Numeracy
Outcome 3
[HIGHER]
Introduction
Section 1: Mean and Standard Deviation
Section 2: Median and Quartiles
Section 3: Transposition of Formulae
Section 4: Correlation and Regression
Section 5: Recurrence Relations
Tutor Assignments
3
5
25
49
87
113
139
In Outcome 3, answers for each section are at the end
of that section.
INTRODUCTION
The following is an extract from the National Unit Specification of standards.
Outcome 3
Apply in combination a wide range of numerical and statistical skills.
Performance criteria
(a)
Work with a mathematical concept.
(b)
Decide the steps to be carried out.
(c)
Carry out a number of sustained complex calculations.
This pack contains five sections. Each section deals with a topic that contains work at the
level required for Numeracy (Higher). The Tutor Assignments for each section are all at the
end of the pack.
You will have received guidance from your tutor as to which topics you should study; it is
doubtful if you will have to do them all.
Equally, your tutor may well give you topics which are not included in this pack.
Whichever topics you do, I hope you find them interesting.
Resources
You will need a scientific calculator.
Assessment
The assessment paper is likely to contain questions on a variety of topics. You will have to
answer a selection of these. Your tutor will give you guidance on this.
Good luck with the Outcome.
OUTCOME 3: NUMERACY/HIGHER
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OUTCOME 3: NUMERACY/HIGHER
MEAN AND STANDARD DEVIATION
SECTION 1
Measures of Central Location and Measures of Dispersion
When analysing data, we frequently have more numbers than we can comfortably cope
with, and very often the first job is to try to cut down the amount of information that we
have.
There are many ways of doing this, but one way is to try to summarise the entire data set by
using just a few numbers to represent it, so that these numbers can be used in further
calculations or quoted by people, such as politicians.
Needless to say, any old numbers won’t do, and over the years several different methods
have evolved. These fall into two main categories: measures of central location (or
tendency) and measures of dispersion.
Measure of central location: the mode
This is a very grandiose title for a very simple idea. If you could represent a whole set of
numbers by just one, what would it be?
Since it’s difficult to talk hypothetically, it will help if we have some actual numbers to play
with.
Suppose I stand outside the gates of a factory and ask the first 15 people who come out
what they earned for the their day’s work. Here is a possible set of answers, the figures all
rounded off to the nearest whole £:
45 52 39 52 63 60 52 58 54 47 41 52 55 58 52
What one number would you pick to represent this set?
Because they’re all jumbled up it’s difficult to make any sense of them at all. So we’ll start by
putting them in order of size, smallest to largest.
39 41 45 47 52 52 52 52 52 54 55 58 58 60 63
Let’s call this set A.
If we take the path of least resistance, and wish to make no calculations at all, the
representative number that jumps out is 52. This is because there are more 52s than there
are of any other number.
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MEAN AND STANDARD DEVIATION
In this case we say that £52 is the mode of the distribution.
• The mode is the member of the data set which occurs more often than any other
number.
The mode is by far the simplest measure of central location, which can be thought of as a
value around which the set of data is located.
Associated with measures of central location are the so-called measures of dispersion.
Measure of dispersion: the range
Measures of dispersion tell us how dispersed, or spread out, the data set is.
Here is another set of figures, giving daily earnings for a different sample of workers at the
same factory, already lined up in order of size:
25 28 35 38 42 47 52 52 52 59 64 67
We’ll call this set B. Although there are only 12 numbers in this data set, that doesn’t stop
us comparing them. The mode is still 52, but the numbers are spread out more.
This leads us to the simplest measure of spread – the range.
• The range of a set of data is the difference between the largest and smallest member of
the data set.
The range of set A is £63 – £39 = £24
The range of set B is £67 – £25 = £42
Set B is thus more spread out than set A, which you presumably had realised already, but at
least we now have a figure to represent the amount of spread.
These two measures, the mode and the range, are very rudimentary and unsophisticated.
But at least statisticians the world over know exactly what they are talking about when they
use these two words – mode and range – and so long as they are aware of the limitations of
these two measures, they can use them to good effect.
And the mode and range don’t need much effort to be found!
When teaching