫ −∞x f(x) dx = ∫5 4 dx
9
x
9
2
= x
8 5
= 81 − 25 = 7.
8 8
(iv)
Var(X ) =
∞
∫−∞x 2 f(x) dx − [E(X )]2 = ∫5 4 dx − 72
9 2
x
9
3
= x − 49
12 5
= 729 − 125 − 49
12 12
= 1.333 to 3 decimal places.
(v)
Standard deviation = variance = 1.333 = 1.155.
So the percentage within 1 standard deviation of the mean is
2 × 1.155 × 100% = 57.7%.
4
?
●
250
What percentage would be within 1 standard deviation of the mean for a normal
distribution? Why is the percentage less in this example?
The mean and variance of the uniform (rectangular) distribution
In the previous example the mean and variance of a particular uniform
distribution were calculated. This can easily be extended to the general uniform
distribution given by:
The uniform (rectangular) distribution
1
b−a
=0
f(x) =
for a x b
otherwise.
f(x)
1
b�a
a
0
S2
10
mean
a�b
2
b
x
Figure 10.12
Mean
By symmetry the mean is
Variance
Var(X) =
a +b
.
2
∞
∫−∞ x 2 f(x) dx − [E(X )]2
=
∫∫a x 2 f(x) dx − [E(X )]2
=
x2
∫a b − a dx − a +2 b
b
∫
( )
b
2
b
3
− 1 (a 2 + 2ab + b 2)
= x
3(b − a) a 4
3
3
= b − a − 1 (a 2 + 2ab + b 2)
3(b − a) 4
= (b − a) (b 2 + ab + a 2) − 1 (a 2 + 2ab + b 2)
3(b − a)
4
= 1 (b 2 – 2ab + a 2)
12
= 1 (b − a)2
12
251
Continuous random variables
S2
10
EXERCISE 10B
1
The continuous random variable X has p.d.f. f(x) where
f(x) = 18x
=0
for 0 x 4
otherwise.
Find
E(X)
(ii) Var(X)
(iii) the median value of X.
(i)
2
The continuous random variable T has p.d.f. defined by
f(t) = 6 − t
18
=0
for 0 t 6
otherwise.
Find
(i) E(T )
(ii) Var(T)
(iii) the median value of T.
3
The continuous random variable Y has p.d.f. f(y) defined by
f(y) = 12y 2(1 – y) for 0 y 1
=0
otherwise.
(i)
(ii)
(iii)
4
Find E(Y ).
Find Var(Y ).
Show that, to 2 decimal places, the median value of Y is 0.61.
The random variable X has p.d.f.
1
f(x) = 6 for –2 x 4
= 0 otherwise.
(i)
(ii)
(iii)
(iv)
5
Sketch the graph of f(x).
Find P(X < 2).
Find E(X).
Find P( X 1).
The continuous random variable X has p.d.f. f(x) defined by
2 x (3 − x) for 0 x 3
f(x) = 9
otherwise.
0
(i)
(ii)
(iii)
(iv)
(v)
252
Find E(X).
Find Var(X).
Find the mode of X.
Find the median value of X.
Draw a sketch graph of f(x) and comment on your answers to parts (i), (iii)
and (iv) in the light of what it shows you.
6
k (3 + x) for 0 x 2
The function f(x) =
otherwise.
0
is the probability density function of the random variable X.
(ii)
(iii)
Show that k = 18.
Find the mean and variance of X.
Find the probability that a randomly selected value of X lies between
1 and 2.
7
A continuous random variable X has a uniform (rectangular) distribution
over the interval (4, 7). Find
(i) the p.d.f. of X
(ii) E(X)
(iii) Var(X)
(iv) P(4.1 X 4.8).
8
The distribution of the lengths of adult Martian lizards is uniform between
10 cm and 20 cm. There are no adult lizards outside this range.
(i)
(ii)
(iii)
9
Write down the p.d.f. of the lengths of the lizards.
Find the mean and variance of the lengths of the lizards.
What proportion of the lizards have lengths within
(a) one standard deviation of the mean
(b) two standard deviations of the mean?
The continuous random variable X has p.d.f. f(x) defined by
a
f(x) = x for 1 x 2
0 otherwise.
(i)
(ii)
(iii)
(iv)
(v)
10
Exercise 10B
(i)
S2
10
Find the value of a.
Sketch the graph of f(x).
Find the mean and variance of X.
Find the proportion of values of X between 1.5 and 2.
Find the median value of X.
The random variable X denotes the number of hours of cloud cover per day at
a weather forecasting centre. The probability density function of X is given by
(x − 18)2
f(x) =
k
0
0 x 24,
otherwise,
where k is a constant.
(i)
(ii)
(iii)
Show that k = 2016.
On how many days in a year of 365 days can the centre expect to have
less than 2 hours of cloud cover?
Find the mean number of hours of cloud cover per day.
253
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q7 June 2005]
Continuous random variables
S2
10
11
The random variable X has probability density function given by
4x k
f(x) =
0
0x1
otherwise.
where k is a positive constant.
(i)
(ii)
(iii)
(iv)
Show that k = 3.
Show that the mean of X is 0.8 and find the variance of X.
Find the upper quartile of X.
Find the interquartile range of X.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q5 June 2006]
12
If Usha is stung by a bee she always developes an allergic reaction. The time
taken in minutes for Usha to develop the reaction can be modelled using the
probability density function given by
k
f(t) = t + 1
0
0 t 4,
otherwise,
where k is a constant.
(i)
(ii)
(iii)
1
.
ln 5
Find the probability that it takes more than 3 minutes for Usha to
develop a reaction.
Find the median time for Usha to develop a reaction.
Show that k =
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q7 June 2008]
13
The time in minutes taken by candidates to answer a question in an
examination has probability density function given by
k(6t − t 2)
f(t) =
0
3 t 6,
otherwise,
where k is a constant.
(i)
(ii)
(iii)
(iv)
1
Show that k = 18
.
Find the mean time.
Find the probability that a candidate, chosen at random, takes longer
than 5 minutes to answer the question.
Is the upper quartile of the times greater than 5 minutes, equal to 5
minutes or less than 5 minutes? Give a reason for your answer.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q5 June 2009]
254
14
The time in hours taken for clothes to dry can be modelled by the continuous
random variable with probability density function given by
f(t) = k t
0
1 t 4,
otherwise,
Exercise10B
where k is a constant.
3
(i) Show that k = 14.
(ii) Find the mean time taken for clothes to dry.
(iii) Find the median time taken for clothes to dry.
(iv) Find the probability that the time taken for clothes to dry is between the
mean time and the median time.
S2
10
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q7 November 2008]
15
The random variable T denotes the time in seconds for which a firework
burns before exploding. The probability density function of T is given by
k e0.2t
f(t) =
0
0 t 5,
otherwise,
where k is a positive constant.
1 .
(i) Show that k =
5(e − 1)
(ii) Sketch the probability density function.
(iii) 80% of fireworks burn for longer than a certain time before they explode.
Find this time.
[Cambridge International AS and A Level Mathematics 9709, Paper 71 Q5 June 2010]
KEY POINTS
1
2
If X is a continuous random variable with p.d.f. f(x)
●
∫ f(x) dx = 1
●
f(x) 0
●
P(c x d) =
●
E(X) = ∫ x f(x) dx
●
Var(X) = ∫ x 2 f(x) dx − [E(X)]2
●
The mode of X is the value for which f(x) has its greatest magnitude.
for all x
d
∫c f(x) dx
The uniform (rectangular) distribution over the interval (a, b)
●
f(x) = 1
b−a
●
E(X) = 12 (a + b)
●
Var(X) =
(b − a)2
12
255
Linear combinations of random variables
S2
11
11
Linear combinations of
random variables
To approach zero defects, you must have statistical control of processes.
David Wilson
Unfair dismissal?
Janice
Just had one of those days. ‘Everything went wrong. First the school bus arrived
5 minutes late to pick up my little boy. Then it was wet and slippery and there were so
many people about that I just couldn’t walk at my normal speed; usually I take 15 minutes
but that day it took me 18 to get to work. And then when I got to work I had to wait
31–2 minutes for the lift instead of the usual 1–2 minute. So instead of arriving my normal
10 minutes early I was one minute late.’
Mrs Dickens just wouldn’t listen. She said she did not employ people to make excuses
and told me to leave there and then.
Do you think I have a case for unfair dismissal?
Like Janice, we all have days when everything goes wrong at once. There were
three random variables involved in her arrival time at work: the time she had to
wait for the school bus, S; the time she took to walk to work, W, and the time she
had to wait for the lift, L.
Her total time for getting to work, T, was the sum of all three: T = S + W + L.
Janice’s case was essentially that the probability of T taking such a large value
was very small. To estimate that probability you would need information about
the distributions of the three random variables involved. You would also need to
know how to handle the sum of two or more (in this case three) random variables.
The expectation (mean) of a function of X, E(g[X])
However, before you can do this, you need to extend some of the work you did
in Chapter 4 on random variables. There you learnt that, for a discrete random
variable X with P(X = xi) = pi ,
Σx × P(X = x ) = Σx p
= E[(X – µ) ] = Σ(x – µ) × P(X = x ) = Σ(x – µ) p
= E(X ) – E[(X)] = Σx × P(X = x ) – µ = Σx p – µ
its expectation = E(X) = µ =
and its variance = σ2
i
2
2
256
i
i i
2
i
2
2
i
i
i
2
i
2
2
i i
i
2
This only finds the expected value and variance of a particular random variable.
Sometimes you will need to find the mean, i.e. the expectation, of a function of a
random variable. That sounds rather forbidding and you may think the same of
the definition given below at first sight. However, as you will see in the next two
examples, the procedure is straightforward and common sense.
If g[X] is a function of the discrete random variable X then E(g[X])
is given by
E(g[X]) = ∑g[xi] × P(X = xi).
i
EXAMPLE 11.1
What is the expectation of the square of the number that comes up when a fair
die is rolled?
SOLUTION
Let the random variable X be the number that comes up when the die is rolled.
g[X ] = X 2
E(g[X ]) = E(X 2) =
∑ xi2 × P(X = xi)
i
=
12
× +
1
6
22
The expectation (mean) of a function of X, E(g[X])
●
S2
11
× 16 + 32 × 16 + 42 × 16 + 52 × 16 + 62 × 16
= 1 × 16 + 4 × 16 + 9 × 16 + 16 × 16 + 25 × 16 + 36 × 16
=
91
6
= 15.17
Note
This calculation could also have been set out in table form as shown below.
x
P(X = xi )
xi2
xi2 × P(X = xi )
1
1
_
6
1
1
_
6
2
1
_
6
4
4
_
6
3
1
_
6
9
9
_
6
4
1
_
6
16
16
_
6
5
1
_
6
25
25
_
6
6
1
_
6
36
36
_
6
Total
1
91
_
6
– = 15.17
E(g[X]) = 91
6
257
Linear combinations of random variables
S2
11
?
●
EXAMPLE 11.2
E(X 2)is not the same as [E(X)]2. In this case 15.57 3.52 which is 12.25. In fact,
the difference between E(X 2)and [E(X)]2 is very important in statistics. Why is this?
A random variable X has the following probability distribution.
Outcome
Probability
(i)
(ii)
(iii)
1
2
3
0.4
0.4
0.2
Calculate E(4X + 5).
Calculate 4E(X) + 5.
Comment on the relationship between your answers to parts (i) and (ii).
SOLUTION
(i)
E(g[X]) = ∑g[xi] × P(X = xi)
with g[X] = 4X + 5
i
xi
1
2
3
g[xi]
9
13
17
0.4
0.4
0.2
P(X = xi)
E(4X + 5) = E(g[X])
= 9 × 0.4 + 13 × 0.4 + 17 × 0.2
= 12.2
(ii)
E(X) = 1 × 0.4 + 2 × 0.4 + 3 × 0.2 = 1.8
and so
4E(X) + 5 = 4 × 1.8 + 5
= 12.2
(iii)
Clearly E(4X + 5) = 4E(X) + 5, both having the value 12.2.
Expectation: algebraic results
In Example 11.2 above you found that E(4X + 5) = 4E(X) + 5.
The working was numerical, showing that both expressions came out to be 12.2,
but it could also have been shown algebraically. This would have been set out
as follows.
258
Proof
Reasons (general rules)
E(4X + 5) = E(4X) + E(5)
= 4E(X) + E(5)
= 4E(X) + 5
S2
11
E(X ± Y ) = E(X) ± E(Y )
E(aX) = aE(X)
E(c) = c
Notice the last one, which in this case means the expectation of 5 is 5. Of course
it is; 5 cannot be anything else but 5. It is so obvious that sometimes people find
it confusing! In general
E(aX + c) = aE(X) + c
Expectation: algebraic results
Look at the general rules on the right-hand side of the page. (X and Y are
random variables, a and c are constants.) They are important but they are also
common sense.
These rules can be extended to take in the expectation of the sum of two
functions of a random variable.
E(f [X ] + g[X ]) = E(f[X ]) + E(g[X ])
where f and g are both functions of X.
EXAMPLE 11.3
The random variable X has the following probability distribution.
x
P(X = x)
1
2
3
4
0.6
0.2
0.1
0.1
Find
(i)
Var(X)
Var(7)
(ii)
(iii)
Var(3X)
(iv)
Var(3X + 7).
What general rule do the answers to parts (ii) and (iv) illustrate?
SOLUTION
(i)
x
1
2
3
4
x2
1
4
9
16
0.6
0.2
0.1
0.1
P(X = x)
E(X) = 1 × 0.6 + 2 × 0.2 + 3 × 0.1 + 4 × 0.1
= 1.7
2
E(X ) = 1 × 0.6 + 4 × 0.2 + 9 × 0.1 + 16 × 0.1
= 3.9
Var(X) = E(X 2) – [E(X)]2
= 3.9 – 1.72
= 1.01
259
(ii)
Linear combinations of random variables
S2
11
(iii)
(iv)
EXERCISE 11A
1
Var(7) = E(72) – [E(7)]2
= E(49) – [7]2
= 49 – 49
=0
General result
Var(c) = 0 for a constant c.
This result is obvious; a constant is
constant and so can have no spread.
Var(3X) = E[(3X)2] – µ2
= E(9X 2) – [E(3X)]2
= 9E(X 2) – [3E(X)]2
= 9 × 3.9 – (3 × 1.7)2
= 35.1 – 26.01
= 9.09
General result
Var(aX) = a2 Var(X).
Notice that it is a2 and not a on the righthand side, but that taking the square root
of each side gives the standard deviation
of (aX) = a × standard deviation (X) as
you would expect from common sense.
Var(3X + 7)
= E[(3X + 7)2]
– [E(3X + 7)]2
= E(9X 2 + 42X + 49)
– [3E(X) + 7]2
= E(9X 2) + E(42X) + E(49)
– [3 × 1.7 + 7]2
= 9E(X 2) + 42E(X)
+ 49 – 12.12
= 9 × 3.9 + 42 × 1.7
+ 49 – 146.41
= 9.09
General result
Var(aX + c) = a2 Var(X).
Notice that the constant c does not
appear on the right-hand side.
The probability distribution of a random variable X is as follows.
x
P(X = x)
(i)
1
2
3
4
5
0.1
0.2
0.3
0.3
0.1
Find
E(X)
(b) Var(X).
Verify that Var(2X) = 4Var(X).
(a)
(ii)
2
The probability distribution of a random variable X is as follows.
x
P(X = x)
(i)
260
(ii)
0
1
2
0.5
0.3
0.2
Find
(a) E(X)
(b) Var(X).
Verify that Var(5X + 2) = 25Var(X).
3
4
Prove that Var(aX – b) = a2 Var(X) where a and b are constants.
A coin is biased so that the probability of obtaining a tail is 0.75. The coin is
tossed four times and the random variable X is the number of tails obtained.
Find
(ii)
5
Exercise 11A
(i)
E(2X)
Var(3X).
A discrete random variable W has the following distribution.
w
P(W = w)
S2
11
1
2
3
4
5
6
0.1
0.2
0.1
0.2
0.1
0.3
Find the mean and variance of
W+7
(ii) 6W – 5.
(i)
6
The random variable X is the number of heads obtained when four unbiased
coins are tossed. Construct the probability distribution for X and find
(i)
(ii)
(iii)
7
The discrete random variable X has probability distribution given by
(i)
(ii)
8
E(X)
Var(X)
Var(3X + 4).
P(X = x) = (4x + 7)
for x = 1, 2, 3, 4.
68
Find
(a) E(X)
(b) E(X 2)
(c) E(X 2 + 5X – 2).
Verify that E(X 2 + 5X – 2) = E(X 2) + 5E(X) – 2.
A bag contains four balls, numbered 2, 4, 6, 8 but identical in all other
respects. One ball is chosen at random and the number on it is denoted by N,
so that P(N = 2) = P(N = 4) = P(N = 6) = P(N = 8) = 14 .
Show that µ = E(N) = 5 and σ2 = Var(N) = 5.
Two balls are chosen at random one after the other, with the first ball being
–
replaced after it has been drawn. Let N be the arithmetic mean of the numbers
–
on the two balls. List the possible values of N and their probabilities of being
–
–
obtained. Hence evaluate E(N ) and Var(N ).
[MEI]
261
The sums and differences of independent random variables
Sometimes, as in the case of Janice in the website forum thread on page 256, you
may need to add or subtract a number of independent random variables. This
process is illustrated in the next example.
The possible lengths (in cm) of the blades of cricket bats form a discrete uniform
distribution:
38, 40, 42, 44, 46.
The possible lengths (in cm) of the handles of cricket bats also form a discrete
uniform distribution:
22, 24, 26.
0.3
0.2
probability
EXAMPLE 11.4
probability
Linear combinations of random variables
S2
11
0
38
40 42 44 46
length of blade (cm)
0
22
24
26
length of handle (cm)
Figure 11.1
The blades and handles can be joined together to make bats of various lengths,
and it may be assumed that the lengths of the two sections are independent.
(i)
(ii)
(iii)
(iv)
(v)
How many different (total) bat lengths are possible?
Work out the mean and variance of random variable X1, the length (in cm)
of the blades.
Work out the mean and variance of random variable X2, the length (in cm)
of the handles.
Work out the mean and variance of random variable X1 + X2, the total length
of the bats.
Verify that
E(X1 + X2) = E(X1) + E(X2)
and
262
Var(X1 + X2) = Var(X1) + Var(X2).
SOLUTION
The number of different bat lengths is 7. This can be seen from the sample
space diagram below.
26
64
66
68
70
72
24
62
64
66
68
70
22
60
62
64
66
68
40
42
44
length of blade (cm)
46
38
total length
of bats
Figure 11.2
(ii)
Length of blade (cm)
38
40
42
44
46
Probability
0.2
0.2
0.2
0.2
0.2
Σ
E(X1) = µ1 = xp
= (38 × 0.2) + (40 × 0.2) + (42 × 0.2) + (44 × 0.2) + (46 × 0.2)
= 42 cm
( 2)
S2
11
The sums and differences of independent random variables
length of handle (cm)
(i)
2
Var(X1) = E X 1 – µ1
( 2)
E X 1 = (382 × 0.2) + (402 × 0.2) + (422 × 0.2) + (442 × 0.2) + (462 × 0.2)
= 1772
Var(X1) = 1772 – 422 = 8
(iii)
Length of handle (cm)
22
24
26
Probability
1
3
1
3
1
3
E(X2) = µ2 = (22 × 13 ) + (24 × 13 ) + (26 × 13 )
= 24 cm
( 2)
2
Var(X2) = E X 2 – µ2
( 2)
E X 2 = (222 × 13 ) + (242 × 13 ) + (262 × 13 )
= 578.667 to 3 d.p.
Var(X2) = 578.667 – 242 = 2.667 to 3 d.p.
263
Linear combinations of random variables
S2
11
(iv)
The probability distribution of X1 + X2 can be obtained from figure 11.2.
Total length of cricket bat (cm)
60
62
64
66
68
70
72
Probability
1
15
2
15
3
15
3
15
3
15
2
15
1
15
(
) (
) (
2 +
+ ( 70 × 15
) (72 × 151 )
) (
) (
1 +
2 +
3 +
3 +
3
E(X1 + X 2) = 60 × 15
62 × 15
64 × 15
66 × 15
68 × 15
)
= 66 cm
Var(X1 + X 2) = E[(X1 + X 2)2] − 662
(
) (
) (
2 +
+ ( 702 × 15
) (722 × 151 )
) (
) (
1 +
2 +
3 +
3
E[(X1 + X 2)2] = 602 × 15
682 × 15
622 × 15
642 × 135 + 662 × 15
)
= 65500 = 4366.667 to 3 d.p.
15
Var(X1 + X 2) = 4366.667 − 662 = 10.667 to 3 d.p.
(v)
E(X1 + X2) = 66 = 42 + 24 = E(X1) + E(X2), as required.
Var(X1 + X2) = 10.667 = 8 + 2.667 = Var(X1) + Var(X2), as required.
Note
You should notice that the standard deviations of X1 and X2 do not add up to the
standard deviation of (X1 + X2).
8+
i.e.
2. 667 ≠
10. 667
2. 828 + 1. 633 ≠ 3. 266
General results
Example 11.4 has illustrated the following general results for the sums and
differences of random variables.
For any two random variables X1 and X2
●
E(X1 + X2) = E(X1) + E(X2)
Replacing X2 by –X2 in this result gives
E(X1 + (–X2)) = E(X1) + E(–X2)
●
E(X1 – X2) = E(X1) – E(X2)
If the variables X1 and X2 are independent then
●
264
Var(X1 + X2) = Var(X1) + Var(X2)
Replacing X2 by –X2 gives
S2
11
Var(X1 + (–X2)) = Var(X1) + Var(–X2)
Var(X1 – X2) = Var(X1) + (–1)2 Var(X2)
Var(X1 – X2) = Var(X1) + Var(X2)
The sums and differences of normal variables
If the variables X1 and X2 are normally distributed, then the distributions of
(X1 + X2) and (X1 – X2) are also normal. The means of these distributions are
E(X1) + E(X2) and E(X1) – E(X2).
You must, however, be careful when you come to their variances, since you may
only use the result that
Var(X1 ± X2) = Var(X1) + Var(X2)
to find the variances of these distributions if the variables X1 and X2 are
independent.
This is the situation in the next two examples.
EXAMPLE 11.5
Robert Fisher, a keen chess player, visits his local club most days. The total time
taken to drive to the club and back is modelled by a normal variable with mean
25 minutes and standard deviation 3 minutes. The time spent at the chess club
is also modelled by a normal variable with mean 120 minutes and standard
deviation 10 minutes. Find the probability that on a certain evening Mr Fisher
is away from home for more than 212 hours.
The sums and differences of independent random variables
●
SOLUTION
Let the random variable X1 N(25, 32) represent the driving time, and the
random variable X2 N(120, 102) represent the time spent at the chess club.
Then the random variable T, where T = X1 + X2 N(145, ( 109)2), represents
his total time away.
So the probability that Mr Fisher is away for more than 212 hours (150 minutes) is
given by
P(T 150) = 1 – Φ
(
150 − 145
109
)
required
area
= 1 – Φ(0.479)
= 0.316.
145 150
X1 � X2
standard deviation � 109
Figure 11.3
265
Linear combinations of random variables
S2
11
EXAMPLE 11.6
In the manufacture of a bridge made entirely from wood, circular pegs have to
fit into circular holes. The diameters of the pegs are normally distributed with
mean 1.60 cm and standard deviation 0.01 cm, while the diameters of the holes
are normally distributed with mean 1.65 cm and standard deviation 0.02 cm.
What is the probability that a randomly chosen peg will not fit into a randomly
chosen hole?
SOLUTION
Let the random variable X be the diameter of a hole:
X N(1.65, 0.022) = N(1.65, 0.0004).
Let the random variable Y be the diameter of a peg:
Y N(1.60, 0.012) = N(1.6, 0.0001)
Let F = X – Y. F represents the gap remaining between the peg and the hole and
so the sign of F determines whether or not a peg will fit in a hole.
E(F ) = E(X) – E(Y ) = 1.65 – 1.60 = 0.05
Var(F ) = Var(X ) + Var(Y ) = 0.0004 + 0.0001 = 0.0005
F N(0.05, 0.0005)
If for any combination of peg and hole the value of F is negative, then the peg will
not fit into the hole.
The probability that F 0 is given by
Φ
(
)
required
area
0 − 0.05
= 1 – Φ(–2.236)
0.0005
= 1 – 0.9873
0
0.05
F
standard deviation � 0.0005
= 0.0127.
Figure 11.4
EXERCISE 11B
1
The menu at a café is shown below.
Main course
Fish and Chips
Spaghetti
Pizza
Steak and Chips
Dessert
$3
$3.50
$4
$5.50
Ice Cream
Apple Pie
Sponge Pudding
$1
$1.50
$2
The owner of the café says that all the main-course dishes sell equally well, as
do all the desserts, and that customers’ choice of dessert is not influenced by
the main course they have just eaten.
266
The variable M denotes the cost of the main course, in dollars, and the variable
D the cost of the dessert. The variable T denotes the total cost of a two-course
meal: T = M + D.
(i)
(iii)
(iv)
(v)
and
2
mean (T ) = mean (M) + mean (D)
variance (T ) = variance (M) + variance (D).
X1 and X2 are independent random variables with distributions N(50, 16) and
N(40, 9) respectively. Write down the distributions of
(i)
(ii)
(iii)
X1 + X2
X1 – X2
X2 – X1.
3
A play is enjoying a long run at a theatre. It is found that the play time may
be modelled as a normal variable with mean 130 minutes and standard
deviation 3 minutes, and that the length of the intermission in the middle of
the performance may be modelled by a normal variable with mean 15 minutes
and standard deviaton 5 minutes. Find the probability that the performance is
completed in less than 140 minutes.
4
The time Melanie spends on her history assignments may be modelled as being
normally distributed, with mean 40 minutes and standard deviation 10 minutes.
The times taken on assignments may be assumed to be independent. Find
(i)
(ii)
(iii)
5
Exercise 11B
(ii)
Find the mean and variance of M.
Find the mean and variance of D.
List all the possible two-course meals, giving the price for each one.
Use your answer to part (iii) to find the mean and variance of T.
Hence verify that for these figures
S2
11
the probability that a particular assignment will last longer than an hour
the time in which 95% of all assignments can be completed
the probability that two assignments will be completed in less than
75 minutes.
The weights of full cans of a particular brand of pet food may be taken to
be normally distributed, with mean 260 g and standard deviation 10 g. The
weights of the empty cans may be taken to be normally distributed, with mean
30 g and standard deviation 2 g. Find
(i)
(ii)
(iii)
the mean and standard deviation of the weights of the contents of the cans
the probability that a full can weighs more than 270 g
the probability that two full cans together weigh more than 540 g.
267
Linear combinations of random variables
S2
11
6
The independent random variables X1 and X2 are distributed as follows:
X1 ∼ N(30, 9);
X2 ∼ N(40, 16).
Find the distributions of the following :
(i)
(ii)
X1 + X2
X1 – X2.
7
In a vending machine the capacity of cups is normally distributed, with mean
200 cm3 and standard deviation 4 cm3. The volume of coffee discharged per
cup is normally distributed, with mean 190 cm3 and standard deviation
5 cm3. Find the percentage of drinks which overflow.
8
On a distant island the heights of adult men and women may both be taken
to be normally distributed, with means 173 cm and 165 cm and standard
deviations 10 cm and 8 cm respectively.
(i)
(ii)
9
Find the probability that a randomly chosen woman is taller than a
randomly chosen man.
Do you think that this is equivalent to the probability that a married
woman is taller than her husband?
The lifetimes of a certain brand of refrigerator are approximately normally
distributed, with mean 2000 days and standard deviation 250 days.
Mrs Chudasama and Mr Poole each buy one on the same date.
What is the probability that Mr Poole’s refrigerator is still working one year
after Mrs Chudasama’s refrigerator has broken down?
10
A random sample of size 2 is chosen from a normal distribution N(100, 25).
Find the probability that
(i)
(ii)
11
the sum of the sample numbers exceeds 215
the first observation is at least 19 more than the second observation.
A mathematics module is assessed by an examination and by coursework.
The examination makes up 75% of the total assessment and the coursework
makes up 25%. Examination marks, X, are distributed with mean 53.2 and
standard deviation 9.3. Coursework marks, Y, are distributed with mean 78.0
and standard deviation 5.1. Examination marks and coursework marks are
independent. Find the mean and standard deviation of the combined mark
0.75X + 0.25Y.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q2 June 2006]
12
The cost of electricity for a month in a certain town under scheme A consists
of a fixed charge of 600 cents together with a charge of 5.52 cents per unit
of electricity used. Stella uses scheme A. The number of units she uses in a
month is normally distributed with mean 500 and variance 50.41.
(i)
268
Find the mean and variance of the total cost of Stella’s electricity in a
randomly chosen month.
Under scheme B there is no fixed charge and the cost in cents for a month is
normally distributed with mean 6600 and variance 421. Derek uses scheme B.
(ii)
S2
11
Find the probability that, in a randomly chosen month, Derek spends
more than twice as much as Stella spends.
More than two independent random variables
The results on pages 264–265 may be generalised to give the mean and variance
of the sums and differences of n random variables, X1, X2, … , Xn .
●
E(X1 ± X2 ± … ± Xn ) = E(X1) ± E(X2) ± … ± E(Xn )
and, provided X1, X2, … , Xn are independent,
●
Var(X1 ± X2 ± … ± Xn ) = Var(X1) + Var(X2) + … + Var(Xn ).
If X1, X2, … , Xn is a set of normally distributed variables, then the distribution of
(X1 ± X2 ± … ± Xn ) is also normal.
EXAMPLE 11.7
More than two independent random variables
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q4 November 2007]
The mass, X, of a suitcase at an airport is modelled as being normally distributed,
with mean 15 kg and standard deviation 3 kg. Find the probability that a random
sample of ten suitcases weighs more than 154 kg.
SOLUTION
The mass X of one suitcase is given by
X N(15, 9).
Then the mass of each of the ten suitcases has the distribution of X; call them X1,
X2, … , X10.
Let the random variable T be the total weight of ten suitcases.
T = X1 + X2 + … + X10.
E(T ) = E(X1) + E(X2) + … + E(X10)
= 15 + 15 + … + 15
= 150
Var(T ) = Var(X1) + Var(X2) + … + Var(X10)
=9+9+…+9
= 90
Similarly
required
area
So T N(150, 90).
The probability that T exceeds 154 is given by
1–Φ
(
)
154 − 150
= 1 – Φ(0.422)
90
= 1 – 0.6635
= 0.3365.
150 154
standard deviation � 90
Figure 11.5
T
269
Linear combinations of random variables
S2
11
EXAMPLE 11.8
The running times of the four members of a 4 × 400 m relay team may all be
taken to be normally distributed, as follows.
Member
Mean time (s)
Standard deviation (s)
Adil
52
1
Ben
53
1
Colin
55
1.5
Dexter
51
0.5
Assuming that no time is lost during changeovers, find the probability that the
team finishes the race in less than 3 minutes 28 seconds.
SOLUTION
Let the total time be T.
E(T) = 52 + 53 + 55 + 51 = 211
Var(T) = 12 + 12 + 1.52 + 0.52 = 4.5
So T N(211, 4.5).
The probability of a total time of less than
3 minutes 28 seconds (208 seconds) is given by
Φ
(
)
208 − 211
= Φ(–1.414)
4.5
required
area
208
211
standard deviation � 4.5
= 1 – 0.9213
= 0.0787.
Figure 11.6
Linear combinations of two or more independent
random variables
The results given on pages 264–265 can also be generalised to include linear
combinations of random variables.
For any random variables X and Y,
●
E(aX + bY) = aE(X) + bE(Y),
where a and b are constants.
If X and Y are independent
●
Var(aX + bY) = a2 Var(X) + b2 Var(Y)
If the distributions of X and Y are normal, then the distribution of (aX + bY) is
also normal.
These results may be extended to any number of random variables.
270
T
EXAMPLE 11.9
What is the probability that a protecting strip 275 cm long will be too short for a
randomly selected work surface?
W
L
W
Figure 11.7
SOLUTION
S2
11
More than two independent random variables
In a workshop joiners cut out rectangular sheets of laminated board, of length
L cm and width W cm, to be made into work surfaces. Both L and W may be
taken to be normally distributed with standard deviation 1.5 cm. The mean of L is
150 cm, that of W is 60 cm, and the lengths L and W are independent. Both of the
short sides and one of the long sides have to be covered by a protective strip (the
other long side is to lie against a wall and so does not need protection).
Denoting the length and width by the independent random variables L and W
and the total length of strip required by T:
T = L + 2W
E(T) = E(L) + 2E(W)
= 150 + 2 × 60
= 270
Var(T) = Var(L) + 22 Var(W)
= 1.52 + 4 × 1.52
= 11.25
The probability of a strip 275 cm long being too short is given by
(
)
1 – Φ 275 − 270 = 1 – Φ(1.491)
11.25
= 1 – 0.932
= 0.068.
Note
You have to distinguish carefully between the random variable 2W, which means
twice the size of one observation of the random variable W, and the random variable
W1 + W2, which is the sum of two independent observations of the random variable W.
In the last example
and
In contrast,
and
E(2W ) = 2E(W ) = 120
Var(2W ) = 22Var(W ) = 4 × 2.25 = 9.
E(W1 + W2) = E(W1) + E(W2) = 60 + 60 = 120
Var(W1 + W2) = Var(W1) + Var(W2) = 2.25 + 2.25 = 4.5.
271
A machine produces sheets of paper the thicknesses of which are normally
distributed with mean 0.1 mm and standard deviation 0.006 mm.
(i)
Linear combinations of random variables
S2
11
EXAMPLE 11.10
(ii)
State the distribution of the total thickness of eight randomly selected sheets
of paper.
Single sheets of paper are folded three times (to give eight thicknesses). State
the distribution of the total thickness.
SOLUTION
Denote the thickness of one sheet (in mm) by the random variable W, and the
total thickness of eight sheets by T.
(i)
Eight separate sheets
In this situation T = W1 + W2 + W3 + W4 + W5 + W6 + W7 + W8
where W1, W2, …, W8 are eight independent observations of the variable W.
The distribution of W is normal with mean 0.1 and variance 0.0062.
So the distribution of T is normal with
mean = 0.1 + 0.1 + … + 0.1 = 8 × 0.1 = 0.8
variance = 0.0062 + 0.0062 + … + 0.0062
= 8 × 0.0062
= 0.000 288
standard deviation = 0.000 288 = 0.017.
The distribution is N(0.8, 0.0172).
(ii)
Eight thicknesses of the same sheet
In this situation T = W1 + W1 + W1 + W1 + W1 + W1 + W1 + W1 = 8W1
where W1 is a single observation of the variable W.
So the distribution of T is normal with
mean = 8 × E(W) = 0.8
variance = 82 × Var(W) = 82 × 0.0062 = 0.002 304
standard deviation = 0.002 304 = 0.048.
The distribution is N(0.8, 0.0482).
?
●
272
Notice that in both cases the mean thickness is the same but for the folded paper
the variance is greater. Why is this?
EXERCISE 11C
A garage offers motorists ‘Road worthiness tests While U Wait’ and claims that
an average test takes only 20 minutes. Assuming that the time taken can be
modelled as a normal variable with mean 20 minutes and standard deviation
2 minutes, find the distribution of the total time taken to conduct six tests in
succession at this garage. State any assumptions you make.
2
A company manufactures floor tiles of mean length 20 cm with standard
deviation 0.2 cm. Assuming the distribution of the lengths of the tiles is
normal, find the probability that, when 12 randomly selected floor tiles are
laid in a row, their total length exceeds 241 cm.
3
The masses of wedding cakes produced at a bakery are independent and
may be modelled as being normally distributed with mean 4 kg and standard
deviation 100 g. Find the probability that a set of eight wedding cakes has a
total mass between 32.3 kg and 32.7 kg.
4
A random sample of 15 items is chosen from a normal population with mean
30 and variance 9. Find the probability that the sum of the variables in the
sample is less than 440.
5
The distributions of four independent random variables X1, X2, X3 and X4 are
N(7, 9), N(8, 16), N(9, 4) and N(10, 1) respectively.
S2
11
Exercise 11C
1
Find the distributions of
(i)
(ii)
(iii)
6
The distributions of X and Y are N(100, 25) and N(110, 36), and X and Y are
independent. Find
(i)
(ii)
7
the probability that 8X + 2Y 1000
the probability that 8X – 2Y 600.
The distributions of the independent random variables A, B and C are
N(35, 9), N(30, 8) and N(35, 9). Write down the distributions of
(i)
(ii)
(iii)
(iv)
8
X1 + X2 + X3 + X4
X1 + X2 – X3 – X4
X1 – X2 – X3 + X4.
A+B+C
5A + 4B
A + 2B + 3C
4A – B – 5C.
The distributions of the independent random variables X and Y are N(60, 4)
and N(90, 9). Find the probability that
(i)
(ii)
(iii)
X – Y –35
3X + 5Y 638
3X 2Y.
273
Linear combinations of random variables
S2
11
9
If X N(60, 4) and Y N(90, 9) and X and Y are independent, find the
probability that
(i)
(ii)
(iii)
(iv)
10
The distribution of the weights of those rowing in a very large regatta may be
taken to be normal with mean 80 kg and standard deviation 8 kg.
(i)
(ii)
11
What total weight would you expect 70% of randomly chosen crews of
four rowers to exceed?
State what assumption you have made in answering this question and
comment on whether you consider it reasonable.
The quantity of fuel used by a coach on a return trip of 200 km is modelled as
a normal variable with mean 45 litres and standard deviation 1.5 litres.
(i)
(ii)
12
when one item is sampled from each population, the one from the Y
population is more than 35 greater than the one from the X population.
the sum of a sample consisting of three items from population X and five
items from population Y exceeds 638.
the sum of a sample of three items from population X exceeds that of two
items from population Y.
Comment on your answers to questions 8 and 9.
Find the probability that in nine return journeys the coach uses between
400 and 406 litres of fuel.
Find the volume of fuel which is 95% certain to be sufficient to cover the
total fuel requirements for two return journeys.
The weekly takings at three cinemas are modelled as independent normally
distributed random variables with means and standard deviations as shown
in the table, in $.
Mean
Standard deviation
Cinema A
6000
400
Cinema B
9000
800
Cinema C
5100
180
(i)
(ii)
(iii)
Find the probability that the weekly takings at cinema A will be less than
those at cinema C.
Find the probability that the weekly takings at cinema B will be at least
twice those at cinema C.
The parent company receives a weekly levy consisting of 12% of the
weekly takings at cinema A, 20% of those at cinema B and 8% of those at
cinema C. Find the probability that this levy exceeds $3000 in any given
week. Hence find the probability that in a 4-week period the weekly levy
exceeds $3000 at least twice.
[MEI, adapted]
274
13
Assume that the weights of men and women may be taken to be normally
distributed, men with mean 75 kg and standard deviation 4 kg, and women
with mean 65 kg and standard deviation 3 kg.
14
Exercise 11C
At a village fair, tug-of-war teams consisting of either five men or six
women are chosen at random. The competition is then run on a knock-out
basis, with teams drawn out of a hat. If in the first round a women’s team is
drawn against a men’s team, what is the probability that the women’s team
is the heavier? State any assumptions you have made and explain how they
can be justified.
S2
11
The four runners in a relay team have individual times, in seconds, which
are normally distributed, with means 12.1, 12.2, 12.3, 12.4, and standard
deviations 0.2, 0.25, 0.3, 0.35 respectively. Find the probability that, in a
randomly chosen race,
(i)
(ii)
the total time of the four runners is less than 48 seconds
runners 1 and 2 take longer than runners 3 and 4.
What assumption have you made and how realistic is the model?
15
Jim Longlegs is an athlete whose specialist event is the triple jump. This is
made up of a hop, a step and a jump. Over a season the lengths of the hop, step
and jump sections, denoted by H, S and J respectively, are measured, from
which the following models are proposed:
H N(5.5, 0.52) S N(5.1, 0.62) J N(6.2, 0.82)
where all distances are in metres. Assume that H, S and J are independent.
(i)
(ii)
(iii)
(iv)
In what proportion of his triple jumps will Jim’s total distance exceed
18 metres?
In six successive independent attempts, what is the probability that at
least one total distance will exceed 18 m?
What total distance will Jim exceed 95% of the time?
Find the probability that, in Jim’s next triple jump, his step will be
greater than his hop.
[MEI]
16
The random variable X has the distribution N(3.2, 1.22). The sum of
60 independent observations of X is denoted by S. Find P(S 200).
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q2 June 2007]
17
Weights of garden tables are normally distributed with mean 36 kg and
standard deviation 1.6 kg. Weights of garden chairs are normally distributed
with mean 7.3 kg and standard deviation 0.4 kg. Find the probability that the
total weight of 2 randomly chosen tables is more than the total weight of 10
randomly chosen chairs.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q3 November 2008]
275
18
Linear combinations of random variables
S2
11
A journey in a certain car consists of two stages with a stop for filling up
with fuel after the first stage. The length of time, T minutes, taken for each
stage has a normal distribution with mean 74 and standard deviation 7.3.
The length of time, F minutes, it takes to fill up with fuel has a normal
distribution with mean 5 and standard deviation 1.7. The length of time
it takes to pay for the fuel is exactly 4 minutes. The variables T and F are
independent and the times for the two stages are independent of each other.
(i)
(ii)
(iii)
Find the probability that the total time for the journey is less than
154 minutes.
A second car has a fuel tank with exactly twice the capacity of the first
car. Find the mean and variance of this car’s fuel fill-up time.
This second car’s time for each stage of the journey follows a normal
distribution with mean 69 minutes and standard deviation 5.2 minutes.
The length of time it takes to pay for the fuel for this car is also exactly
4 minutes. Find the probability that the total time for the journey taken
by the first car is more than the total time taken by the second car.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q7 November 2005]
KEY POINTS
1
2
For any discrete random variable X and constants a and c :
●
E(c) = c
●
E(aX) = aE(X)
●
E(aX + c) = aE(X) + c
●
Var(c) = 0
●
Var(aX) = a2 Var(X)
●
Var(aX + c) = a2 Var(X).
For two random variables X and Y, whether independent or not, and
constants a and b,
●
E(X ± Y) = E(X) ± E(Y)
●
E(aX + bY ) = aE(X ) + bE(Y)
and, if X and Y are independent,
3
●
Var(X ± Y) = Var(X) + Var(Y)
●
Var(aX ± bY) = a2 Var(X) + b 2 Var(Y).
For a set of n random variables, X1, X2, …, Xn ,
●
E(X1 ± X2 ± … ± Xn) = E(X1) ± E(X2) ± … ± E(Xn )
and, if the variables are independent,
●
4
276
Var(X1 ± X2 ± … ± Xn) = Var(X1) + Var(X2) + … + Var(Xn).
If random variables are normally distributed so are the sums, differences
and other linear combinations of them.
12
Sampling
S2
12
G. Polya
PoliticsNow.com
Independent set to become member of parliament
Terms and notation
If you wish to learn swimming you have to go into the water.
Next week’s local election looks set to produce the
first independent member of parliament for many
years, according to an opinion poll conducted by the
PoliticsNow.com team.
When 30 potential voters were asked who they
thought would make the best member of parliament,
12 opted for Independent candidate Mrs Chalashika.
The other three candidates attracted between 3 and
9 votes.
Mrs Grace Chalashika is
taking the polls by storm.
Assuming that the figures quoted in the article are true, does this really mean that
Independent Mrs Chalashika will be elected to Parliament next week?
Only time will tell that, but meanwhile the newspaper report raises a number of
questions that should make you suspicious of its conclusion.
Was the sample large enough? Thirty seems a very small number.
Were those interviewed asked the right question? They were asked who they thought
would make the best member of parliament, not who they intended to vote for.
How was the sample selected? Was it representative of the whole electorate?
Before addressing these questions you will find it helpful to be familiar with the
language and notation associated with sampling.
Terms and notation
PoliticsNow.com took a sample of size 30. Taking samples and interpreting them
is an essential part of statistics. The populations in which you are interested are
often so large that it would be quite impractical to use every item; the electorate
for that area might well number 70 000.
A sample provides a set of data values of a random variable, drawn from all such
possible values, the parent population. The parent population can be finite, such
as all professional footballers, or infinite, such as the points where a dart can land
on a dart board.
277
A representation of the items available to be sampled is called the sampling
frame. This could, for example, be a list of the sheep in a flock, a map marked
with a grid or an electoral register. In many situations no sampling frame exists
nor is it possible to devise one, for example, for the cod in the North Atlantic.
The proportion of the available items that are actually sampled is called the
sampling fraction.
Sampling
S2
12
A parent population, often just called the population, is described in terms of its
parameters, such as its mean, µ, and variance, σ2. By convention Greek letters are
used to denote these parameters.
A value derived from a sample is written in Roman letters: mean, x–, variance, s2,
etc. Such a number is the value of a sample statistic (or just statistic).When sample
statistics are used to estimate the parent population parameters they are called
estimates.
Thus if you take a random sample in which the mean is x–, you can use x– to
estimate the parent mean, µ. If in a particular sample x– = 23.4, then you can use
23.4 as an estimate of the population mean. The true value of µ will generally be
somewhat different from your estimated value.
Upper case letters, X, Y, etc., are used to represent the random variables, and
lower case letters, x, y, etc., to denote particular values of them. In the example
of PoliticsNow.com’s survey of voters, you could define X to be the percentage of
voters, in a sample of size 30, showing support for Mrs Chalashika. The particular
value from this sample, x , is 12
× 100 = 40%.
30
( )
Sampling
There are essentially two reasons why you might wish to take a sample:
●
●
to estimate the values of the parameters of the parent population
to conduct a hypothesis test.
There are many ways you can interpret data. First you will consider how sample
data are collected and the steps you can take to ensure their quality.
An estimate of a parameter derived from sample data will in general differ from
its true value. The difference is called the sampling error. To reduce the sampling
error, you want your sample to be as representative of the parent population as
you can make it. This, however, may be easier said than done.
Here are a number of questions that you should ask yourself when about to take
a sample.
1 Are the data relevant?
278
It is a common mistake to replace what you need to measure by something else
for which data are more easily obtained.
2 Are the data likely to be biased?
S2
12
Sampling
You must ensure that your data are relevant, giving values of whatever it is that
you really want to measure. This was clearly not the case in the example of the
PoliticsNow.com survey, where the question people were asked, ‘Who would make
the best member of parliament?’, was not the one whose answer was required. The
question should have been ‘Which person do you intend to vote for?’.
Bias is a systematic error. If, for example, you wished to estimate the mean time
of young women running 100 metres and did so by timing the members of a
hockey team over that distance, your result would be biased. The hockey players
would be fitter and more athletic than most young women and so your estimate
for the time would be too low.
You must try to avoid bias in the selection of your sample.
3 Does the method of collection distort the data?
The process of collecting data must not interfere with the data. It is, for example,
very easy when designing a questionnaire to frame questions in such a way as to
lead people into making certain responses. ‘Are you a law-abiding citizen?’ and
‘Do you consider your driving to be above average?’ are both questions inviting
the answer ‘Yes’.
In the case of collecting information on voting intentions another problem arises.
Where people put the cross on their ballot papers is secret and so people are
being asked to give away private information. There may well be those who find
this offensive and react by deliberately giving false answers.
People often give the answer they think the questioner wants to receive.
4 Is the right person collecting the data?
Bias can be introduced by the choice of those taking the sample. For example, a
school’s authorities want to estimate the proportion of the students who smoke,
which is against the school rules. Each class teacher is told to ask five students
whether they smoke. Almost certainly some smokers will say ‘No’ to their
teacher for fear of getting into trouble, even though they might say ‘Yes’ to a
different person.
5 Is the sample large enough?
The sample must be sufficiently large for the results to have some meaning. In
this case the intention was to look for differences of support between the four
candidates and for that a sample of 30 is totally inadequate. For opinion polls, a
sample size of about 1000 is common.
279
The sample size depends on the precision required in the results. For example, in
the opinion polls for elections a much larger sample is required if you want the
estimate to be reliable to within 1% than if 5% will do.
S2
12
Sampling
6 Is the sampling procedure appropriate in the circumstances?
The method of choosing the sample must be appropriate. Suppose, for example,
that you were carrying out the survey of people’s voting intentions in the
forthcoming election for PoliticsNow.com. How would you select the sample of
people you are going to ask?
If you stood in the town centre in the middle of one morning and asked
passers-by, you would probably get an unduly high proportion of those who,
for one reason or another, were not employed. It is quite possible that this
group has different voting intentions from those in work.
If you selected names from the telephone directory, you would automatically
exclude those who do not have telephones, those who do not have landlines and
those who are ex-directory.
It is actually very difficult to come up with a plan which will yield a fair sample,
one that is not biased in some direction or another. There are, however, a
number of established sampling techniques and these are described in the next
section of this chapter.
?
●
Each of the situations below involves a population and a sample. In each case
identify both, briefly but precisely.
1
A member of parliament is interested in whether her constituents support
proposed legislation to make convicted drug dealers do hard physical work
every day while they are in prison. Her staff report that letters on the proposed
legislation have been received from 361 constituents of whom 309 support it.
2
A flour company wants to know what proportion of households in Karachi
bake some or all of their own bread. A sample of 500 residential addresses in
Karachi is taken and interviewers are sent to these addresses. The interviewers
are employed during regular working hours on weekdays and interview only
during these hours.
3
The Chicago Police Department wants to know how black residents of
Chicago feel about police service. A questionnaire with several questions about
the police is prepared. A sample of 300 postal addresses in predominantly
black areas of Chicago is taken and a police officer is sent to each address to
administer the questionnaire to an adult living there.
Each sampling situation contains a serious source of probable bias. In each case
give the reason that bias may occur and also the direction of the bias.
280
Sampling techniques
●
sample size
Sampling fraction = –––––––––––––––
population size
S2
12
Sampling techniques
In considering the following techniques it is worth repeating that a key aim when
taking a sample is to obtain a sample that is representative of the parent population
being investigated. It is assumed that the sampling is done without replacement,
otherwise, for example, one person could give an opinion twice, or more. The
fraction of the population which is selected is called the sampling fraction.
Random sampling
In a random sampling procedure, every member of the population may be
selected; there is a non-zero probability of this happening (and, of course, the
probability is less than 1). In many random sampling procedures, for example,
drawing names out of a hat, every member of the population has an equal
probability of being selected.
In a simple random sampling procedure, every possible sample of a given size is
equally likely to be selected. It follows that in such a procedure every member of
the parent population is equally likely to be selected. However, the converse is
not true. It is possible to devise a sampling procedure in which every member is
equally likely to be selected but some samples are not permissible.
?
●
1
A school has 20 classes, each with 30 students. One student is chosen at
random from each class, giving a sample size of 20. Why is this not a simple
random sampling procedure?
2
If you write the name of each student in the school on a slip of paper, put all
the slips in a box, shake it well and then take out 20, would this be a simple
random sample?
Simple random sampling is fine when you can do it, but you must have a
sampling frame. The selection of items within the frame is often done using
tables of random numbers.
Using random numbers
Usually, each element in the frame is given a number, starting at 1. You then
select elements for the sample using random number tables or the random
number generator on a calculator or computer.
Suppose that you need to select a sample of 15 houses from a numbered list of
483 houses. Using random number tables, you choose a random starting position
and take the digits in groups of three. If the first set of three digits is 247, you put
281
house number 247 from the list into your sample. If the next number is 832, you
ignore it because it does not correspond to a house in the list. You continue in
this way until you have a sample of 15 houses. (If any number occurs more than
once, you still only include it once in your sample.)
Sampling
S2
12
In some circumstances, you might choose to assign random numbers in a less
wasteful way. For example, you could subtract 500 from any random numbers
above 500, so instead of discarding 832 you would choose house (832 – 500) = 332.
Whether this is worthwhile depends on the sample size and the method being used
to link the numbers to the elements in the sampling frame.
When using a random number generator on a calculator, you use the same
procedure. If the calculator only provides three digits and you need five, you can
generate two sets of three digits and discard the last digit.
ACTIVITY 12.1
Using the random numbers below, which items would you choose from a
numbered list of the 17 841 inhabitants of a town if you want a random sample of
size 10? Start with the top left random number and work along each row in order.
54
12
45
85
51
25
91
60
66
16
32
98
63
95
55
87
35
71
26
46
71
65
88
82
88
83
37
56
95
04
14
35
98
94
19
50
36
59
82
35
91
22
89
71
36
80
48
45
45
44
27
07
17
16
48
45
92
57
02
65
77
59
94
08
12
43
77
33
53
21
38
44
47
43
14
63
40
43
34
37
Other sampling techniques
There are many other sampling techniques. Survey design, the formulation of the
most appropriate sampling procedure in a particular situation, is a major topic
within statistics.
Stratified sampling
282
You have already thought about the difficulty of conducting a survey of people’s
voting intentions in a particular area before an election. In that situation it is
possible to identify a number of different sub-groups which you might expect
to have different voting patterns: low, medium and high income groups; urban,
suburban and rural dwellers; young, middle-aged and elderly voters; men and
women; and so on. The sub-groups are called strata. In stratified sampling, you
would ensure that all strata were sampled. You would need to sample from high
income, suburban, elderly women; medium income, rural young men; etc. In this
example, 54 strata (3 × 3 × 3 × 2) have been identified. If the numbers sampled in
the various strata are proportional to the size of their populations, the procedure
is called proportional stratified sampling. If the sampling is not proportional, then
appropriate weighting has to be used.
The selection of the items to be sampled within each stratum is usually done by
simple random sampling. Stratified sampling will usually lead to more accurate
results about the entire population, and will also give useful information about
the individual strata.
Cluster sampling also starts with sub-groups, or strata, of the population, but
in this case the items are chosen from one or several of the sub-groups. The
sub-groups are now called clusters. It is important that each cluster should be
reasonably representative of the entire population. If, for example, you were
asked to investigate the incidence of a particular parasite in the puffin population
of Northern Europe, it would be impossible to use simple random sampling.
Rather you would select a number of sites and then catch some puffins at each
place. This is cluster sampling. Instead of selecting from the whole population
you are choosing from a limited number of clusters.
Exercise 12A
Cluster sampling
S2
12
Systematic sampling
Systematic sampling is a method of choosing individuals from a sampling frame.
If you were surveying telephone subscribers, you might select a number at
random, say 66, and then sample the 66th name on every page of the directory.
If the items in the sampling frame are numbered 1, 2, 3, ..., you might choose a
random starting point like 38 and then sample numbers 38, 138, 238 and so on.
When using systematic sampling you have to beware of any cyclic patterns within
the frame. For example, suppose a school list is made up class by class, each of
exactly 25 children, in order of merit, so that numbers 1, 26, 51, 76, 101, ..., in the
frame are those at the top of their class. If you sample every 50th child starting
with number 26, you will conclude that the children in the school are very bright.
Quota sampling
Quota sampling is the method often used by companies employing people to
carry out opinion surveys. An interviewer’s quota is always specified in stratified
terms: how many males and how many females, etc. The choice of who is
sampled is then left up to the interviewer and so is definitely non-random.
EXERCISE 12A
1
Alan wishes to choose one child at random from the eleven children in his
music class. The children are numbered 2, 3, 4, and so on, up to 12. Alan then
throws two fair dice, each numbered from 1 to 6, and chooses the child whose
number is the sum of the scores on the two dice.
(i)
(ii)
Explain why this is an unsatisfactory method of choosing a child.
Describe briefly a satisfactory method of choosing a child.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q1 November 2008]
283
2
S2
12
Identify the sampling procedures that would be appropriate in the following
situations.
Sampling
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
A local education officer wishes to estimate the mean number of
children per family on a large housing estate.
A consumer protection body wishes to estimate the proportion of
trains that are running late.
A marketing consultant wishes to investigate the proportion of
households in a town that have a personal computer.
A local politician wishes to carry out a survey into people’s views on
capital punishment within your area.
A health inspector wishes to investigate what proportion of people wear
spectacles.
Ministry officials wish to estimate the proportion of cars with bald tyres.
A television company wishes to estimate the proportion of householders
who have not paid their television licence fee.
The police want to find out how fast cars travel in the outside lane of a
motorway.
A sociologist wants to know how many girlfriends the average
18-year-old boy has had.
The headteacher of a large school wishes to estimate the average number
of hours of homework done per week by the students.
KEY POINTS
1
2
3
There are essentially two reasons why you might wish to take a sample:
●
to estimate the values of the parameters of the parent population
●
to conduct a hypothesis test.
When taking a sample you should ensure that:
●
the data are relevant
●
the data are unbiased
●
the data are not distorted by the act of collection
●
a suitable person is collecting the data
●
the sample is of a suitable size
●
a suitable sampling procedure is being followed.
In a random sample, every member of the population has a non-zero
probability of being selected. In many random sampling procedures, every
member of the population has an equal probability of being selected.
In simple random sampling, every possible sample of a given size has an
equal probability of being selected.
284
Other sampling procedures include stratified sampling, cluster sampling,
systematic sampling and quota sampling.
When we spend money on testing an item, we are buying confidence in
its performance.
Tony Cutler
Interpreting sample data using the normal distribution
Sydney set to become greenhouse?
from our Science Correspondent Ama Williams
On a recent visit to a college in Sydney, I was
intrigued to find experiments being conducted
to measure the level of carbon dioxide in the
air we are all breathing. Readers will of course
know that high levels of carbon dioxide are
associated with the greenhouse effect.
Lecturer Ray Peng showed me round his
laboratory. ‘It is delicate work, measuring parts
per million, but I am trying to establish what is
the normal level in this area. Yesterday we took
ten readings and you can see the results for
yourself: 336, 334, 332, 332, 331, 331, 330,
330, 328, 326.’
When I commented that there seemed to be
a lot of variation between the readings, Ray
assured me that that was quite in order.
‘I have taken hundreds of these
measurements in the past,’ he said. ‘There is always a standard deviation
of 2.5. That’s just natural variation.’
I suggested to Ray that his students should test whether these results are
significantly above the accepted value of 328 parts per million. Certainly they
made me feel uneasy. Is the greenhouse effect starting here in Australia?
S2
13
Interpreting sample data using the normal distribution
13
Hypothesis testing and
confidence intervals using
the normal distribution
Ray Peng has been trying to establish the carbon dioxide level in Sydney. How do
you interpret his figures? Do you think the correspondent has a point when she
says she is worried that the greenhouse effect is already happening in Australia?
If suitable sampling procedures have not been used, then the resulting data may
be worthless, indeed positively misleading. You may wonder if that is the case
with Ray’s figures, and about the accuracy of his analysis of the samples too.
His data are used in subsequent working in this chapter, but you may well feel
there is something of a question mark hanging over them. You should always be
prepared to treat data with a healthy degree of caution.
285
Estimating the population mean, µ
Ray’s data were as follows.
336, 334, 332, 332, 331, 331, 330, 330, 328, 326.
His intention in collecting them was to estimate the mean of the parent population,
the population mean.
The mean of these figures, the sample mean, is given by
x = 336 + 334 + 332 + 332 + 331 + 331 + 330 + 330 + 328 + 326
10
= 331.
What does this tell you about the population mean, µ?
It tells you that it is about 331 but it certainly does not tell you that it is definitely
and exactly 331. If Ray took another sample, its mean would probably not be 331
but you would be surprised (and suspicious) if it were very far away from it. If
he took lots of samples, all of size 10, you would expect their means to be close
together but certainly not all the same.
If you took 1000 such samples, each of size 10, the distribution of their means
might look like figure 13.1. You will notice that this distribution looks rather like
the normal distribution and so may well wonder if this is indeed the case.
frequency density
Hypothesis testing and confidence intervals using the normal distribution
S2
13
Putting aside any concerns about the quality of the data, what conclusions can
you draw from them?
329
286
Figure 13.1
330
331
332
sample mean (parts per million)
333
The distribution of sample means
sampling distribution of the means, or often just the sampling distribution, and is
σ 2
denoted by N µ, . This is illustrated in figure 13.2. It is a special case of the
n
Central Limit Theorem which you will meet later in this chapter, on page 298.
distribution of sample means
2
N µ, σn
( )
distribution of individual
(
items N µ, σ2
)
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13
Interpreting sample data using the normal distribution
In this chapter, it is assumed that the underlying population has a normal
distribution with mean µ and standard deviation σ so it can be denoted by
N(µ, σ2). In that case the distribution of the means of samples is indeed
σ
. This is called the
normal; its mean is µ and its standard deviation is
n
µ
Figure 13.2
A hypothesis test for the mean using the normal distribution
If your intention in collecting sample data is to test a theory, then you should set
up a hypothesis test.
Ray Peng was mainly interested in establishing data on carbon dioxide levels for
Sydney. The correspondent, however, wanted to know whether levels were above
normal, and so she could have set up and conducted a test.
Here is the relevant information, given in a more condensed format.
EXAMPLE 13.1
Ama Williams believes that the carbon dioxide level in Sydney has risen above
the usual level of 328 parts per million. A sample of 10 specimens of Sydney air
are collected and the carbon dioxide level within them is determined. The results
are as follows.
336, 334, 332, 332, 331, 331, 330, 330, 328, 326.
Extensive previous research has shown that the standard deviation of the levels
within such samples is 2.5, and that the distribution may be assumed to be normal.
Use these data to test, at the 0.1% significance level, Ama’s belief that the level of
carbon dioxide at Sydney is above normal.
287
Hypothesis testing and confidence intervals using the normal distribution
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13
SOLUTION
As usual with hypothesis tests, you use the distribution of the statistic you are
measuring, in this case the normal distribution of the sample means, to decide
which values of the test statistic are sufficiently extreme as to suggest that the
alternative hypothesis, not the null hypothesis, is true.
Null hypothesis, H0:
µ = 328
The level of carbon dioxide in
Sydney is normal.
Alternative hypothesis, H1:
µ > 328
The level of carbon dioxide in
Sydney is above normal.
One-tail test
The significance level is 0.1%. This is the probability of a Type I error for this test.
Method 1: Using critical regions
2
Since the distribution of sample means is N µ, σ , critical values for a test on
n
the sample mean are given by µ ± k ×
σ
.
n
In this case, if H0 is true, µ = 328; σ = 2.5; n = 10.
The test is one-tail, for µ > 328, so only the right-hand tail applies. This gives a
value of k = 3.090 since normal distribution tables give Φ(3.090) = 0.999 and so
1 − Φ(3.090) = 0.001.
The critical value is thus 328 + 3.09 ×
2.5
= 330.4, as shown in figure 13.3.
10
critical value
330.4
328
331
x = 331
Figure 13.3
However, the sample mean x = 331, and 331 > 330.4.
Therefore the sample mean lies within the critical region, and so the null
hypothesis is rejected in favour of the alternative hypothesis: that the mean
carbon dioxide level is above 328, at the 0.1% significance level.
288
Method 2: Using probabilities
σ 2
The distribution of sample means, X, is N µ, .
n
According to the null hypothesis, µ = 328 and it is known that σ = 2.5 and n = 10.
This area represents
the probability of
a result at least as
extreme as that found.
328
µ
331
x
= 331
Figure 13.4
The probability of the mean, X, of a randomly chosen sample being greater than
the value found, i.e. 331, is given by
331 − 328
P(X 331) = 1 − Φ
2.5
10
Interpreting sample data using the normal distribution
2
So this distribution is N 328, 2.5 ; see figure 13.4.
10
S2
13
The figure 0.999 93
comes from normal
distribution tables for
suitable values of z.
= 1 − Φ(3.79)
= 1 − 0.999 93 = 0.000 07
Since 0.000 07 < 0.001, the required significance level (0.1%), the null hypothesis
is rejected in favour of the alternative hypothesis.
Method 3: Using critical ratios
observed value – expected value
.
standard deviation
z = 331 − 328 = 3.79
In this case
2.5
10
This is now compared with the critical value for z given in your tables.
The critical ratio is given by z =
p
0.75
0.90
0.95
0.975
0.99
0.995
0.9975
0.999
0.9995
z
0.674
1.282
1.645
1.960
2.326
2.576
2.807
3.090
3.291
Figure 13.5 Critical values for the normal distribution
So the critical value is z = 3.090.
Since 3.79 > 3.09, H0 is rejected.
289
Notes
Hypothesis testing and confidence intervals using the normal distribution
S2
13
1 A hypothesis test should be formulated before the data are collected and not
after. If sample data lead you to form a hypothesis, then you should plan a
suitable test and collect further data on which to conduct it. It is not clear whether
or not the test in the previous example was being carried out on the same data
which were used to formulate the hypothesis.
2 If the data were not collected properly, any test carried out on them may be
worthless.
EXAMPLE 13.2
Observations over a long period of time have shown that the mass of adult males
of a type of bat is normally distributed with mean 110 g and standard deviation
10 g. A scientist has a theory that in one area these bats are becoming smaller,
possibly as an adaptation to changes in their environment. He plans to trap 20
adult male bats, weigh them and then release them. He will then use the data to
carry out a suitable hypothesis test at the 5% significance level.
(i)
(ii)
(iii)
State the null and alternative hypotheses.
Find the critical value for the test.
Find the probability of a Type I error.
In fact the mean mass of the bats has reduced to 108 g but the standard deviation
has remained unaltered.
(iv)
Calculate the probability that the test will produce a Type II error.
The mean mass of the scientist’s sample of bats is 107 g.
(v)
Carry out the hypothesis test and state what type of error, if any, results.
SOLUTION
(i)
(ii)
The hypotheses are:
Null hypothesis H0:
µ = 110
Alternative hypothesis H1:
µ 110
The mean mass of the bats is still
110 g.
The mean mass of the bats is less
than 110 g.
This is a one-tail test at the 5% significance level so the critical value is:
X = 110 – 1.645 × 10 = 106.3 to 1 d.p.
20
where X is the sample mean.
The null hypothesis will be rejected if X 106.3.
290
acceptance
region
(0.95)
106.3
critical value
110
X
Figure 13.6
(iii)
A Type I error occurs when a true null hypothesis is rejected. In this case,
the probability of this happening is represented by the dark pink area in
figure 13.6. It is just the same as the significance level of the test and so is
5% or 0.05.
(iv)
A Type II error will occur if X 106.3 because in that case the null
hypothesis, which is false, will be accepted.
In fact µ = 108 and so the probability that X 106.3 is given by
106
3
108
.
–
= 0.77 (to 2 s.f.)
1– Φ
10
20
(v)
Interpreting sample data using the normal distribution
rejection
region
(0.05)
S2
13
Since 107 106.32, the null hypothesis is accepted. The evidence does not
support the scientist’s theory.
However, this is the wrong result so a Type II error has occurred. The answer
from part (iv) shows that with the test set up as it was, a Type II error is quite
likely to occur.
Known and estimated standard deviation
Notice that you can only use this method of hypothesis testing if you already
know the value of the standard deviation of the parent population, σ. Ray Peng
had said that from taking hundreds of measurements he knew it to be 2.5.
It is more often the situation that you do not know the population standard
deviation or variance and so have to estimate it from your sample data. For
such estimates the estimated standard deviation, s, is worked out using slightly
differently formulae from those you met in Chapter 1. In certain places n – 1 is
used instead of n.
To calculate an unbiased estimate of the population mean and variance from a
sample you should use the following formulae for these estimators:
291
Hypothesis testing and confidence intervals using the normal distribution
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13
Estimated mean,
x =
∑x
n
Estimated variance, s 2 =
2
x
1
2 − (∑ )
x
n − 1 ∑
n
An alternative notation to s for the estimated standard deviation, which is
sometimes used, is σ̂.
EXAMPLE 13.3
An IQ test, established some years ago, was designed to have a mean score of 100.
A researcher puts forward a theory that people are becoming more intelligent
(as measured by this particular test). She selects a random sample of 150 people,
all of whom take the test. The results of the tests, where x represents the score
obtained, are n = 150, Σx = 15 483, Σx 2 = 1 631 680.
Carry out a suitable hypothesis test on the researcher’s theory, at the 1%
significance level. You may assume that the test scores are normally distributed.
SOLUTION
H0: The parent population mean is unchanged, i.e. µ = 100.
H1: The parent population mean has increased, i.e. µ > 100.
One-tail test
The significance level is 1%. This is the probability of Type I error for this test.
From the sample, unbiased estimates for the mean and standard deviation are:
x =
s2 =
So
∑ x = 15483 = 103.22
n
150
2
x
1 1631680 − 154832 = 224.998…
1
2 – (∑ )
x
=
∑
150
n – 1
n 149
s = 15.0 (to 3 s.f.)
The standardised z value corresponding to x = 103.22 is calculated using µ = 100
and approximating σ by s = 15.0.
z=
x − µ 103.22 − 100
=
= 2.629
15
σ
150
n
For the 1% significance level, the critical value is z = 2.326.
The test statistic is compared with the critical value and since 2.629 > 2.326 the
null hypothesis is rejected.
292
The evidence supports the view that scores on this IQ test are now higher; see
figure 13.7.
Exercise 13A
critical value
2.326
critical region
standardised value
0
actual value
100
µ
2.326
2.629
z
103.22
x
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13
test statistic
2.629
Figure 13.7
EXERCISE 13A
1
A magazine conducted a survey about the sleeping time of adults. A random
sample of 12 adults was chosen from the adults travelling to work on a train.
(i)
(ii)
Give a reason why this is an unsatisfactory sample for the purposes of the
survey.
State a population for which this sample would be satisfactory.
A satisfactory sample of 12 adults gave numbers of hours of sleep as shown
below.
4.6 6.8 5.2 6.2 5.7 7.1 6.3 5.6 7.0 5.8 6.5 7.2
(iii)
Calculate unbiased estimates of the mean and variance of the sleeping
times of adults.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q1 June 2008]
2
For each of the following, the random variable X N(µ, σ2), with known
standard deviation. A random sample of size n is taken from the parent
population and the sample mean, x , is calculated.
Carry out hypotheses tests, given H0 and H1, at the significance level indicated.
σ
n
x
H0
H1
Sig. level
(i)
8
6
195
µ = 190
µ > 190
5%
(ii)
10
10
47.5
µ = 55
µ < 55
1%
(iii)
15
25
104.7
µ = 100
µ ≠ 100
10%
(iv)
4.3
15
34.5
µ = 32
µ > 32
2%
(v)
40
12
345
µ = 370
µ ≠ 370
5%
293
Hypothesis testing and confidence intervals using the normal distribution
S2
13
3
A machine is designed to make paperclips with mean mass 4.00 g and standard
deviation 0.08 g. The distribution of the masses of the paperclips is normal. Find
(i)
(ii)
(iii)
A quality control officer weighs a random sample of 25 paperclips and finds
their total mass to be 101.2 g.
(iv)
4
Conduct a hypothesis test at the 5% significance level of whether this
provides evidence of an increase in the mean mass of the paperclips. State
your null and alternative hypotheses clearly.
It is known that the mass of a certain type of lizard has a normal distribution
with mean 72.7 g and standard deviation 4.8 g. A zoologist finds a colony of
lizards in a remote place and is not sure whether they are of the same type. In
order to test this, she collects a sample of 12 lizards and weighs them, with the
following results.
80.4
(i)
(ii)
(iii)
5
67.2
74.9
78.8
76.5
75.5
80.2
81.9
79.3
70.0
69.2
69.1
Write down, in precise form, the zoologist’s null and alternative
hypotheses, and state whether a one-tail or two-tail test is appropriate.
Carry out the test at the 5% significance level and write down your
conclusion.
Would your conclusion have been the same at the 10% significance level?
Observations over a long period of time have shown that the mid-day
temperature at a particular place during the month of June is normally
distributed with a mean value of 23.9 °C with standard deviation 2.3 °C. An
ecologist sets up an experiment to collect data for a hypothesis test of whether
the climate is getting hotter. She selects at random 20 June days over a five-year
period and records the mid-day temperature. Her results (in °C) are as follows.
(i)
(ii)
(iii)
294
the probability that an individual paperclip, chosen at random, has mass
greater than 4.04 g
the standard error of the mass for random samples of 25 paperclips
the probability that the mean mass of a random sample of 25 paperclips is
greater than 4.04 g.
20.1
26.2
23.3
28.9
30.4
28.4
17.3
22.7
25.1
24.2
15.4
26.3
19.3
24.0
19.9
30.3
32.1
26.7
27.6
23.1
State the null and alternative hypotheses that the ecologist should use.
Carry out the test at the 10% significance level and state the conclusion.
Calculate an unbiased estimate of the population variance and comment
on it.
6
(i)
(ii)
(iii)
(iv)
7
35
21
12
7
2
1.5
1.5
1
18
20
16
11
8
8
9
0.25
5
11
28
35 35
16
0.25
0.25
15
17
17 35
35
4
0.25
2
0.5
0.5
1
1
S2
13
Exercise 13A
The keepers of a lighthouse were required to keep records of weather
conditions. Analysis of their data from many years showed the visibility at
mid-day to have a mean value of 14 nautical miles with standard deviation
5.4 nautical miles. A new keeper decided he would test his theory that the air
had become less clear (and so visibility reduced) by carrying out a hypothesis
test on data collected for his first 36 days on duty. His figures (in nautical
miles) were as follows.
Write down a distributional assumption for the test to be valid.
Write down suitable null and alternative hypotheses.
Carry out the test at the 2.5% significance level and state the conclusion
that the lighthouse keeper would have come to.
Criticise the sampling procedure used by the keeper and suggest a better one.
A chemical is packed into bags by a machine. The mean weight of the bags is
controlled by the machine operator, but the standard deviation is fixed at 0.96
kg. The mean weight should be 50 kg, but it is suspected that the machine has
been set to give underweight bags. If a random sample of 36 bags has a total
weight of 1789.20 kg, is there evidence to support the suspicion? (You must
state the null and alternative hypotheses and you may assume that the weights
of the bags are normally distributed.)
[MEI]
8
Archaeologists have discovered that all skulls found in excavated sites in
a certain country belong either to racial group A or to racial group B. The
mean lengths of skulls from group A and group B are 190 mm and 196 mm
respectively. The standard deviation for each group is 8 mm, and skull lengths
are distributed normally and independently.
A new excavation produced 12 skulls of mean length x and there is reason to
believe that all these skulls belong to group A. It is required to test this belief
statistically with the null hypothesis (H0) that all the skulls belong to group A
and the alternative hypothesis (H1) that all the skulls belong to group B.
(i)
(ii)
State the distribution of the mean length of 12 skulls when H0 is true.
Explain why a test of H0 versus H1 should take the form:
‘Reject H0 if x > c ’,
(iii)
(iv)
where c is some critical value.
Calculate this critical value c to the nearest 0.1 mm when the probability of
rejecting H0 when it is in fact true is chosen to be 0.05.
Perform the test, given that the lengths (in mm) of the 12 skulls are as
follows.
204.1 201.1 187.4 196.4 202.5 185.0
192.6 181.6
194.5
183.2
200.3
295
202.9
[MEI]
Hypothesis testing and confidence intervals using the normal distribution
S2
13
9
The packaging on a type of electric light bulb states that the average lifetime
of the bulbs is 1000 hours. A consumer association thinks that this is an
overestimate and tests a random sample of 64 bulbs, recording the lifetime,
x hours, of each bulb. You may assume that the distribution of the bulbs’
lifetimes is normal.
The results are summarised as follows.
n = 64,
(i)
(ii)
(iii)
10
11
Σx = 63 910.4,
Σx 2 = 63 824 061
Calculate unbiased estimates for the population mean and variance.
State suitable null and alternative hypotheses to test whether the statement
on the packaging is overestimating the lifetime of this type of bulb.
Carry out the test, at the 5% significance level, stating your conclusions
carefully.
A sample of 40 observations from a normal distribution X gave Σx = 24 and
Σx 2 = 596. Performing a two-tail test at the 5% level, test whether the mean
of the distribution is zero.
A random sample of 75 eleven-year-olds performed a simple task and
the time taken, t minutes, was noted for each. You may assume that the
distribution of these times is normal.
The results are summarised as follows.
n = 75,
(i)
(ii)
(iii)
12
Σt = 1215,
Σt 2 = 21 708
Calculate unbiased estimates for the population mean and variance.
State suitable null and alternative hypotheses to test whether there is
evidence that the mean time taken to perform this task is greater than
15 minutes.
Carry out the test, at the 1% significance level, stating your conclusions
carefully.
Bags of sugar are supposed to contain, on average, 2 kg of sugar. A quality
controller suspects that they actually contain less than this amount, and so
90 bags are taken at random and the mass, x kg, of sugar in each is measured.
You may assume that the distribution of these masses is normal.
The results are summarised as follows.
n = 90,
(i)
(ii)
(iii)
296
Σx = 177.9,
Σx 2 = 353.1916
Calculate unbiased estimates for the population mean and variance.
State suitable null and alternative hypotheses to test whether there is any
evidence that the sugar is being sold ‘underweight’.
Carry out the test, at the 2% significance level, stating your conclusions
carefully.
13
A machine produces jars of skin cream, filled to a nominal volume of 100 ml.
The machine is actually supposed to be set to 105 ml, to ensure that most jars
actually contain more than the nominal volume of 100 ml. You may assume
that the distribution of the volume of skin cream in a jar is normal.
The results are summarised as follows.
n = 80,
(i)
(ii)
(iii)
14
Σx = 8376,
Σx2 = 877 687
Exercise 13A
To check that the machine is correctly set, 80 jars are chosen at random, and
the volume, x ml, of skin cream in each is measured.
S2
13
Calculate unbiased estimates for the population mean and standard
deviation.
State suitable null and alternative hypotheses for a test to see whether the
machine appears to be set correctly.
Carry out the test, at the 10% significance level, stating your conclusions
carefully.
A study of a large sample of books by a particular author shows that the
number of words per sentence can be modelled by a normal distribution with
mean 21.2 and standard deviation 7.3. A researcher claims to have discovered
a previously unknown book by this author. The mean length of 90 sentences
chosen at random in this book is found to be 19.4 words.
(i)
(ii)
Assuming the population standard deviation of sentence lengths in this
book is also 7.3, test at the 5% level of significance whether the mean
sentence length is the same as the author’s. State your null and alternative
hypotheses.
State in words relating to the context of the test what is meant by a Type I
error and state the probability of a Type I error in the test in part (i).
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q4 June 2005]
15
The number of cars caught speeding on a certain length of motorway is
7.2 per day, on average. Speed cameras are introduced and the results shown
in the following table are those from a random selection of 40 days after this.
Number of cars caught speeding
4
5
6
7
8
9
10
Number of days
5
7
8
10
5
2
3
(i)
(ii)
(iii)
Calculate unbiased estimates of the population mean and variance of the
number of cars per day caught speeding after the speed cameras were
introduced.
Taking the null hypothesis H0 to be µ = 7.2, test at the 5% level whether
there is evidence that the introduction of speed cameras has resulted in a
reduction in the number of cars caught speeding.
State what is meant by Type I error in words relating to the context of
the test in part (ii). Without further calculation, illustrate on a suitable
diagram the region representing the probability of this Type I error.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q7 June 2006]
297
Hypothesis testing and confidence intervals using the normal distribution
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13
16
A machine has produced nails over a long period of time, where the length
in millimetres was distributed as N(22.0, 0.19). It is believed that recently
the mean length has changed. To test this belief a random sample of 8 nails
is taken and the mean length is found to be 21.7 mm. Carry out a hypothesis
test at the 5% significance level to test whether the population mean has
changed, assuming that the variance remains the same.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q3 June 2007]
17
In summer the growth rate of grass in a lawn has a normal distribution with
mean 3.2 cm per week and standard deviation 1.4 cm per week. A new type
of grass is introduced which the manufacturer claims has a slower growth
rate. A hypothesis test of this claim at the 5% significance level was carried
out using a random sample of 10 lawns that had the new grass. It may be
assumed that the growth rate of the new grass has a normal distribution with
standard deviation 1.4 cm per week.
(i)
(ii)
Find the rejection region for the test.
The probability of making a Type II error when the actual value of the
mean growth rate of the new grass is m cm per week is less than 0.5. Use
your answer to part (i) to write down an inequality for m.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q2 November 2007]
The Central Limit Theorem
organicveg.com
The perfect apple grower
Fruit buyer Tom Sisulu writes:
Fruit grower, Rose Ncune, believes that, after years of trials, she has developed trees that
will produce the perfect supermarket apple. ‘There are two requirements’ Rose told me. ‘The
average weight of an apple should be 100 grams and they should all be nearly the same size. I
have measured hundreds of mine and the standard deviation is a mere 5 grams.’
Rose invited me to take any ten apples off the shelf and weigh them for myself. It was
quite uncanny; they were all so close to the magic 100 grams: 98, 107, 105, 98, 100, 99, 104, 93,
105, 103.
Rose is calling her apple the ‘Cape Pippin’.
What can you conclude from the weights of Tom’s sample of ten apples?
298
Before going any further, it is appropriate to question whether his sample was
random. Rose invited Tom to ‘take any ten apples off the shelf’. That is not
necessarily the same as taking any ten off the tree. The apples on the shelf could
all have been specially selected to impress him. So what follows is based on the
assumption that Rose has been honest and the ten apples really do constitute a
random sample.
The sample mean is
x = 98 + 107 + 105 + 98 + 100 + 99 + 104 + 93 + 105 + 103 = 101.2
10
What does that tell you about the population mean, µ?
To estimate how far the value of µ is from 101.2, you need to know something
about the spread of the data; the usual measure is the standard deviation, σ. In
the blog for organicveg.com you are told that σ = 5.
The result that if repeated samples of size n are drawn from a population with a
normal distribution with mean µ and standard deviation σ, the distribution of
the sample means is also normal; its mean is µ and its standard deviation is σ is
n
proved in Appendix 6 on the CD.
The Central Limit Theorem
?
●
S2
13
This is actually a special case of a more general result called the Central Limit
Theorem. The Central Limit Theorem covers the case where samples are drawn
from a population which is not necessarily normal.
●
The Central Limit Theorem states that for samples of size n drawn from any
distribution with mean µ and finite variance σ2, the distribution of the
σ 2
sample means is approximately N µ, n for sufficiently large n.
This theorem is fundamental to much of statistics and so it is worth pausing to
make sure you understand just what it is saying.
It deals with the distribution of sample means. This is called the sampling
distribution (or more correctly the sampling distribution of the means). There are
three aspects to it.
1
2
The mean of the sample means is µ, the population mean of the original
distribution. That is not a particularly surprising result but it is extremely
important.
σ
. This is often called the
The standard deviation of the sample means is
n
standard error of the mean.
Within a sample you would expect some values above the population mean,
others below it, so that overall the deviations would tend to cancel each other
out, and the larger the sample the more this would be the case. Consequently
the standard deviation of the sample means is smaller than that of individual
items, by a factor of n .
3
The distribution of sample means is approximately normal.
299
Hypothesis testing and confidence intervals using the normal distribution
S2
13
This last point is the most surprising part of the theorem. Even if the underlying
parent distribution is not normal, the distribution of the means of samples of
a particular size drawn from it is approximately normal. The larger the sample
size, n, the closer this distribution is to the normal. For any given value of n the
sampling distribution will be closest to normal where the parent distribution is
not unlike the normal.
In many cases the value of n does not need to be particularly large. For most
parent distributions you can rely on the distribution of sample means being
normal if n is about 20 or 25 (or more).
distribution of sample means
2
N µ, σn
(
)
distribution of individual
items; it has mean µ and
standard deviation σ but
in this case is not normal
µ
Figure 13.8
Confidence intervals
Returning to the figures on the Cape Pippin apples, you would estimate the
population mean to be the same as the sample mean, namely 101.2.
You can express this by saying that you estimate µ to lie within a range of values,
an interval, centred on 101.2:
101.2 − a bit < µ < 101.2 + a bit.
Such an interval is called a confidence interval.
Imagine you take a large number of samples and use a formula to work out
the interval for each of them. If you catch the true population mean in 90% of
your intervals, the confidence interval is called a 90% confidence interval. Other
percentages are also used and the confidence intervals are named accordingly.
The width of the interval is clearly twice the ‘bit’.
Finding a confidence interval involves a very simple calculation but the reasoning
behind it is somewhat subtle and requires clear thinking. It is explained in the
next section, but you may prefer to make your first reading of it a light one. You
should, however, come back to it at some point; otherwise you will not really
understand the meaning of confidence intervals.
300
The theory of confidence intervals
It is now that the strength of the Central Limit Theorem becomes apparent. This
states that the distribution of the means of samples of size n drawn from this
σ
.
population is approximately normal with mean µ and standard deviation
n
S2
13
Confidence intervals
To understand confidence intervals you need to look not at the particular sample
whose mean you have just found, but at the parent population from which it was
drawn. For the data on the Cape Pippin apples this does not look very promising.
All you know about it is its standard deviation σ (in this case 5). You do not
know its mean, µ, which you are trying to estimate, or even its shape.
In figure 13.9 the central 90% region has been shaded leaving the two 5% tails,
corresponding to z values of ±1.645, unshaded. So if you take a large number of
samples, all of size n, and work out the sample mean x for each one, you would
expect that in 90% of cases the value of x would lie in the shaded region between
A and B.
5%
5%
.05
A
.05
µ
B
standard deviation nσ
µ − 1.645σ
n
x
µ + 1.645σ
n
Figure 13.9
For such a value of x to be in the shaded region
it must be to the right of A:
it must be to the left of B:
σ
n
σ
x < µ + 1.645
n
x > µ − 1.645
➀
➁
Rearranging these two inequalities:
σ
σ
➀
> µ or µ < x + 1.645
x + 1.645
n
n
σ
➁
<µ
x − 1.645
n
Putting them together gives the result that in 90% of cases
x − 1.645
σ
σ
< µ < x + 1.645
n
n
and this is the 90% confidence interval for µ.
301
Hypothesis testing and confidence intervals using the normal distribution
S2
13
The numbers corresponding to the points A and B are called the 90% confidence
limits and 90% is the confidence level. If you want a different confidence level, you
use a different z value from 1.645.
This number is often denoted by k ; commonly used values are:
Confidence level
90%
95%
99%
k
1.645
1.960
2.576
and the confidence interval is given by
x − k σ to x + k σ .
n
n
Note
Notice that this is a two-sided symmetrical confidence interval for the mean, µ.
Confidence intervals do not need to be symmetrical and can be one-sided. The term
‘confidence interval’ is a general one, applying not just to the mean but to other
population parameters, like variance and skewness, as well. All these cases,
however, are outside the scope of this book.
The P % confidence interval for the mean is an interval constructed from sample
data in such a way that P % of such intervals will include the true population
mean. Figure 13.10 shows a number of confidence intervals constructed from
different samples, one of which fails to catch the population mean.
x
µ
Figure 13.10
In the case of the data on the Cape Pippin apples,
x = 101.2,
302
σ = 5,
n = 10
and so the 90% confidence interval is
5
101.2 − 1.645 ×
to
10
98.6
to
101.2 + 1.645 ×
103.8.
5
10
Known and estimated standard deviation
Notice that you can only use this procedure if you already know the value of the
standard deviation of the parent population, σ. In this example, Rose Ncune said
that she knew, from hundreds of measurements of her apples, that its value is 5.
Confidence intervals
It is more often the situation that you do not know the population standard
deviation or variance, and have to estimate it from your sample data. If that is
the case, the procedure is different in that you use the t distribution rather than
the normal provided that the parent population is normally distributed, and this
results in different values of k. However, the use of the t distribution is beyond
the scope of this book.
S2
13
However, if the sample is large, for example over 50, confidence intervals worked
out using the normal distribution will be reasonably accurate even though the
standard deviation used is an estimate from the sample. So it is quite acceptable
to use the normal distribution for large samples whether the standard deviation is
known or not.
EXPERIMENTS
These experiments are designed to help you understand confidence intervals,
rather than to teach you anything new about dice.
When a single die is thrown, the possible outcomes, 1, 2, 3, 4, 5, 6, are all equally
1
likely with probability 6 . Consequently the expectation or mean score from
throwing a die is
1
1
1
µ = 1 × 6 + 2 × 6 + … + 6 × 6 = 3.5.
Similarly the standard deviation is
σ = (12 × 16 + 22 × 16 + … + 62 × 16) – 3.52 = 1.708.
Imagine that you know σ but don’t know µ and wish to construct a 90%
confidence interval for it.
Converging confidence intervals
Start by throwing a die once. Suppose you get a 5. You have a sample of size 1,
namely {5}, which you could use to work out a sort of 90% confidence interval
(but see the warning overleaf).
This confidence interval is given by
5 − 1.645 × 1.708 to 5 + 1.645 × 1.708
1
1
2.19 to 7.81.
303
!
Hypothesis testing and confidence intervals using the normal distribution
S2
13
So far the procedure is not valid. The sample is small and the underlying
distribution is not normal. However, things will get better. The more times you
throw the die, the larger the sample size and so the more justifiable the procedure.
Now throw the die again. Suppose this time you get a 3. You now have a sample of
size 2, namely {5, 3}, with mean 4, and can work out another confidence interval.
The confidence interval is given by
4 − 1.645 × 1.708 to 4 + 1.645 × 1.708
2
2
2.79 to 5.21.
Now throw the die again and find a third confidence interval, and a fourth, fifth
and so on. You should find them converging on the population mean of 3.5;
but it may take some time to get close, particularly if you start with, say, two 6s.
This demonstrates that, the larger the sample you take, the narrower the range of
values within the confidence interval.
Catching the population mean
Organise a group of friends to throw five dice (or one die five times), and do this
100 times. Each of these gives a sample of size 5 and so you can use it to work out
a 90% confidence interval for µ.
You know that the real value of µ is 3.5 and it should be that this is caught within
90% of 90% confidence intervals.
?
●
Out of your 100 confidence intervals, how many actually enclose 3.5?
How large a sample do you need?
You are now in a position to start to answer the question of how large a sample
needs to be. The answer, as you will see in Example 13.4, depends on the
precision you require, and the confidence level you are prepared to accept.
304
EXAMPLE 13.4
A trading standards officer is investigating complaints that a coal merchant is
giving short measure. Each sack should contain 25 kg but some variation will
inevitably occur because of the size of the lumps of coal; the officer knows from
experience that the standard deviation should be 1.5 kg.
SOLUTION
The 95% confidence interval for the mean is given by
x − 1.96 σ
to
x + 1.96 σ
n
n
and so, since σ = 1.5, the inspector’s requirement is that
⇒
How large a sample do you need?
The officer plans to take, secretly, a random sample of n sacks, find the total
weight of the coal inside them and thereby estimate the mean weight of the coal
per sack. He wants to present this figure correct to the nearest kilogram with 95%
confidence. What value of n should he choose?
S2
13
1.96 × 1.5 0.5
n
1.96 × 1.5 n
0.5
⇒
n 34.57.
So the inspector needs to take 35 sacks.
Large samples
Given that the width of a confidence interval decreases with sample size, why is it
not standard practice to take very large samples?
The answer is that the cost and time involved have to be balanced against the
quality of information produced. Because the width of a confidence interval
depends on 1 and not on 1 , increasing the sample size does not produce a
n
n
proportional reduction in the width of the interval. You have, for example, to
increase the sample size by a factor of 4 to halve the width of the interval. In the
previous example the inspector had to weigh 35 sacks of coal to achieve a class
interval of 2 × 0.5 = 1 kg with 95% confidence. That is already quite a daunting
task; does the benefit from reducing the interval to 0.5 kg justify the time, cost
and trouble involved in weighing another 105 sacks?
305
Confidence intervals for a proportion
In this chapter you have seen how to calculate confidence intervals for the
population mean. Confidence intervals can also be found for other population
parameters, like the variance or, in a binomial situation, the proportion of the
population with a particular characteristic.
Hypothesis testing and confidence intervals using the normal distribution
S2
13
In Example 13.5 a confidence interval for a population proportion is found. The
method assumes that the sample taken is large, and so the normal approximation
to the binomial distribution may be used.
In an experiment where a (large) sample of size n has been taken and m items are
found to have the characteristic under investigation, the population proportion is
estimated to be p = m .
n
For all samples of size n, the following estimates may then be made:
Mean number of occurrences per sample = np
● Variance of number of occurrences per sample = npq = np(1 – p)
np(1 − p) p(1 – p)
=
● Variance of estimated proportion =
n
n2
p(1 − p)
● Standard deviation of estimated proportion =
n
So the confidence interval for the proportion is given by
●
p−k
p(1 – p)
p(1 – p)
< population proportion < p + k
,
n
n
where the values of k are taken from the normal distribution: 1.96 for a 95%
(two-sided) confidence interval, 1.645 for a 90% interval, etc.
ExAMPLE 13.5
A certain type of moth is found in two colours, brown and white. In an
experiment, 100 moths from a particular region are captured. 30 of them are
found to be brown, the remainder white.
Calculate the 95% confidence interval for the population proportion of brown
moths.
SOLUTION
30
Estimated population proportion of brown moths, p = 100
= 0.3.
So the 95% confidence interval is given by
0.3 – 1.96
0.3(1 – 0.3) < population
population proportion
proportion < 0.3 + 1.96
100
0.3(1 – 0.3)
100
giving
0.210 < population proportion < 0.390 (to 3 s.f.).
306
EXERCISE 13B
1
A biologist studying a colony of beetles selects and weighs a random sample
of 20 adult males. She knows that, because of natural variability, the weights
of such beetles are normally distributed with standard deviation 0.2 g. Their
weights, in grams, are as follows.
(ii)
2
5.4
4.9
5.0
4.8
5.7
5.2
5.2
5.4
5.1
5.6
5.0
5.2
5.1
5.3
5.2
5.1
5.3
5.2
5.2
Find the mean weight of the beetles in this sample.
Find 95% confidence limits for the mean weight of such beetles.
Exercise 13B
(i)
5.2
S2
13
An aptitude test for deep-sea divers has been designed to produce scores which
are approximately normally distributed on a scale from 0 to 100 with standard
deviation 25. The scores from a random sample of people taking the test were
as follows.
23 35
89
35
12
45
60
78
34
66
Find the mean score of the people in this sample.
(ii) Construct a 90% confidence interval for the mean score of people taking
the test.
(iii) Construct a 99% confidence interval for the mean score of people taking the
test. Compare this confidence interval with the 90% confidence interval.
In a large city the distribution of incomes per family has a standard deviation
of $5200.
(i)
3
(i)
(ii)
For a random sample of 400 families, what is the probability that the
sample mean income per family is within $500 of the actual mean income
per family?
Given that the sample mean income was, in fact, $8300, calculate a 95%
confidence interval for the actual mean income per family.
[MEI, adapted]
4
A manufacturer of women’s clothing wants to know the mean height of the
women in a town (in order to plan what proportion of garments should be of
each size). She knows that the standard deviation of their heights is 5 cm. She
selects a random sample of 50 women from the town and finds their mean
height to be 165.2 cm.
(i)
(ii)
(iii)
Use the available information to estimate the proportion of women in the
town who were
(a) over 170 cm tall
(b) less than 155 cm tall.
Construct a 95% confidence interval for the mean height of women in the
town.
Another manufacturer in the same town wants to know the mean height
of women in the town to within 0.5 cm with 95% confidence. What is the
minimum sample size that would ensure this?
307
Hypothesis testing and confidence intervals using the normal distribution
S2
13
5
An examination question, marked out of 10, is answered by a very large
number of candidates. A random sample of 400 scripts are taken and the
marks on this question are recorded.
Mark
0
1
2
3
4
5
6
7
8
9
10
Frequency
12
35
11
12
3
20
57
87
20
14
129
(i)
(ii)
6
An archaeologist discovers a short manuscript in an ancient language which
he recognises but cannot read. There are 30 words in the manuscript and they
contain a total of 198 letters. There are two written versions of the language.
In the early form of the language the mean word length is 6.2 letters with
standard deviation 2.5; in the late form words were given prefixes, raising the
mean length to 7.6 letters but leaving the standard deviation unaltered. The
archaeologist hopes the manuscript will help him to date the site.
(i)
(ii)
7
Calculate the sample mean and the sample standard deviation.
Assuming that the population standard deviation has the same value
as the sample standard deviation, find 90% confidence limits for the
population mean.
Construct a 95% confidence interval for the mean word length of the
language.
What advice would you give the archaeologist?
A football boot manufacturer did extensive testing on the wear of the front studs
of its Supa range. It found that, after 30 hours’ use, the wear (i.e. the amount by
which the length was reduced) was normally distributed with standard deviation
1.3 mm. However, the mean wear on the studs of the boot on the dominant foot
of the player was 4 mm more than on the studs of the other boot.
(i)
Using the manufacturer’s figure, find the standard deviation of the
differences in wear between a pair of boots after 30 hours’ use.
The coach of a football team accepted the claim for the standard deviation
but was suspicious of the claim about the mean difference. He chose ten of his
squad at random. He fitted them with new boots and measured the wear after
30 hours of use with the following results.
1
2
3
4
5
6
7
8
9
10
Dominant foot
6.5
8.3
4.5
6.7
9.2
5.3
7.6
8.1
9.0
8.4
Other foot
4.2
4.6
2.3
3.8
7.0
4.7
1.4
3.8
8.4
5.7
Player
(ii)
(iii)
308
Using the value found in part (i) for the population standard deviation of
the differences, calculate 95% confidence limits for the mean difference in
wear based on the sample data.
Use these limits to explain whether or not you consider the coach’s
suspicions were justified.
8
S2
13
Exercise 13B
A school decided to introduce a new P.E. programme for its new students
to try to improve the fitness of the students. In order to see whether the
programme was effective, several tests were done. For one of these, the
students were timed on a run of 1 kilometre in their first week in the school
and again ten weeks later. A random sample of 100 of the students did both
runs. The differences of their mean times, subtracting the time of the second
run from that of the first, were calculated. The mean and standard deviation
were found to be 0.75 minutes and 1.62 minutes respectively.
Calculate a 90% confidence interval for the population mean difference. You
may assume that the differences are distributed normally. What assumption
have you made in finding this confidence interval?
The organiser of the programme considers that it should lead to an
improvement of at least half a minute in the average times. Explain whether
or not this aim has been achieved.
9
In an experiment to see if reaction times were affected by whether or not
individuals are hungry, 2000 randomly chosen soldiers were tested before
and after they had eaten a substantial lunch. The test used was to drop a
metre rule, which was held vertically so that its lower end was level with
the thumb and first finger of each person, and to measure how far the rule
fell before it was caught. For each person, the difference, d, of the distance
measured after lunch minus the distance measured before lunch was found.
From these it was calculated that Σd = 1626 and Σd 2 = 258 632.
Use these data to provide a 98% confidence interval for the population mean
difference, stating any assumptions you have made.
What does your confidence interval suggest about reaction times before and
after a meal?
10
The weights in grams of oranges grown in a certain area are normally
distributed with mean µ and standard deviation σ. A random sample of 50 of
these oranges was taken, and a 97% confidence interval for µ based on this
sample was (222.1, 232.1).
(i)
(ii)
Calculate unbiased estimates of µ and σ2.
Estimate the sample size that would be required in order for a 97%
confidence interval for µ to have width 8.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q2 June 2009]
11 (i)
(ii)
Give a reason why, in carrying out out a statistical investigation, a sample
rather than a complete population may be used.
Rose wishes to investigate whether men in her town have a different
life-span from the national average of 71.2 years. She looks at
government records for her town and takes a random sample of the ages
of 110 men who have died recently. Their mean age in years was 69.3 and
the unbiased estimate of the population variance was 65.61.
309
(a)
Hypothesis testing and confidence intervals using the normal distribution
S2
13
(b)
Calculate a 90% confidence interval for the population mean and
explain what you understand by this confidence interval.
State with a reason what conclusion about the life-span of men in her
town Rose could draw from this confidence interval.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q4 November 2005]
12
Diameters of golf balls are known to be normally distributed with mean µ cm
and standard deviation σ cm. A random sample of 130 golf balls was taken
and the diameters, x cm, were measured. The results are summarised by
Σx = 555.1 and Σx 2 = 2371.30.
(i)
(ii)
(iii)
Calculate unbiased estimates of µ and σ2.
Calculate a 97% confidence interval for µ.
300 random samples of 130 golf balls are taken and a 97% confidence
interval is calculated for each sample. How many of these intervals would
you expect not to contain µ?
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q4 November 2008]
13
A random sample of n people were questioned about their internet use.
87 of them had a high-speed internet connection. A confidence interval
for the population proportion having a high-speed internet connection is
0.1129 < p < 0.1771.
(i)
(ii)
Write down the mid-point of this confidence interval and hence find the
value of n.
This interval is an α% confidence interval. Find α.
[Cambridge International AS and A Level Mathematics 9709, Paper 71 Q2 June 2010]
14 (i)
Explain what is meant by the term ‘random sample’.
In a random sample of 350 food shops it was found that 130 of them had
Special Offers.
(ii)
(iii)
Calculate an approximate 95% confidence interval for the proportion of
all food shops with Special Offers.
Estimate the size of a random sample required for an approximate 95%
confidence interval for this proportion to have a width of 0.04.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q3 November 2007]
15
A survey of a random sample of n people found that 61 of them read The
Reporter newspaper. A symmetric confidence interval for the true population
proportion, p, who read The Reporter is 0.1993 < p < 0.2887.
(i)
(ii)
Find the mid-point of this confidence interval and use this to find the
value of n.
Find the confidence level of this confidence interval.
[Cambridge International AS and A Level Mathematics 9709, Paper 7 Q3 June 2005]
310
KEY POINTS
1
Distribution of sample means
●
2
2
and variance σn , i.e. x N µ, σn .
● The standard error of the mean (i.e. the standard deviation of the sample
σ
means) is given by
.
n
Hypothesis testing
●
●
Key points
2
For samples of size n drawn from a normal distribution with mean µ and
finite variance σ2, the distribution of sample means is normal with mean µ
S2
13
Sample data may be used to carry out a hypothesis test on the null
hypothesis that the population mean has some particular value, µ0,
i.e. H0: µ = µ0.
x − µ0
The test statistic is z = σ
and the normal distribution is used.
n
For situations where the population mean, µ, is unknown but the population
variance, σ2 (or standard deviation, σ), is known:
3
The Central Limit Theorem
For samples of size n drawn from any distribution with mean µ and finite
variance σ2, the distribution of the sample means is approximately
σ 2
N µ, n for sufficiently large n.
4
The standard error of the mean
The standard error of the mean (i.e. the standard deviation of the sample
means) is given by σ .
n
5
Confidence intervals
●
Two-sided confidence intervals for µ are given by
x −k σ
n
●
to
x +k σ .
n
The value of k for any confidence level can be found using normal
distribution tables.
Confidence level
90%
95%
99%
k
1.645
1.960
2.576
311
Answers
S1
Answers
Neither University of Cambridge International Examinations nor OCR bear any responsibility for the example
answers to questions taken from their past question papers which are contained in this publication.
Chapter 1
?
●
Squeezed
The data is rounded to one decimal place, so 3.02
becomes 3.0 etc.
(Page 3)
The editor has explained clearly why the investigation
is worth doing: there is growing concern about cycling
accidents involving children.
Good quality data is data that best represents the
research topic: in this case it is to establish whether or
not the number of accidents is significant.
?
●
(Page 4)
Another thing he might consider is to investigate
accidents in a similar community in order to be able to
make comparisons.
(Page 7)
Not all the branches have leaves. However, all the
branches must be shown in order to show correctly the
shape of the distribution.
?
●
(Page 9)
If the basic stem-and-leaf diagram has too many lines,
you may squeeze it as shown below. In doing this you
lose some of the information but you should get a better
idea of the shape of the distribution.
Unsqueezed
312
30
31
32
33
34
35
36
37
38
39
40
41
42
43
2
4
0
3
3
0
0
0
3
0
2
0
30 2 represents 3.02
6
3* 0 1 1
3t
2 3 3
3f
4 5 5 5 5 5 5
Zeros and ones go in this line. The
numbers were 3.02, 3.06, 3.14
Twos and threes go in this line.
Fours and fives go in this line.
3s 6 6 6 6 6 6 7 7 7 7 7 7
The reporter is focusing on two aspects of the
investigation: he is looking at cycling accidents in the
area over a period of time and he is considering the
distribution of ages of accident victims. Both of these
sources are relevant to the investigation.
?
●
3* 0 represents 3.0
3
8 8 8 8 9 9 9 9
Sixes and sevens go
in this line.
Eights and nines go in this line.
4* 0 1 1 1 1 1 1
4t 3 3 3
?
●
(Page 9)
Positive and negative data can be represented on a stemand-leaf diagram in the following way.
Data set:
–36 –32 –28 –25 –24 –20 –18 –15
–12 –6 5
8
12 13 18 26
n = 16
–3 2 represents –32
–3
–2
–1
–0
0
1
2
6
8
8
6
5
2
6
2
5 4 0
5 2
8
3 8
5
6
3
0
1
3
0
7
4
4
1
3
4
Exercise 1A (page 10)
4 8
4 4
3 3 4 8
5
0 1 1 4 4
0 2 4
1
3.27, 3.32, 3.36, 3.43, 3.45, 3.49, 3.50, 3.52, 3.56, 3.56,
3.58, 3.61, 3.61, 3.64, 3.72
2
0.083, 0.086, 0.087, 0.090, 0.091, 0.094, 0.098, 0.102,
0.103, 0.105, 0.108, 0.109, 0.109, 0.110, 0.111, 0.114,
0.123, 0.125, 0.131
3
n = 13
21
2 represents 0.212
5
2
3
0
1
3
2
The stretched stem-and-leaf diagram shows a
clear bimodal spread to the distribution. The
first peak (20s) may indicate first marriages
and the second peak (40s) may indicate second
marriages.
6
3 7
2 8
3 9
8 (i)
n = 10
780
1 represents 78.01
780
790
800
810
820
1
4
4
3
0
(ii)
7 (i)
(ii)
(iii)
2
6
8
7 9
5
(ii)
9
8 represents 28 years of age
0
1
2
3
4
5
6
7
8
5
9
2
2
1
2
0
6
3
4
5
1
6
3
5
9
2
8 8 9 9
4 5 5 5 6 6 7 7 7 8 8 8 9 9
5 6
3 6
1
The distribution has positive skew.
83 years
1*
1
2*
2
3*
3
4*
4
5*
5
6*
6
7*
7
8*
6
0
0
3
1
8
1
0
3
2
9
1
0
4
2
9
1 2 3 4 5 6 6 7 8 8 9
1 2 2 3 5 9
5 5 6 6 7 8 9
7
3
6
0
5
0
5
3
5
1
7
3
8
1
6
0
9
3
5
2
1 represents 21 °C
9
1
6
0
9
1 2 3 4
7 8 8 9
1 2 2 3
4
6 6 7 8 9
2
5
9
1
0
1
2
4
7
2
1
3
5
1
4
3
4
6
2
3 4 7 8 8
4 5 8 9 9
7 9
The distribution is approximately symmetrical
and unimodal.
1
2
S1
n = 87
n = 40
1
2
3
4
5
6
7
8
n = 30
–1
–0
0
1
2
3
0.013, 0.089, 1.79, 3.43, 3.51, 3.57, 3.59, 3.60, 3.64,
3.66, 3.68, 3.71, 3.71, 3.73, 3.78, 3.79, 3.80, 3.85, 3.94,
7.45, 10.87
6 (i)
The stem-and-leaf diagram with steps of 10
suggests a slight positive skew.
Chapter 1
21
22
23
24
25
26
4
(iv)
8 represents 18 marks
1
2
3
4
5
6
7
8
9
10
7
2
0
0
0
0
0
0
0
0
8
5
0
0
1
1
0
2
0
0
6
2
3
1
2
1
7
1
6
4
4
1
4
3
7
2
9
4
4
2
5
4
5
4
2
6
4
6
5
3
6
5
7
5
4
6
5
7
6
4
6
6
7
8
4
7
6
9
9
5
8
7
9
9
6 7 9
8 9 9
4 5 5 6 7 7 8
The distribution is symmetrical apart from a peak
in the 90s. There is a large concentration of marks
between 30 and 80.
10 (i)
n = 40
1
9 represents 19 years of age
1
2
3
4
5
6
7
7
0
0
0
7
5
2
(ii)
The distribution is bimodal. This is possibly
because those who hang-glide are the reasonably
young and active (average age about 25 years)
and those who are retired and have taken it up as
a hobby (average age about 60).
9
1
0
5
8
5
9 9
1 2 2 3 3 3 4 5 6 6 6 6 8 8 8 9
1 4 5 7
8
5 6 7 9
313
Answers
S1
n = 50
11
(iii) (a)
Untreated
8
764444222
99522
32
88532
86532
32
8
1
2
1
1
1
1
5
0
0
0
0
1
1
7 5 means 57 kg (untreated)
Treated
5 0
0 1 889
0 2 013334455567
0 3 001123333555888
0 4 112244479
0 5 0233449
1 6 1123
GRO seems to have improved the yield of lime trees,
though there are a significant number of untreated
trees that are matching the yield of the treated trees.
11
12
13
14
15
16
4
1
0
0
4
5
5
4
0
1
(b)
1.46, 1.18
(iii)
1.30
(iv)
The median 1.30 is close to the mean value of this
range (1.32) so the median seems a reasonable
estimate of the length.
(Page 18)
The median as it is not affected by the extreme values.
Exercise 1B (page 18)
1 (i)
bimodal, 116 and 132, mean = 122.5,
median = 122
(iii)
mode = 6, mean = 5.3, median = 6
(a)
(ii) (a)
(b)
314
Small data set so mode is inappropriate. You
would expect all the students in one class to
be uniformly spread between 14 years
0 months and 15 years, so any of the other
measures would be acceptable.
mode = 0, mean = 52.8, median = 58
The median. Small sample makes the mode
unreliable and the mean is influenced by
extreme values.
7
4
8
0
8
5
(ii)
13
11
12
13
14
15
16
2 means 13.7 minutes
and 13.2 minutes
9-year-olds
0
2
0
0
2 7
4
1 9
1 4 7
15.6 minutes
Exercise 1C (Page 22)
1 (i)
mode = 2
(ii)
median = 3
(iii)
mean = 3.24
2 (i)
mode = 39
(ii)
median = 40
(iii)
mean = 40.3
(iv)
The sample has a slight positive skew. Any of the
measures would do; however, the mean allows
one to calculate the total number of matches.
3 (i)
mode = 19
(ii)
median = 18
(iii)
mean = 17.9
(iv)
The outliers affect the mean. As the distribution
is, apart from the extremes, reasonably
symmetrical, the median or mode are acceptable.
The median is the safest for a relatively small
data set.
median = 14 years 6 months
(b)
Anything but the mode will do. The
distribution, uniform in theory, means that
mean = median. This sample reflects that well.
7
9
7
mode = 14 years 8 months,
mean = 14 years 5.5 months,
no unique mode, mean = 3.45, median = 3.5
16-year-olds
mode = 45, mean = 39.6, median = 41
(ii)
Small sample so the mode is inappropriate;
the mean is affected by outliers, so the
median is the best choice.
3 (i)
8
6 6 8 9 9
0 0 0 2 2 3 3
2 6
(ii)
2 (i)
(iv) (a)
n = 25
11 4 represents 1.14 metres
12 (i)
?
●
(b)
mode = 0 and 21 (bimodal), mean = 29.4,
median = 20
4 (i)
(ii)
5 (i)
(ii)
mode = 1, mean = 1.4, median = 1
the median
mode = 1, mean = 2.1, median = 2
the median
?
●
(Page 25)
The upper boundaries are not stated. 0– could mean 0–9
or it could mean at least 0.
(Page 27)
Mid-class values: 114.5, 124.5, 134.5, 144.5, 154.5,
164.5, 174.5, 184.5
1
S1
Mean = 161.5 cm
Mid-class values: 24.5, 34.5, 44.5, 54.5, 64.5,
79.5, 104.5
2 (i)
The mode can be estimated as in this example for a
unimodal histogram.
Mean = 48.5 minutes
(ii)
Chapter 1
?
●
Exercise 1D (page 31)
The second value seems significantly higher.
To make the comparison valid the method of
data collection would have to be similar, as
would the target children sampled.
Mid-interval values: 4.5, 14.5, 24.5, 34.5, 44.5,
54.5, 64.5, 74.5, 84.5
3 (i)
Mean (stated age) = 29.7
The median can be estimated by interpolation of the
interval containing the median or by use of a cumulative
frequency curve.
The estimated mean age once adjusted compares
well with the actual mean of 30.4 years.
Mid-class values: 25, 75, 125, 175, 250, 400, 750,
3000
5 (i)
Mean = 950.8 m
(ii)
(Page 30)
Robert needs to increase his estimate by 0.5 cm
(162.32 cm becomes 162.82 cm). The mean of the raw
data is 162.86 cm. The estimated value is very close.
(ii)
?
●
The way in which this data is grouped seems
to have a marked effect on the mean. This is
probably because the distribution is so skewed.
59.5 x 99.5
6 (i)
(Page 29)
The fairest answer is there is not enough information.
Ignoring the journalistic prose, ‘ . . . our town council
rate somewhere between savages and barbarians . . . ’,
the facts given are correct. However, to say whether or
not the council is negligent one would need to compare
accident statistics with other similar communities. Also,
one would need to ask who is responsible for a cyclist’s
risk of having or not having an accident? Perhaps parents
should ensure there is adequate training given to their
children, and so on.
?
●
(iii)
(Page 28)
The median, as it is least affected by extreme values.
?
●
Add 0.5; mean age = 30.2 years
Mean = 43.1 cm
4
The mean of the original data is 39.73. The estimate is
reasonably close at 39.855.
?
●
(ii)
138.5 g
(Page 35)
With the item $90 removed the mean is $15.79, compared
to $19.50. The extreme value ‘dragged’ the value of the
mean towards it.
?
●
(Page 35)
Each deviation is by definition the data value − the mean.
As the mean is central to the data, some deviations will
be negative (below the mean) and some will be positive
(above the mean). The total deviation above the mean
cancels out the total deviation below the mean.
?
●
(Page 40)
Using 656 instead of the accurate value of 655.71…
results in
variance = 430 041.14… – 430 336
= –294.85…
which, being negative, is impossible.
315
Answers
S1
?
●
(Page 41)
(ii)
With the value 96 omitted mean = 54.2, standard
deviation = 7.9. The value 96 is more than five standard
deviations above the new mean value.
Exercise 1E (page 41)
1 (i)
(ii)
2
Mean = 2.36
Standard deviation = 1.49
(ii)
Mahmood: mean = 1.03, standard deviation = 1.05
Raheem: mean = 1.03, standard deviation = 0.55
Mean = 1.1, standard deviation = 1.24
5
Mean = 0.4, standard deviation = 0.4
6 (i)
(ii)
(iii)
7 (i)
(ii)
8 (i)
(iii)
Overall mean rainfall = 1.62 cm, overall standard
deviation = 0.135 cm
(iv)
84.4 cm
10 (i)
(ii)
13 (i)
No. The harvest was less than two standard
deviations above the expected value.
Mean = 25.8 g, standard deviation = 3.98 g
Mode = 1
(ii)
The distribution is unimodal and has positive
skew.
(iii)
Mean = 2, standard deviation = 1.70; the value 8
may be regarded as an outlier as it is more than
two standard deviations above the mean.
(iv) (a)
(b)
(v)
14 (i)
(ii)
15 (i)
Yes. The value is more than two standard
deviations above the mean rainfall.
No. The value is less than one standard deviation
below the mean rainfall.
(ii)
(iv)
Mean = 24.8 °C, standard deviation = 1.05 °C
Both routes have the same mean time but the
country route is less variable or more consistent.
Total weight = 3147.72 g
Σ x 2 = 84 509.9868
(iii) n = 200, Σ x = 5164.84, Σ x 2 =136 549.913
Thermostat A is better. The smaller standard
deviation shows it is more consistent in its
settings.
Town route: mean time = 20 minutes, standard
deviation = 4.60 minutes
Country route: mean time = 20 minutes,
standard deviaiton = 1.41 minutes
The value 166 cm is less than two standard
deviations above 162.82 (Robert’s data) and is
less than two standard deviations below 170.4
(Asha’s data). Consequently it is impossible to
say, without further information, which data set
it belongs to.
(ii)
A: mean = 25 °C, standard deviation = 1.41 °C
B: mean = 25 °C, standard deviation = 2.19 °C
(ii)
9 (i)
12 (i)
On average they scored the same number of goals
but Raheem was more consistent.
4
Standard deviation = 2.54 cm
The value 171 is more than three standard
deviations above 162.82 and less than one
standard deviation above 170.4 so it seems likely
that this value is from Asha’s data set.
Mean = 6.04, standard deviation = 1.48
(i)
316
11 (i)
(ii)
Exclude it since a difference of 8 is impossible.
Check the validity and include if valid.
Mean = 1.88, standard deviation = 1.48
44.1 years
14.0 years
Since the standard deviation is $0, all Fei’s rides
must cost the same. Since the mean is $2.50, it
follows that both the roller coaster and the water
slide cost $2.50.
$1.03
Exercise 1F (page 49)
1
Mean =252.34 g, standard deviation = 5.14 g
2 (i)
x– = –1.7, standard deviation = 3.43
The higher yield was probably the result of the
underlying variability but that is likely to be
connected to different weather patterns.
(ii)
Mean = 94.83 mm, standard deviation = 0.343 mm
(iii)
–18 is more than four standard deviations below
the mean value.
Sample mean = 2.075 mm, standard deviation
= 0.185 mm
(iv)
New mean = 94.863 mm,
new standard deviation = 0.255 mm
The desired mean is less than 0.5 standard
deviations from the observed mean so the
machine setting seems acceptable.
3 (i)
(ii)
Mean = 6.19, standard deviation = 0.484
6.68
–2.05, 10.24
a = 7, standard deviation = 2.51
Mean = 33.75 minutes,
standard deviation = 2.3 minutes
The modal class is that with the largest frequency density
and so it has the highest bar on a histogram.
2 (i)
0.4
? (Page 58)
●
0.3
0.2
0.1
20
0
0
10
The first interval has width 9.5, the last 10.5. All the
others are 10. The reason for this is that the data can
neither be negative nor exceed 70. So even if part marks
were given, and so a mark such as 22.6 was possible, a
student still could not obtain less than 0 or more than
the maximum of 70.
50
For Mensah’s Wood there is a reasonably even
spread of trees with diameter from 0.5 cm to
30.5 cm. For Ashanti Forest the distribution is
centred about trees with diameter in the 16–20 cm
interval. Neither wood has many trees with
diameter greater than 30 cm.
frequency density
(Page 54)
40
(iv)
(Page 52)
170 friends; 143 friends (to 3 s.f.)
20
30
diameter (cm)
Mensah’s Wood: 21–30; Ashanti Forest: 16–20
Standard deviation = 8.88
Chapter 2
10
(iii)
a = 12
(ii)
?
●
0
(ii)
0
00
10
10
a = 5, standard deviation = 3.93
1
0
Mean = 4, standard deviation = 6.39
9
2
90
8
3
0
Mean = 6, standard deviation = 3
50
Ashanti Forest
0
Mean = 67.07; standard deviation = 9.97
40
4
50
6 7 8 8
2 3 4
4
7
12 (i)
20
30
diameter (cm)
9
Some negative skew, but otherwise a fairly
normal shape.
(ii)
10
0
7
9
2
0
0
frequency density
4
5
6
7
8
?
●
0
n = 14
4 7 represents 47
6 (i)
11
1
40
(iii)
–47 is many more than two standard deviations
from the mean.
0
(ii)
S1
Mensah’s Wood
2
30
5 (i)
(ii)
80
10 & 12, 10, 10
x– = –4.3 cm, standard deviation = 13.98 cm
0
(iv)
70
15 & 16, 15, 15
0.5 d 10.5, 10.5 d 15.5, 15.5 d 20.5,
20.5 d 30.5, 30.5 d 50.5
1 (i)
0
(iii)
Exercise 2A (page 59)
60
50 & 60, 50, 50
Chapter 2
(ii)
frequency density
No unique mode (5 & 6), mean = 5, median = 5
4 (i)
The distribution has strong positive skew.
317
S1
The data are discrete. ‘Number of pages’ can only
take integer values.
7 (i)
3 (i)
4
The data are positively skewed. Even though
the data are discrete (suggesting a stem-and-leaf
diagram or vertical line graph) the data are very
spread out with most of the data values less than
200. A histogram will show the distribution
properties best.
Answers
frequency density
(ii)
3
2
1
0
10
0
12
0
14
0
16
0
18
0
20
0
22
0
60
80
0.2
138.4 g
4 (i)
frequency density
15
0
10
100 200 300 400 500 600 700 800 900 1000
number of pages
8 (i)
5
0
(ii)
0.1
frequency density
(ii)
frequency density
mass (g)
20
30
40
50
length (cm)
60
70
42.1 cm
(ii)
40
30
20
10
0
1
2
3
time (hours)
4
5
2.1 hours
5 (i)
frequency density
Activity 2.1 (Page 63)
0.25
(
0
(ii)
100
200 300 400
song length (s)
500
600
6.0 volts, 0.067 volts
?
●
60
40
mean – 2sd
5.866
20
mean
6.0
mean + 2sd
6.134
5.
90
5.
95
6.
00
6.
05
6.
10
6.
15
6.
20
85
5.
5.
80
0
318
(
)
)
Whilst the quartiles Q1 and Q3 differ from those obtained
with a graphical calculator, either method is acceptable.
264 seconds
frequency density
(per 0.05 volts)
6
For a list of n items of data, an Excel spreadsheet uses
the ‘method of hinges’. It places the median, Q2, at
position n + 1, the lower quartile, Q1, at position
2
1
n
+
1
= 21 + n + 1 and the upper quartile, Q3, at
2 1+ 2
4
3(n + 1) − 1
1 n +1
+n =
position 2
2.
4
2
potential difference (volts)
(Page 63)
The data are a sample from a parent population. The
true values for the quartiles are those of the parent
population, but these are unknown.
Exercise 2B (page 71)
5 (i)
7
(b)
6
(c)
4.5, 7.5
(d)
3
(e)
none
(a)
50
1
60
6
(c)
8, 14
(d)
6
70
13
(e)
none
80
17
23
(b)
26
(c)
23, 28
(d)
5
90
19
(e)
14, 37
100
20
36
(b)
118
(c)
115, 125
(d)
10
(e)
141
(ii)
20
5, 5
(ii)
35, 5
(iii)
50, 50
(iv)
80, 50
74
(ii)
73, 76
(iii)
3
S1
Chapter 2
11
(iv) (a)
3 (i)
Cumulative
frequency
(b)
(iii) (a)
2 (i)
y
15
(ii) (a)
cumulative frequency
1 (i)
15
10
5
0
(iv)
(iii)
50
60
70
80
yield (kg)
90
100
65 kg, 16 kg
(iv)
70
82
73 74
(v)
76
58
On average the golfers played better in the
second round; their average score (median) was
four shots better. However, the wider spread of
data (the IQR for the second round was twice
that for the first) suggests some golfers played
very much better but a few played less well.
4 (i)
65
74
(v)
66 kg, 18.5 kg; the estimated values are quite close
to these figures.
(vi)
Grouping allows one to get an overview of the
distribution but in so doing you lose detail.
6 (i)
0
7
1
(ii)
2
4
The vertical line graph as it retains more data for
this small sample.
cumulative frequency
200
150
100
50
Q1 Q2
0
5
(ii) (a)
11 s
(iii)
Q3
10
15
20
interval (seconds)
(b)
25
30
35
10.5 s
13.1 s
319
cumulative frequency
Answers
S1
(iii)
13.5 to 14.6 minutes
800
(iv)
m = 18.2 minutes, s = 14.2 minutes
700
(v)
155 to 170 people
600
10 (i)
7
0
1
2
3
500
400
2
2
1
1
(ii)
Median = 19, Q1 = 10, Q3 = 24, IQR = 14
200
(iii)
The median is preferable. The mode (23) is not
near the centre of the data.
0
Q1 Q2 Q3
40
45
50
55 60
mass (g)
65
Chapter 3
70
Exercise 3A (page 86)
(i)
7.5%
(ii)
54 g
(iii)
7g
2 (i)
1
6
12.2 s, 6.11 s
3 (i)
12
98
=
(v) 56
98
=
8 (i)
cumulative frequency
(ii)
66
1 534, assuming each faulty torch has only one fault.
50
40
4 (i)
70%
30
20
0
5.5
3
6
6
49
(ii)
53
98
4
7
(vi)
5
98
They might draw.
0.45
(iv)
0.45
18 E 12
2
6
20
14.5 seconds
a = 494, b = 46
=
1
2
(iv) 3
6
45
98
(iv)
14
4
1
8
11
1
2
(b)
4
20
(e)
12
20
=
3
5
(f)
0
(h)
P(E ∪ S) = P(E) + P(S) – P(E ∩ S)
(i)
P(E ∪ O) = P(E) + P(O) – P(E ∩ O)
1
100
10
20
30
40
time (minutes)
50
60
=
2
20
1
= 10
3
7
=
17
=
200
0
42
98
19
7
1
5
(c)
10
20
(g)
1
=
1 (d)
2
Exercise 3B (page 90)
300
1
2
5
9
16
10
20
400
=
15 �
13 3 O
S
(ii) (a)
500
320
(iii) 3
6
(iii)
(iii)
10
(ii)
1
2
(ii)
10.5 15.5 20.5 25.5 30.5 35.5
x (seconds)
600
=
(ii)
0.35
5 (i)
10
0
9 (i)
5 6 8 8
4 6 7 7 9
2 3 3 3 5 6 7
5
300
100
cumulative frequency
1 2 represents 12 people
(i)
0.2401
(ii)
0.5002
(iii)
0.4998
0.51
Boy
0.49
Girl
0.51
Boy
0.49
Girl
0.51
Boy
0.49
Girl
2
(ii)
0.1732
(iii)
0.1816
0.88
Car not
stolen
0.12
Car stolen
0.88
Car not
stolen
No
break-in
3
1
12
Correct
11
12
1
12
11
12
Correct
1
12
Wrong
1
12
11
12
11
12
1
12
Correct
Wrong
11
12
Wrong
1
12
11
12
1
12
11
12
(i)
(a)
0.000 58
(b)
0.77
(c)
0.020
(ii) (a)
4
7
8 (i)
First die
Correct
Wrong
Correct
Wrong
0.52
(c)
0.094
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
1
= 12
(iii)
7
Wrong
(iv)
The different outcomes are not all equally
probable.
Correct
Wrong
9
0.31
11
12
10 (i)
(iii)
0.382
(iv)
0.618
1
4
(ii)
81
256
(iii)
81
1024
(iv)
1
16
(v)
3
4
(vi)
3
32
11 (i)
?
●
S1
(Page 98)
P(T S) = 109
= 0.645
169
5
0.992
Not
colour-blind
(i)
0.0016
(ii)
0.0064
(iii)
0.2064
(iv)
0.7936
2
27
or 00741
.
125
(ii) 216
0.2
Left-handed
43
= 0.494
P(T S ′) = 87
0.8
Not
left-handed
So P(T S) P(T S ′)
0.2
Left-handed
0.8
Not
left-handed
Colour-blind
0.008
5
(iii)
9
2
Correct
0.93
6 (i)
1
(ii) 3
36
0.0052
(b)
For a sequence of events you multiply the
probabilities. However, 61 × 61 × 61 × 61 × 61 × 61 gives
the probability of six 6s in six throws. To find the
probability of at least one 6 you need 1 – P(no 6s)
and that is 1 − 65 × 65 × 65 × 65 × 65 × 65 = 0.665.
Chapter 3
0.93
0.0084
Car stolen
Second die
0.07
(i)
0.12
Break-in
or 05787
.
or 05556
.
?
●
(Page 98)
T represents those who had training; T ′ those with
no training: S those who stayed in the company; all
employees. S ′ is inside the box but not in the S region.
For example, in part (i), the answer is 152
256. 152 is in T (but
not in T ′), 256 is everyone.
?
●
(Page 100)
The first result was used in answering part (i) and the
second result in answering part (iii).
321
Answers
S1
Exercise 3C (page 100)
(ii)
0.372
0.6
(ii)
0.556
(iii)
0.395
(iii)
0.625
(iv)
0.047
(iv)
8
(v)
0.321
(vi)
0.075
(vii)
0.028
(viii)
0.0021
1 (i)
(ix)
0.000 95
35
100
1
3 (i) 6
7
4 (i) 15
5
(iii) 21
2 (i)
5 (i)
(ii)
=
(x)
42
100
5
(ii) 12
11
(ii) 21
5
(iv) 11
7
20
(ii)
=
9 (i)
J
21
50
15
65
2
(iii) 5
(iii)
3
= 13
3
P(B A) P(B) and P(A B) P(A) so the events
A and B are not independent.
Hunter dies
Hunter lives
Total
Quark dies
1
12
5
12
1
2
Quark lives
2
12
1
3
1
2
Totals
1
4
3
4
1
(ii)
Milk
No sugar
1
4
Lemon
Sugar
No sugar
3
4
0.1
Neither
0.805
(iii)
0.42; 0.8
2010
8 (i)
0.7
R
0.3
S
0.5
0.3
R
S
0.5
S
=
1
(b) 6
19
(e) 24
0.7
2012
R
0.3
S
0.5
R
0.5
S
0.7
R
0.3
S
0.5
R
0.5
S
28
(c) 45
10
(f) 39
206
427
717
P(E) = 1281
=
(iv)
11 (i)
(ii)
(ii)
358
564
=
239
427
179
282
0.4375
0.3
0.252
0.440
13 (i)
1st set of lights
2011
R
0.7
322
No sugar
9
20
0.525
(ii)
Sugar
LH
412
618
717
and so M and E are not
However, 1281
≠ 1281
× 1281
independent events.
12 (i)
11
20
29
If M and E are independent events then
P(M and E) = P(M) × P(E).
Sugar
2
5
0.7
0.2
618
1281
412
(ii) 1281
(iii)
3
5
1
(a) 4
4
(d) 5
10 (i)
5
(iv)
6
7 (i)
6
Key: J = Juniors
M = Males
LH = Left-handed players
6 (i)
5
(iii) 12
5
28
4
0.5 and 0.875
1
(ii) 12
M
5
10
0.48
�
(90)
2nd set of lights
0.8
0.2
0.8
0.6
Stop
0.7
Not stop
0.3
Stop
0.7
Not stop
0.3
Stop
0.7
Not stop
0.3
Stop
0.7
Not stop
Not stop
Stop
Not stop
0.2
0.3
Stop
Stop
0.4
3rd set of lights
Not stop
(ii)
0.224
(iii)
0.392
(iv)
0.633
5
(iii) (a) 18
2
(i)
?
●
(ii)
(Page 105)
?
●
0
1
2
3
4
5
P(Y = r)
3
18
5
18
4
18
3
18
2
18
1
18
The distribution has positive skew.
r)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
(Page 105)
A discrete frequency distribution is best illustrated by a
vertical line chart.
Using such a diagram you can see that the frequency
distribution is positively skewed, see figure 4.1.
P(X = r)
2
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
P(X = r)
3
4
5
6
7
8
9
10
11
12
(ii)
4 (i)
P(X
(ii)
5 (i)
The distribution is symmetrical.
P(X
r)
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
(ii)
P(X
2
4
6
8
10
12
r
3
4
r
5
k = 0.4
(ii) (a)
r
2
1
(b) 2
r
Exercise 4A (Page 111)
1 (i)
1
2
(iii) (a) 3
3 (i)
S1
r
P(Y
You could conduct a traffic survey at peak times, over
fixed periods of time, for example, 1 hour in the morning
and 1 hour in the evening, over a period of a working
week. You could count both the number of vehicles and
the number of people travelling in each vehicle.
2
(c) 3
Chapter 4
Chapter 4
1
(b) 2
2
4
6
8
0.1
0.2
0.3
0.4
0.3
(b)
0.35
3
4
k = 20
49
r)
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
5
6
7
r
0.248 (to 3 s.f.)
r
0
1
2
3
P(X = r)
1
8
3
8
3
8
1
8
The distribution is symmetrical.
r)
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
r
323
S1
(iii)
0.5
61
, P(X = 5) = 216
;
P(X 5) = 125
216
(iv)
No, an equal number of heads and tails is
possible, so probability 0.5.
91
P(X 6) = 1, P(X = 6) = 216
(iv)
P(X
Answers
6 (i)
r
P(X = r)
1
1
16
2
16
2
16
3
16
2
16
2
16
1
16
2
16
1
16
2
3
4
6
8
9
12
16
The distribution has negative skew.
r)
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
4
10
P(X = r)
0
1
2
3
4
0.2
0.24
0.32
0.24
0
6
10
0
1
2
3
0.351 04
0.449 28
0.184 32
0.015 36
3
9
R
6
9
B
4
9
R
5
9
B
2
8
1
k = 35
(iii)
Since the probability distributions look quite
different, the model is not a good one.
r)
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
(iii)
?
●
observed
model
0
1
2
1
P(X = 1) = 216
(ii)
8
=
P(X 2) = 216
(iii)
27
19
P(X 3) = 216
, P(X = 3) = 216
;
1
27
64
37
P(X 4) = 216
, P(X = 4) = 216
;
3
4
R
6
8
3
8
B
5
8
3
8
B
5
8
4
8
B
4
8
B
R
R
R
0
1
2
3
P(X = r)
1
6
1
2
3
10
1
30
3
11
11 (i)
(ii)
r
6
r
a = 0.42
P(X
324
5
B
(ii)
P(Y = r)
9 (i)
4
R
Let Y represent the number of chicks.
r
8 (i)
3
k = 0.08
r
(ii)
2
10 (i)
1
(ii) 4
7 (i)
1
r
0
1
2
3
P(X = r)
14
55
28
55
12
55
1
55
(Page 114)
Carshare.com could have based their claim on the
results of traffic surveys. These could be used to calculate
summary statistics such as the mean and standard
deviation of the number of people per vehicle, as well
as the number of vehicles per hour.
5
r
?
●
(Page 114)
By comparing measures of central tendency and spread,
it is possible to infer whether or not there is a significant
difference between their values. There are many different
tests of statistical inference; the ideas involved are
introduced in Chapter 8 if you go on to study Statistics 2.
It is also possible to compare statistically the proportion
of vehicles with a single occupant.
Activity 4.1 (Page 114)
k = 0.1
7 (i)
Mean = 2
Variance = 0.86
1.25
(iii)
0.942 (to 3 s.f.)
8 (ii)
(iii)
The increase in average occupancy, together with a
significant reduction in the proportion of vehicles with
a single occupant, could be used to infer that the scheme
has been successful.
9 (i)
The two measures of spread are almost the same.
?
●
(Page 117)
If the expectation, E(X), is not exact in decimal form,
then calculations by hand using the definition of Var(X)
may be tedious and/or prone to arithmetic errors by
premature approximation of E(X). The alternative
formulation of Var(X) may be more appropriate in
such cases.
r
0
1
2
3
4
6
P(X = r)
1
4
1
3
1
9
1
6
1
9
1
36
41
E(X) = 53, Var(X) = 18
r
0
80
120
160
P(X = r)
5
28
20
28
2
28
1
28
(ii)
E(X) = $71.43, Var(X) = 1412.245 (to 3 s.f.)
(iii)
E(W) = $500, Var(X) = 3600 or 13 200
10 (i) P(X
Exercise 4B (Page 118)
1
E(X) = 3
2 (i)
(ii)
3 (i)
(ii)
3
4
5
6
r
Mean = 1.92, standard deviation = 0.987 (to 3 s.f.)
(iii)
0.25
p = 0.8
(iv)
0.431
4
5
0.8
0.2
r
50
100
P(Y = r)
0.4
0.6
11
a = 0.2, b = 0.25
12 (i)
Attempt 1
0.5
0.5
0.5
0.5
(b)
1
6
(c)
17
18
(b)
Attempt 4
Answers
Does not
answer
Answers
0.5
0.5
E(X) = 1.944; Var(X) = 2.052
5
9
Attempt 3
Answers
Does not
answer
Var(X) = 5.833
5
12
Attempt 2
Answers
0.5
E(X) = 7
(ii) (a)
6 (i)
2
(ii)
(iii) (a)
5 (i)
1
P(X 3.125) = 0.5625
P(X = r)
4 (i)
r)
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
E(X) = 3.125
r
(ii)
Chapter 4
Activity 4.2 (Page 116)
Mean = 1.7
Variance = 0.87
S1
(ii)
Does not
answer
1
18
Does not
answer
0.5
E(X) = 1.5
(ii)
Var(X) = 0.75
(iii)
E(Y) = 5, Var(Y) = 2.5
(ii)
(iii)
x
0
1
2
3
4
P(X = x)
1
2
1
4
1
8
1
16
1
16
15
16
325
S1
13 (ii)
E(X) = 3.75, Var(X) = 2.1875
8
14 (i)
2q – p = 0.39, p + q = 0.42; p = 0.15, q = 0.27
9 (i)
(ii)
Answers
2.5875
(ii)
1
120
14!
10 (i)
E(X) = 43 , Var(X) = 7 59
1
14!
(ii)
0.1951
16 (i)
1
10!
11 (i)
(ii)
(iii)
r
2
3
4
5
6
7
P(X = r)
1
36
0
1
18
0
5
36
1
9
r
8
9
10
11
12
P(X = r)
1
9
1
9
1
9
2
9
1
9
?
●
Exercise 5A (Page 127)
1 (i)
40 320
(ii)
56
(iii)
5
7
2 (i)
(ii)
3 (i)
(ii)
4 (i)
(n + 3)(n + 2)
n(n – 1)
8! × 2!
5! × 5!
16!
(ii)
14! × 4!
(n + 1)!
(iii)
(n − 2)! × 4!
5
(i)
9 × 7!
(ii)
n!(n + 2)
6
24
7
40 320
(OPTS, POST, POTS, SPOT, STOP, TOPS)
(ii)
120
(iii)
362 880
(iv)
12
(v)
420
(vi)
50 400
2520
14 (i)
(ii)
360
(iii)
720
(iv)
1800
Investigations (Page 129)
1
n > 6, n > 7, n > 8
2
Examples are 5! , 3! + 3! + 3! + 3! + 3! + 3! + 3!,
7!
4! + 4! − 3!, etc.
3
(i)
44
(ii) (a)
1
n
n–1
1
4
24
13 (i)
(Page 124)
Oscar could have put the bricks in order by chance.
1
A probability of 120 is small but not very small. What
would really be convincing is if he could repeat the task
whenever he was given the bricks.
=
12 6
24
8 23
Chapter 5
1.38 (i.e. 1 or 2 people)
(ii)
(iv) 5
9
326
120
(ii)
1
6
15 (i)
720
(b)
?
●
12
4
(Page 130)
No, it does not matter.
?
●
(Page 131)
Multiply top and bottom by 43!
49 × 48 × 47 × 46 × 45 × 44 × 43! = 49!
6!
43! 6 × 43!
?
●
49C
6
?
●
(Page 131)
49
= = 13 983 816 ≈ 14 million
6
(Page 131)
By following the same argument as for the UK National
Lottery example but with n for 49 and r for 6.
?
●
if 0! = 1
? (Page 135)
●
1
2
3 628 800
(iii)
0.986
(ii)
900
(iii)
126
16 (i)
(a)
1
The probability is 24 , assuming the selection is done
at random, so RChowdhry is not justified in saying
‘less than one in a hundred’.
1 1 1
As a product of probabilities 3 × 2 × 4 =
1
24
(a)
30
(b)
1680
(c)
5040
(ii) (a)
15
(b)
70
(c)
210
1 (i)
1
2 2730
1
3 593 775
4
715
5
280
6 (i)
(ii)
7 (i)
11 (i)
1
14
3
(b) 7
1
126
45
(b) 126
15
90 720
4.94 ×
1011
(ii)
79 833 600
(iii)
21
12 (i)
(ii)
13 (i)
(b)
517
(Page 142)
2
0.271
3
0.214
4
0.294
5 (i)
(ii)
2 177 280
90
86 400
(iii)
288
Poor visibility might depend on the time of day,
or might vary with the time of year. If so, this
simple binomial model would not be applicable.
1
8
3
(ii) 8
3
(iii) 8
1
(iv) 8
(b)
75
(b)
120
7 (i)
(ii)
8 (i)
(ii)
9 (i)
33 033 000
(ii)
0.146
6 (i)
1
(ii) 7893600
(ii) (a)
1316
She should really try to improve her production
process so as to reduce the probability of a bulb being
substandard.
126
(a)
216
Chapter 6
210
1
9 (i) 120
10 (i)
(b)
15
1 64
3000
(ii) (a)
60
Exercise 6A (Page 145)
31 824
(ii) (a)
8 (i)
(ii) (a)
?
●
Exercise 5B (Page 137)
S1
831 600
15 (i)
again if 0! = 1
n
(ii)
Chapter 6
()
n!
C = (n ) =
=1
n n !(n − n)!
n
n!
0 = 0 = 0! × n ! = 1
nC
n
259 459 200
14 (i)
(Page 132)
(ii)
0.246
Exactly 7 heads
(a)
0.058
(b)
0.198
(c)
0.296
(d)
0.448
(a)
0.264
(b)
0.368
(c)
0.239
(d)
0.129
2
Assumed the probability of being born in
31
January = 365
. This ignores the possibility of leap
years and seasonal variations in the pattern of
births throughout the year.
327
Answers
S1
10
The three possible outcomes are not equally likely:
‘one head and one tail’ can arise in two ways (HT or
TH) and is therefore twice as probable as ‘two heads’
(or ‘two tails’).
Assumption: the men and women in the office are
randomly chosen from the population (as far as their
weights are concerned).
4 (i)
(ii)
Activity 6.1 (Page 147)
5 (i)
n
Expectation of X =
∑r × P(X = r)
r =0
Since the term with r = 0 is zero
2.7
(iii)
0.267
(iii)
0.167
r =1
(iv)
0.723
7 (i)
(n − 1)! pr −1qn −r
n ! prqn −r = np ×
(r − 1)!(n − r)!
r !(n − r)!
(n − 1)!
= np ×
(r − 1)!((n − 1) − (r − 1))!
(ii)
8 (i)
× pr −1q(n −1)−(r −1)
= np × n–1Cs psq(n–1)–s
where s = r – 1
In the summation, np is a common factor and s runs
from 0 to n – 1 as r runs from 1 to n. Therefore
Expectation of X = np × ∑
s
since q + p = 1.
2
3
328
0.000 129
(a)
(ii)
0 and 1 equally likely
(i)
2
(ii)
0.388
(iii)
0.323
(i)
0.240
(ii)
0.412
(iii)
0.265
(iv)
0.512
(v)
0.384
(vi)
0.096
(vii)
0.317
(b)
0.0322
(i)
(b) 0.299
0.003 22
(iii)
0.0386
(v) 405
8192
p sq(n −1)− s
(c)
0.402
12
0.99; 0.93
(ii)
0.356; expected number of boxes with no broken
eggs = 100 × 0.356 = 35.6, which agrees well with
the observed number, 35.
(iii)
0.0237
11 (i)
Exercise 6B (Page 150)
1
32
81
3 correct answers most likely; 0.260
0.0193
= np(q + p)n −1
= np
(a) 0.195
(ii)
10 (i)
s =0
(d)
1
64
15
(ii)
64
45
(iii)
512
45
(iv)
2048
= np × n–1Cr–1pr–1q(n–1)–(r–1)
n −1C
24
81
9 (i)
Using (n – 1) – (r – 1) = n – r
n −1
(c)
0.002 69
n
The typical term of this sum is
8
81
0.0735
0.121
= ∑r × nCr prqn −r .
(b)
2 min 40 s
(ii)
r =1
r×
1
81
(ii)
6 (i)
n
Expectation of X = ∑r × P(X = r)
(a)
0.132
(ii)
0.0729
(iii)
0.0100
(iv)
Mean = 53 , variance = 10
9
0.212
Chapter 7
Exercise 7A (Page 166)
1 (i)
0.841
(ii)
0.159
(iii)
0.5
(iv)
0.341
2 (i)
0.726
0.29
15 (i)
S1
0.274
(ii)
4340
(iii)
0.0548
(iii)
(iv)
0.219
Because the probability of scoring between 0 and
10 is about 0.99.
0.159
(iv)
0.0258
(ii)
0.841
(v)
113 or more
(iii)
0.841
16 (i)
(iv)
0.683
3 (i)
4 (i)
Chapter 7
(ii)
0.8413
(ii)
0.0228
(iii)
0.1359
19%
5%
5 (i)
0.0668
0.6915
(iii)
0.2417
(ii)
µ = 4650
0.0668
(iii)
0.44 (2 s.f.)
(ii)
0.1587
(iv)
3700 hours
(iii)
0.7745
(v)
0.76 (2 s.f.)
6 (i)
7 (i)
31, 1.98
(ii)
2.1, 13.3, 34.5, 34.5, 13.4, 2.1
(iii)
More data would need to be collected to
say reliably that the weights are normally
distributed.
8 (i)
5.48%
(ii) (a)
(b)
9 (i)
(ii)
10 (i)
1843 km
(ii)
5.254 cm, 0.054 cm
0.2216
(iii)
0.4315
0.077
(ii)
0.847
(iii)
0.674
(iv)
1.313 m
12
20.05, 0.024 77, 0.7794, 22.6%
13
0.0312
(i)
0.93
(ii)
0.080
11.09%, 0.0123, 0.0243
0.136
(iii)
0.370
7.24
(ii)
546
8.75
20 (i)
(ii)
0.546
0.595
21 (i)
(ii)
0.573
0.0276
22 (i)
(ii)
7.72
0.484
23 (i)
(ii)
?
●
1.71 bars, 2.09 bars
(ii)
19 (i)
0.0765
lifetime
7.29
18 (i)
78.65%
5000
0.004 29
17 (i)
25 425 km
(ii)
11 (i)
14
4000
(ii)
96.9 minutes, 103.1 minutes
(Page 175)
One possibility is that some people, knowing their
votes should be secret, resented being asked who they
had supported and so deliberately gave wrong answers.
Another is that the exit poll was taken at a time of
day when those voting were unrepresentative of the
electorate as a whole.
329
Answers
S1
Exercise 7B (Page 175)
106
1 (i)
(ii)
75 and 125
(iii)
39
0.282
2 (i)
(ii)
0.526
(iii) (a)
(b)
(iv)
3
(ii)
Yes. The test was set up before the data were known, the
data are random and independent and it is indeed testing
the claim.
Exercise 8A (Page 187)
1
0.057
2
0.0547 5%
4.33
3
0.102
(a)
0.315
(b)
0.307; assuming the answer to part (i) (a) is
correct, there is a 2.5% error; worse
0.104 Accept H0
Insufficient evidence at the 5% significance level that
the machine needs servicing.
6 (i)
(ii)
0.5245
(iii)
0.677
0.126
0.281
0.748
9 (i)
(ii)
0.887
(iv)
7 (i)
P(2 defectives in 10) = 0.302
In 50 samples of 10, the expected number of
samples with two defectives is 15.1, which agrees
well with the observed 15.
H0: P(mug defective) = 0.2
H1: P(mug defective) 0.2
n = 20. P(0 or 1 defective mug) = 0.0692
Accept H0 since 0.0692 5%
It is not reasonable to assume that the
proportion of defective mugs has been reduced.
Opposite conclusion since 0.0692 10%
0.590
(ii)
0.044
(ii)
0.118
(iii)
0.000 071 2
(iii)
13
(iv)
0.0292
0.311
(v)
11 (i)
(ii)
Not appropriate because np < 5.
(iii)
0.181
(vi)
8
Chapter 8
330
2; 1.183
0.298
10 (i)
?
●
H0: probability that toast lands butter-side down = 0.5
H1: probability that toast lands butter-side down 0.5
0.015
Accept H0
5
86; more (96)
(ii)
(Page 182)
Assuming both types of parents have the same fertility,
boys born would outnumber girls in the ratio 3 : 1. In a
generation’s time there would be a marked shortage of
women of child-bearing age.
Accept H0
0.047
Reject H0
There is evidence that the complaints are justified
at the 5% significance level, though Mr McTaggart
might object that the candidates were not randomly
chosen.
( )( )
8 (i)
Accept H0
4
1
5 3 ; 6.667; 4.444; 13
1 N 29 2000 – N
6 2000C
30
N 30
7
(Page 186)
25
0.246, 0.0796, 0.0179
The normal distribution is used for continuous
data; the binomial distribution is used for discrete
data. If a normal approximation to the binomial
distribution is used then a continuity correction
must be made. Without this the result would not
be accurate.
4 (i)
?
●
9
H0: P(long question right) = 0.5
H1: P(long question right) 0.5
No
H0: P(car is red) = 0.3
H1: P(car is red) 0.3
Accept H0 (Isaac’s claim) since 0.060 5%
H0: P(support for Citizens Party) = 31
H1: P(support for Citizens Party) 31
Accept H0 since 0.0604 2.5%.
There is insufficient evidence to suggest that support
has decreased.
?
●
(Page 191)
Rejection region at 10% significance level is X 4.
4
0 correct or 6 correct
5
Critical region is 3 or 13 letter Zs
6 (i)
Exercise 8B (Page 192)
(iii)
0.0046
(iv)
H0: p = 0.9, H1: p 0.9
(v)
2 (i)
(ii)
0.0417
(iii)
0.0833
n = 17; P(X 13) = 0.0826 5%;
(iv)
0.1184
not sufficient evidence to reject H0.
(v)
Critical region is X 12, since
P(X 12) = 0.0221.
(a)
0.0278
(b)
0.0384
Let p = P(blackbird is male)
H0: p = 0.5, H1: p 0.5
(iv)
You would be more reluctant to accept H1.
Although H0 is still p = 0.5, the sampling method
is likely to give a non-random sample.
(iii)
Complaint justified at the 10% significance level
0.0592
Result is significant at the 5% significance level.
Critical region is X 12.
(ii)
(iii)
(ii)
(iii)
3 (i)
0.196
7 (i)
0.9619
(vi)
(ii)
0.430
(ii)
5
(a)
0.0991
(b)
0.1391
(vi)
1 (i)
(b)
When p = 0.8, he reaches the wrong
conclusion if he rejects H0, i.e. if X 17, with
probability 0.0991.
1
P(X 3) = 0.073 5%
2
P(X 13) = 0.0106 22 %
0.0480
0.0480
(iii)
0.601
(ii)
3 (i)
(ii)
4 (i)
If H0 is wrongly rejected because there were only
0 or 1 red chocolate beans in the sample although
20% of the population were actually red.
0.167
H0: P(pass on 1st attempt) = 0.36
H1: P(pass on 1st attempt) 0.36
Reject H0 (accept the driving instructor’s claim)
since 0.013 5%
Type I error; 0.0293
H0: P(six) = 61
H1: P(six) 61
When p = 0.82, he reaches the wrong
conclusion if he fails to reject H0, i.e. if
X 16, with probability 1 – 0.1391 = 0.8609.
Exercise 8C (Page 195)
4 w 11
(ii)
Critical region is X 17, since
P(X 17) = 0.0991 10% but
P(X 16) = 0.2713 10%.
(iv) (a)
Let p = P(man selected)
H0: p = 0.5, H1: p ≠ 0.5
P(X 4 or X 11) = 0.1184 5%
There is not sufficient evidence to reject H0, so
it is reasonable to suppose that the process is
satisfactory.
Exercise 8D (Page 200)
2 (i)
Let p = P(seed germinates)
H0: p = 0.8, H1: p 0.8, since a higher
germination rate is suspected.
Chapter 8
1 (i)
S2
Accept H0 since 0.225 10%
There is no evidence that the die is biased.
(ii)
P(4 or more sixes) = 0.0697
(iii)
Concluding that the die is fair when it is biased.
Accept H0
1
1
3 P(X 9) = 0.0730 22 %
Reject H0
Accept H0
331
S2
Chapter 9
Exercise 9A (Page 207)
Answers
1 (i)
(ii)
2 (i)
0.266
(iii)
0.815
4 (i)
(iv)
Yes, there seems to be reasonable agreement
between the actual data and the Poisson
predictions.
112
(iv)
288
Assume that mistakes occur randomly, singly,
independently and at a constant mean rate.
(Page 213)
It is not necessarily the case that a car or lorry
passing along the road is a random event. Regular
users will change both Poisson parameters which in
turn will affect the solution to the problem.
2
With so few vehicles they probably are independent.
3
They are more likely in the day than the night. This
raises serious doubts about the test associated with
this model.
4
It could be that their numbers are negligible or it
might be assumed they do not damage the cattle grid.
25
(iii)
6 (i)
239.0, 176.4, 65.1, 16.0, 3.0, 0.4
1
0.099
75
(ii)
(iii)
?
●
0.058
(ii)
5 (i)
0.478, 0.353, 0.130, 0.032, 0.006, 0.0009
0.224
0.185
(ii)
(ii)
0.826
(ii)
3 (i)
0.738
12 (i)
0.111, 0.244, 0.268, 0.377
3
Exercise 9B (Page 213)
3
1 (i)
(ii)
27.5
(iii)
460 or 461
7 (i)
(ii)
(iii)
8
The mean is much greater than the variance
therefore X does not have a Poisson distribution.
Yes because now the values of the mean and
variance are similar.
2 (i)
0.012
Some bottles will contain two or more hard particles.
This will decrease the percentage of bottles that have
to be discarded.
13.9%
Assume the hard particles occur singly,
independently and randomly.
9 (i)
(ii)
0.144
(ii)
0.819
(iii)
2.88
0.175
(b)
0.973
(c)
0.031
0.125; 0.249
(a)
0.180
(b)
0.264
(c)
0.916
(ii)
0.296
(iii)
0.549
3 (i)
(ii)
4 (i)
(a)
0.007
(b)
0.034
(c)
0.084
T ∼ Po(5.0)
(a)
0.134
(b)
0.848
0.209
(ii)
0.086
(ii)
0.219
(iii)
0.673
(iii)
6
10 (i)
11 (i)
(ii)
(a)
0.257
0.184
(b)
0.223
0.125
(c)
0.168
5 (i)
(ii)
332
(a)
0.340
6 (i)
0.531
6 (i)
0.082
(ii)
0.065
(ii)
0.891
(iii)
0.159
(iii)
0.287
0.161
0.270 (3 s.f.)
(b)
0.001 44
(ii)
0.554
(c)
0.962
(iii)
8
λ = −ln(0.2) = 1.6094 ≈ 1.61
(iv)
0.016
(iii)
0.1896
(v)
0.119
(iv)
0.369
(ii)
8 (i)
(ii)
9 (i)
(ii)
(a)
0.485 (3 s.f.)
(b)
0.116 (3 s.f.)
(c)
0.087 (2 s.f.)
The Poisson parameter is unlikely to be the
same for each team and there is a lack of
independence.
7 (i)
8 (i)
(ii)
9 (i)
Independence of arrival and random distribution
through time or uniform average rate of
occurrence or mean and variance approximately
equal or n is large and p is small.
26.9 days
(iv)
0.410 (3 s.f.)
(v)
0.0424
Exercise 9C (Page 221)
0.135 768, 0.140 374, 3.4%
0.140 078, 0.140 374, 0.2%
(iii)
0.140 346, 0.140 374, 0.02%
The agreement between the values improves as n
increases and p decreases.
2 (i)
(ii)
3 (i)
0.224
0.616
0.104
(ii)
0.560
(iii)
0.762
4 (i)
0.0378
X ∼ B(108, 0.05); it must be assumed that
whether or not each person turns up is
independent of whether or not any other person
turns up.
0.12 (2 s.f.)
(b)
0.37 (2 s.f.)
(c)
0.29 (2 s.f.)
X ∼ B(500, 0.01)
(ii)
0.7419
(iii)
0.049 (to 2 s.f.)
Exercise 9D (Page 227)
0.180; 60, 7.75, 0.9124
2 (i)
(ii)
0.559
(ii) (a)
10 (i)
1
1 (i)
1
X ∼ B(150, 80
); X ∼ Po(1.875);
the Poisson distribution is a suitable
approximating distribution because n is large
and p is small.
Mean = 1.84, variance = 1.75 (3 s.f.)
(iii)
2.5; assume that service calls occur singly,
independently and randomly.
(ii)
0.918, 0.358
(iii)
0.158
3 (i)
0.4928
(ii)
0.0733
(iii)
0.6865
4 (i)
0.7620
(ii)
0.0329
(iii)
0.3536
5 (i)
0.188
0.0671
(ii)
0.362
(iii) (a)
(ii)
0.544
(b)
(iii)
0.214
(iv)
0.558
(ii)
5 (i)
Chapter 9
(a)
7 (i)
S2
0.9906
0.9629
1.0000
Four lots of 50 is only one of many ways to make
200, so you would expect the probability in part
(b) to be higher than that in part (a).
333
6 (i)
Answers
S2
0.6144
(b)
0.8342
(ii)
You must assume that the same number of
emails will be received, on average, in the future.
(iii)
For longer time periods, there are more and
more different ways in which the total can be
reached, so the probability increases.
7 (i)
(ii)
8 (i)
(a)
0.617
(b)
0.835
(iv)
The null hypothesis is not rejected. The evidence
does not support the astronomer’s theory.
0.122
(ii)
0.532
(iii)
0.0135
(iv)
0.229
0.113
(iii)
0.0211
0.143
(ii)
0.118
(iii)
0.0316
(Page 235)
It is reasonable to regard the height of a wave as random.
No two waves are exactly the same and in a storm some
are much bigger than others.
Exercise 10A (Page 240)
2
k = 35
1 (i)
(ii)
f(x)
12
35
6
35
0
0.0916
(iv)
1 is in the rejection region so there is evidence
that the new guitar string lasts longer.
2 (i)
(ii)
0.972
(iii)
0.0311
There is enough evidence to accept at the 10%
significance level that ploughing has increased
the number of pieces of metal found.
(ii)
There is not enough evidence to accept at the 5%
significance level that ploughing has increased
the number of pieces of metal found.
(iii)
0.395
2
3
4
5
6
1
k = 12
f(x)
5
12
4
12
3
12
2
12
1
12
0.0202
(ii)
1
(iii) 11
35
(iv) 1
7
0 or 1
(iii)
14 (i)
0.915
People can be expected to call randomly,
independently and at an average uniform rate.
(ii)
13 (i)
(iii)
?
●
0.0824
12 (ii)
Accept H0, there is insufficient evidence to
say that the number of white blood cells has
decreased.
H0: µ = 5.6 where µ is the mean number of
shooting stars
H1: µ 5.6
(iii)
11 (i)
(ii)
Chapter 10
X 2 where X is the number of shooting stars
10 (i)
A Type I error is made if you find that the
number of white blood cells has decreased when,
in fact, the number of white blood cells has not
decreased; 0.0342
15 (i)
0.0593
(ii)
9 (i)
334
(a)
0
(iii)
0.207
1
2
3
4
x
x
a=
(ii)
4
81
8 (i)
f(x)
60
2
50
1
0
1
2
3
4
x
40
30
20
10
(iii) 16
81
4 (i)
S2
Chapter 10
frerquency density
3 (i)
c = 81
0
0.5
1
1.5
2
2.5
weight of fish (kg)
f(x)
(ii)
3
Negative skew
1
8
(ii) f1(w)
0.5
0.4
0.3
0.2
0.1
1
16
–4 –3 –2 –1
0
1
2
3
4
5
6
x
0
(iii) 1
4
(iv) 3
8
5 (i)
(ii)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
f(x)
0.2
0
0.1
0
6 (i)
(ii)
7 (i)
(ii)
1
2
3
4
5
6
x
0.248
k = 29
0.067
1
k = 100
0
(v)
3
w
3
w
3
w
2
f2(w) � 10w (3 � w)2
81
1
2
0.6
0.5
0.4
0.3
0.2
0.1
4w2
f3(w) � 27 (3 � w)
1
2
f4(w)
0.1
(iv)
2
f3(w)
0
f(x)
0.2
(iii)
1
f2(w)
k = 0.048
0.3
(iii)
2w
f1(w) � 9 (3 � w)
2
4
6
8
length of stay (hours)
10
x
19, 17, 28, 36
Yes
Further information needed about the group
4–10 hours. It is possible that many of these stay
all day and so are part of a different distribution.
0.6
0.5
0.4
0.3
0.2
0.1
0
f4(w) � 4w (3 � w)2
27
1
2
f3 matches the data most closely.
3 w
335
S2
(iii)
1.62, 9.49, 20.14, 28.01, 27.55, 13.19, 0
(iv)
Model seems good.
(ii)
0.6
0.04
4 (i)
0.803
9 (i)
f(x)
0.456
(ii)
Answers
3 (i)
1
6
0, 0.1, 0.21, 0.12, 0.05, 0.02, 0
10 (i)
0.1
(b)
0.33
(c)
0.33
(d)
0.16
(e)
0.07
(f)
0.02
k=
1
1728
(b)
0.275
0.095
(c)
0.280
0.016
(ii) (a)
(iii)
(iv) (a)
(d)
(v)
11 (i)
0.132
0.201
(e)
(f)
�2 �1 0
(ii) 2
3
5 (i)
(iii)
(ii)
0.45
k = 41
(iii)
1.5
(iv)
1.5
f(x)
(iv)
2
3
4
x
1
3
1.5
Model quite good. Both positively skewed.
(ii)
1
1
f(x)
(v)
1
2
1
2
1
4
0
1
2
3
4
1
4
x
(iii) 27
32
5
a = 12
12 (i)
0
(ii)
5
12
6 (ii)
3
12
(iii)
0
1
2
3
4
x
7 (i)
(ii)
0.292
(iv)
8 (i)
(Page 247)
(ii)
Distributions (b) and (d)
1.083, 0.326
0.5625
f(x) = 31 for 4 x 7
5.5
0.233
1
f(x) = 10
for 10 x 20
15, 8.33
57.7%
(iii) (a)
?
●
3
(iii) 3
4
(iv) 7
12
?
●
2
(Page 250)
100%
(b)
68%. The normal distribution has a greater proportion
of values near the mean, as can be seen from its shape.
9 (i)
a = 1.443
(ii)
Exercise 10B (Page 252)
1 (i)
336
0.89
(iii)
2.828
2
f(x)
2
2.67
(ii)
2 (i)
x
The graph is symmetrical and peaks when x = 1.5
thus E(X) = mode of X = median value of X = 1.5.
f(x)
(iii)
1
1
(ii)
2
(iii)
1.76
0
1
2
x
(iii)
1.443, 0.083
6 (i)
2
(iv)
41.5%
7 (i)
(a)
2.79
1.414
(b)
8.97
(v)
(c)
20.94
(iii)
11 (ii)
104 or 105
Var(X) = 0.0267
0.931
(iv)
0.223
12 (ii)
0.139
(iii)
2
3
4
5
6
7
8
Probability
1
16
2
16
3
16
4
16
3
16
2
16
1
16
5, 2.5
Exercise 11B (Page 266)
1.24 minutes
4.125 minutes
(iii)
0.148
(iv)
Less than 5 minutes, since 0.148 0.25
14 (ii)
(iii)
2.73 hours
(iv)
0.0241
1.5, 0.167
(iii)
Main course
Dessert
Price
Fish and chips
Fish and chips
Fish and chips
Spaghetti
Spaghetti
Spaghetti
Pizza
Pizza
Pizza
Steak and chips
Steak and chips
Steak and chips
Ice cream
Apple pie
Sponge pudding
Ice cream
Apple pie
Sponge pudding
Ice cream
Apple pie
Sponge pudding
Ice cream
Apple pie
Sponge pudding
$4
$4.50
$5
$4.50
$5
$5.50
$5
$5.50
$6
$6.50
$7
$7.50
f(t)
0.4
0.3
0.2
0.1
(iv)
–3 –2 –1
(iii)
0
1
2
3
4
5
6
7
t
2 (i)
1.48 seconds
Chapter 11
3
(Page 258)
It is the variance of X.
1 (i)
2 (i)
4 (i)
(ii)
5 (i)
(ii)
(a)
E(X) = 3.1
(b)
Var(X) = 1.29
(a)
E(X) = 0.7
(b)
Var(X) = 0.61
E(2X) = 6
(iii)
N(−10, 25)
0.196
Var(3X) = 6.75
10.9, 3.09
18.4, 111.24
0.0228
(ii)
56.45 minutes
(iii)
0.362
230 g, 10.2 g
(ii)
0.1587
(iii)
0.0787
6 (i)
(ii)
7
N(90, 25)
N(10, 25)
5 (i)
Exercise 11A (Page 260)
Mean of T = 5.5, variance = 1.042
(ii)
4 (i)
?
●
4, 0.875
(ii)
2.66 hours
15 (ii)
S2
N
1 (i)
13 (ii)
9
8
5.14
(iii)
(iii)
Chapter 11
10 (ii)
1
(ii)
N(70, 25)
N(−10, 25)
5.92%
8 (i)
(ii)
0.266
No, people do not choose their spouses at
random: the heights of a husband and wife may
not be independent.
337
9
S2
0.151
0.0170
10 (i)
(ii)
Answers
11
311.6 kg
10 (i)
(ii)
0.0037
Mean = 59.4, standard deviation = 7.09
Mean = 3360, variance = 1540 (to 3 s.f.)
12 (i)
(ii)
?
●
0.0693
(Page 272)
With folded paper it is not possible for pieces of paper
that are thicker to be offset by others that are thinner,
and vice versa.
0.4546
11 (i)
(ii)
1
N(120, 24)
Assume times are independent and no time is spent
on changeovers between vehicles.
2
0.0745
3
0.1377
4
0.1946
5 (i)
N(−4, 20)
(iii)
N(0, 20)
(ii)
7 (i)
0.316
(iii)
N(200, 122)
(iv)
N(−65, 377)
8 (i)
0.0827
(ii)
0.3103
(iii)
0.5
9 (i)
0.1446
(iii)
0.5
The situations in 8(i) and 9(i) are the same.
8(ii) considers 3X + 5Y whereas 9(ii) considers
X1 + X2 +X3 + Y1 + … + Y5, so the probabilities
are different.
In both 8(iii) and 9(iii) the mean is zero, so the
probability is 0.5, independent of the variance.
338
0.3338, 0.4082
0.9026
Assume weights of participants are independent
since told teams were chosen at random.
0.037
14%
(ii)
0.6
(iii)
15 m
(iv)
0.3043
16
0.195
17
0.350
0.387
18 (i)
(ii)
Mean = 10, variance = 11.56
(iii)
0.647
0.0827
(ii)
(iv)
(iii)
15 (i)
N(100, 26)
N(295, 353)
0.0856
(ii) 0.238
Assume that no time is lost during baton
changeovers and that the runners’ times are
independent, i.e. that no runners are influenced by
the performance of their team mates or competitors.
The model does not seem entirely realistic in this.
0.316
(ii)
(ii)
14 (i)
N(34, 20)
(ii)
6 (i)
13
93.49 litres
0.0202
12 (i)
Exercise 11C (Page 273)
Assume that the composition of each crew is
selected randomly so that the weights of each
of the four individual rowers are independent
of each other. This assumption may not be
reasonable since there may be some light-weight
and some heavy-weight crews; also men’s and
women’s crews. If this is so it will cast doubt on
the answer to part (i).
Chapter 12
?
●
1
(Page 280)
The population is made up of the member of
parliament’s constituents. The sample is a part
of that population of constituents. Without
information relating to how the constituents’ views
were elicited, the views obtained seem to be biased
towards those constituents who bother to write to
their member of parliament.
2
The population is made up of black residents in
Chicago. The sample is made up of black people
(and possibly some white people as the areas are
‘predominantly black’) from a number of areas in
Chicago.
?
●
(v)
Depends on method of data collection. If survey
is, say, via a postal enquiry, then a random
sample may be selected from a register of
addresses.
(vi)
Cluster sampling. Routes and times are chosen
and a traffic sampling station is established to
randomly stop vehicles to test tyres.
(vii)
Cluster sampling. Areas are chosen and
households are then randomly chosen.
(ix)
Cluster sampling. Meeting places for 18-yearolds are identified and samples of 18-year-olds
are surveyed, probably via a method to maintain
privacy. This might be a questionnaire to
ascertain required information.
(x)
Random sampling. The school student list is
used as a sampling frame to establish a random
sample within the school.
given that police officers are carrying out the
survey they are unlikely to obtain negative views.
(Page 281)
1
Each student is equally likely to be chosen but
samples including two or more students from the
same class are not permissible so not all samples are
equally likely.
2
Yes
S2
chosen to sample and speeds are surveyed.
the areas may not be representative of all
residential areas and therefore of all black people
living in Chicago and
(ii)
Stratified sample as in part (iii).
(viii) Cluster sampling. A period (or periods) is
The survey may be biased in two ways:
(i)
(iv)
Chapter 13
3
The population is made up of households in Karachi.
We are not told how the sample is chosen. Even if a
random sample of households were chosen the views
obtained are still likely to be biased as the interview
timing excludes the possibility of obtaining views of
most of those residents in employment.
Chapter 13
Exercise 13A (Page 293)
Activity 12.1 (Page 282)
1 (i)
There is no single answer since there are several ways
you could use the given random numbers to generate the
sample. This is one possible answer.
14 592
16 371
12 471
17 775
16 718
2595
2771
4598
7107
16 592
2
Exercise 12A (Page 283)
1 (i)
(ii)
2 (i)
Not all the totals have the same probability.
Possible method: writing each child’s name on a
piece of paper, folding them, putting them in a
hat and then drawing out one at random.
3
Cluster sampling. Choose representative streets
or areas and sample from these streets or areas.
(ii)
Stratified sample. Identify routes of interest and
randomly sample trains from each route.
(iii)
Stratified sample. Choose representative areas in
the town and randomly sample from each area as
appropriate.
Commuters are not representative of the whole
population.
(ii)
Adults who travel to work on that train
(iii)
Mean = 6.17 hours, variance = 0.657 hours
(i)
z = 1.53, not significant
(ii)
z = −2.37, significant
(iii)
z = 1.57, not significant
(iv)
z = 2.25, significant
(v)
z = –2.17, significant
(i)
0.3085
(ii)
0.016
(iii)
0.0062
(iv)
H0: µ = 4.00 g, H1: µ > 4.00 g
z = 3, significant
4
(i)
H0: µ = 72.7 g, H1: µ ≠ 72.7 g; two-tail test.
(ii)
z = 1.84, not significant
(iii)
No, significant
339
Answers
S2
5
6
(i)
H0: µ = 23.9°, H1: µ > 23.9°
(ii)
z = 1.29, significant
(iii)
20.6; the standard deviation 4.54 is much greater
than 2.3 so the ecologist should be asking
whether the temperature has become more
variable.
(i)
(ii)
(iv)
8
(ii)
The skulls in group B have greater mean lengths
and so a one-tail test is required.
(iii)
193.8
(iv)
x = 194.3, significant (194.3 193.8)
H0: µ = 1000, H1: µ < 1000
(iii)
z = –1.59, not significant
H0: µ = 0, H1: µ ≠ 0; z = 0.98, not significant
(ii)
H0: µ = 7.2, H1: µ < 7.2;
rejection region is z 6.76;
z = 6.525, significant (6.525 6.76)
The evidence supports the hypothesis that there
has been a reduction in the number of cars
caught speeding.
(iii)
A Type I error would have occurred if you say
that there had been a reduction in the number of
cars caught speeding when such a reduction had
not occurred.
5%
Type I error
16
16.2, 27.36
(ii)
H0: µ = 15, H1: µ > 15
(iii)
z = 1.986, not significant
H0: µ = 2, H1: µ < 2
(iii)
z = –1.68, not significant
17
104.7, 3.019
(ii)
H0: µ = 105, H1: µ ≠ 105
(iii)
z = −0.89, not significant
mean = 7.2
H0: µ = 22.0, H1: µ ≠ 22.0;
rejection region is z 21.698;
z = 21.7, not significant (just, 21.7 21.698)
There is not enough evidence to say that the mean
length has changed.
1.977, 0.017 33
(ii)
13 (i)
Mean = 6.525, variance = 2.871
15 (i)
998.6, 49.77
(ii)
12 (i)
A Type I error would have occurred if you say
that the sentence length is not the same (or the
book is not by the same author) when it is.
Probability = 5%
Choosing 36 consecutive days to collect data is
not a good idea because weather patterns will
ensure that the data are not independent. A
better sampling procedure would be to choose
every tenth day. In this way the effects of weather
patterns over the year would be eliminated.
Yes: z = –1.875, significant at the 5% level
.
(i) N(190, 5.3)
11 (i)
340
z = −2.284, significant
H0: µ = 50 kg, H1: µ < 50 kg;
10
(ii)
H0: µ = 14 nautical miles,
7
9 (i)
There is significant evidence to suggest that the
sentence length is not the same (or the book is
not by the same author).
You must assume that the visibilities are
normally distributed.
H1: µ < 14 nautical miles
(iii)
H0: µ = 21.2, H1: µ ≠ 21.2;
rejection region is z 19.7 or z 22.7;
z = 19.4, significant (19.4 19.7)
14 (i)
?
●
H0: µ = 3.2, H1: µ < 3.2, where µ is the growth rate of
the new grass in cm per week
(i)
z < 2.47
(ii)
m < 2.47
(Page 299)
It tells you that µ is about 101.2 but it does not tell you
what ‘about’ means, how close to 101.2 it is reasonable to
expect µ to be.
?
●
(Page 304)
You would expect about 90 out of the 100 to enclose 3.5.
Exercise 13B (Page 307)
(ii)
2 (i)
5.117 g, 5.293 g
47.7
(ii)
34.7 to 60.7
(iii)
27.3 to 68.1
3 (i)
(ii)
4 (i)
(a)
0.1685
(b)
0.0207
(iii)
385
6 (i)
(ii)
7 (i)
(ii)
11 (i)
µ = 227.1, σ2 = 265
78
Possible answers: cheaper, less time-consuming,
not all population destroyed if sampling is
destructive
(ii) (a)
$7790–$8810
163.8–166.6 cm
(ii)
10 (i)
0.9456
(ii)
5 (i)
8
5.205 g
0.223 to 1.403
Assumptions: the Central Limit Theorem applies and
s 2 is a good approximation for σ2.
The confidence interval suggests that reaction times
are slower after a meal.
6.83, 3.05
6.58, 7.08
5.71 to 7.49
It is more likely that the short manuscript was
written in the early form of the language.
1.838 mm
(ii)
1.63 to 3.91
(iii)
The coach’s suspicions seem to be confirmed as
4 mm is not in the confidence interval.
0.484 to 1.016
Assume that the sample standard deviation is an
acceptable approximation for σ.
The aim has not been achieved as the interval
contains values below 0.5.
(b)
12 (i)
68.0 to 70.6
90% of random samples give rise to
confidence intervals which contain the
population mean.
71.2 is not in the confidence interval so there
is a significant difference in the life-span
from the national average.
µ = 4.27, σ2 = 0.007 93
(ii)
4.253 to 4.287
(iii)
9
13 (i)
(ii)
14 (i)
mid-point = 0.145, n = 600
97.4
A random sample is one in which every item in
the population has an equal probability of being
selected.
(ii)
0.321 to 0.422
(iii)
2240
15 (i)
(ii)
S2
Chapter 13
1 (i)
9
0.244, 250
90%
341
Index
S1
S2
Index
Page numbers in black are in Statistics 1. Page numbers in blue are in Statistics 2.
alternative hypothesis 182, 183,
185–186, 288–291
arithmetic mean see mean
arrangement of objects 123–127,
140, 141–143
assumed mean 45–48, 51
average 14
back-to-back stem-and-leaf
diagrams 9
bar charts 57, 59
bias 279, 280
bimodal distribution 6, 17
binomial coefficients 132–136
binomial distribution 107, 143–145
expectation 146–147, 153
mean 146–147, 153
normal approximation
173–175, 178
Poisson approximation
216–221, 232
probability 144–145, 153
standard deviation 147, 153
using 147–150, 153
variance 147, 153
box-and-whisker plots 64, 65, 68, 76
342
categorical data 13, 51
Central Limit Theorem 287,
298–300, 301, 311
certainty 80
class boundaries
continuous random variables 235
grouped data 24–25, 29–30, 58
class intervals 24
cluster sampling 283
combinations 130–136, 140
complement of an event 81
conditional probability 94–100
confidence intervals 300–304, 311
converging 303–304
for proportions 306
sample size and 303, 305
theory 301–302
continuity corrections 172–173, 178
continuous data 51
grouping 29–30
representation 53–56
continuous random variables
class boundaries 235
mean 244–245, 247, 255
median 246–247, 248
mode 247–248, 255
probability density function
235–240
standard deviation 245
uniform (rectangular) distribution
249–251, 255
variance 244–245, 255
continuous variables 14
critical ratios 289
critical values/regions 189–191, 288,
290–291
cumulative frequency curves 65–67,
76
data
collection 3–4
continuous 29–30, 51, 53–56
discrete see discrete data
extreme values 5
grouped see grouped data
representation 4–9, 52–59
types 13–14
dependent events 89–90, 96–100
discrete data 51
grouping 25–26
mode 17
representation 56–59, 106
discrete random variables 106–111,
122
expectation 114–118, 122, 256, 276
functions of 257–260
mean 114–118, 122, 244, 256, 276
notation 107
outcomes 106
Poisson distribution 204–205
probability distribution 107–111,
114–118, 122
representation 107–110, 122
variance 114–118, 122, 244,
256, 276
discrete variables 13
normal distribution 172–173, 178
distribution 4
of sample means 287, 299–300, 311
shape 6–7
see also binomial distribution;
Poisson distribution; etc.
errors
hypothesis testing 196–199, 201,
227, 291
sampling 278
standard error of means 299, 311
Type I 196–199, 201, 227, 291
Type II 196–199, 201, 291
estimates 278, 291–292, 303
events 78
complement 81
dependent 89–90, 96–100
independent 88–89, 90, 96, 97, 104
mutually exclusive 84–85, 104
expectation 81–82
binomial distribution 146–147, 153
discrete random variables
114–118, 122, 256, 276
function of a random variable
257–259
see also mean
experiments 78
factorials 124–127, 140
frequency 24
frequency charts 54–55
frequency density 54–56, 59, 76
frequency distributions 19–22,
105–106
frequency tables 15, 19–22
grouped data 24–30
assumed mean 48
class boundaries 24–25, 29–30, 58
continuous 29–30
discrete 25–26
mean 26–28, 48, 51
median 27
representation 57–58
impossibility 80
independent events 88–89, 90, 96,
97, 104
independent random variables
addition/subtraction of 262–266
linear combinations 270
mean of sums 262–266,
269–272, 276
normally distributed 265–266, 276
variance of sums 262–266,
269–272, 276
interquartile range 63–64, 76
lower quartile 62–65, 76
mathematical models 106
mean 14–16, 18, 51
assumed 45–48, 51
binomial distribution 146–147, 153
continuous random variables
244–245, 247, 255
discrete random variables
114–118, 122, 244, 256, 276
estimated 26–27, 291–292
frequency distribution 21–22
function of a random variable
257–259
grouped data 26–28, 48, 51
hypothesis testing for 226–227,
287–293
normal distribution 155,
156–157, 161–165, 172–173,
174, 178, 224
notation 14–15
Poisson distribution 203,
206–207, 224, 226–227
population mean 278, 286–287,
291–292
sample mean 278, 286–287,
299–300
standard error of 299, 311
sums of random variables
262–266, 269–272, 276
uniform (rectangular) distribution
250, 251, 255
see also expectation
mean absolute deviation 35–36
measures of central tendency
14–18, 51
see also mean; median; mode
measures of spread 34–41, 51
see also standard deviation;
variance
median 16–17, 18, 51, 62, 76
continuous random variables
246–247, 248
frequency distribution 20
grouped data 27
modal class 51
modal group 6
mode 6, 17–18, 51
continuous random variables
247–248, 255
frequency distribution 20
mutually exclusive events 84–85, 104
normal distribution 154–165, 178
approximating binomial
distribution 173–175, 178
approximating Poisson
distribution 224–225, 232
discrete variables 172–173, 178
hypothesis testing using 287–293
interpretation of sample data
285–293
mean 155, 156–157, 161–165,
172–173, 174, 178, 224
normal curve 161–165, 178
probability 154–165, 178
standard deviation 155, 156–157,
161–165, 178, 224
standardised form 156, 161, 178
sums of random variables
265–266, 276
tables 155–160, 164
variance 174, 178, 224
null hypothesis 182, 183, 185–186,
288–291
numerical data 13–14
S1
S2
Index
histograms 6, 53–59, 76
hypothesis testing 180–199
alternative hypothesis 182, 183,
185–186, 288–291
checklist 183–184, 201
critical ratios 289
critical values/regions 189–191,
288, 290–291
errors 196–199, 201, 227, 291
mean of a Poisson distribution
226–227
null hypothesis 182, 183,
185–186, 288–291
one-tail tests 193, 195, 226, 288, 290
rejection regions 189–191, 291
sample data and 287–293, 311
significance level 182–183, 184,
185–186
steps 184, 201
two-tail tests 193–195, 226
Type I errors 196–199, 201,
227, 291
Type II errors 196–199, 201, 291
using normal distribution
287–293
one-tail tests 193, 195, 226, 288, 290
outcomes 78
outliers 5, 40–41, 51, 64–65, 76
parameters 278
parent populations see populations
Pascal’s triangle 132–133
permutations 129–130, 140
Poisson distribution 107, 202–232
approximating binomial
distribution 216–221, 232
conditions required 204, 207, 232
hypothesis test for mean 226–227
mean 203, 206–207, 224, 226–227
model suitability 202–203, 207
normal approximation 224–225,
232
probability 205–207, 232
recurrence relations 206
sum of distributions 210–213, 232
time intervals 206–207
variance 203, 207, 224
Poisson, Simeon 231
population mean 278, 286–287,
291–292
populations 277, 278, 280
probability 77–78, 104
binomial distribution 144–145, 153
conditional 94–100
discrete random variables
107–111, 114–118, 122
estimating 79–81
measuring 78
normal distribution 154–165, 178
of one event or another 82–85
Poisson distribution 205–207, 232
using binomial coefficients
133–136
probability density 233–234
probability density function
235–240
proportional stratified sampling 282
proportions, confidence intervals
for 306
343
Index
S1
S2
qualitative data 13
quantitative data 13–14
quartile spread 63–64, 76
quartiles 62–71
box-and-whisker plots 64, 65,
68, 76
interquartile range 63–64, 76
outliers 64–65
small data sets 62–63, 76
quota sampling 283
random numbers 281–282
random sampling 281–282, 284
random variables
continuous 233–255
discrete see discrete random
variables
functions of 257–260
independent 262–266,
269–272, 276
linear combinations 256–276
mean of sums 262–266,
269–272, 276
normally distributed 265–266, 276
standard deviation of sums 264
variance of sums 264–265,
269–272, 276
range 34–35, 51
raw data 4, 53
recurrence relations 206
rejection regions 189–191, 291
relative frequency 106
sample mean 278, 286–287, 299–300
sample statistic 278
samples 277, 280
sampling
bias 279, 280
cluster sampling 283
344
considerations 278–280, 284
hypothesis testing 287–293, 311
interpretation using normal
distribution 285–293
large samples 305
notation 278
quota sampling 283
random sampling 281–282, 284
reasons for use 278, 284
sample size 279–280, 303,
304–305
stratified sampling 282–283
systematic sampling 283
techniques 281–283, 284
terms 277–278
sampling distribution of means 287,
299–300, 311
sampling errors 278
sampling fraction 278, 281
sampling frame 278, 281
significance levels 91, 182–183, 184,
185–186
simple random sampling 281, 284
skewness 6–7, 8–9, 58
standard deviation 36–39, 51
binomial distribution 147, 153
continuous random variables 245
estimated 291–292, 303
normal distribution 155, 156–157,
161–165, 178, 224
outliers and 40–41
sample means 299, 311
sums of random variables 264
uniform (rectangular)
distribution 250
standard error of means 299, 311
stem-and-leaf diagrams 7–9, 51, 57
stratified sampling 282–283
systematic sampling 283
tallying 5, 6
tree diagrams 88–90, 98–99
trials 78, 143
two-tail tests 193–195, 226
Type I errors 196–199, 201, 227, 291
Type II errors 196–199, 201, 291
uniform distribution 6, 249
uniform (rectangular) distribution
249–251, 255
mean 250, 251, 255
standard deviation 250
variance 250, 251, 255
unimodal distribution 6
upper quartile 62–65, 76
variables 13
variance 36–37, 38–39, 51
binomial distribution 147, 153
continuous random variables
244–245, 255
discrete random variables
114–118, 122, 244, 256, 276
estimated 291–292
function of a random variable
259–260
normal distribution 174, 178, 224
Poisson distribution 203, 207, 224
sums of random variables
262–266, 269–272, 276
uniform (rectangular) distribution
250, 251, 255
Venn diagrams 81, 83, 84, 98
vertical line charts 106, 107–110,
122, 204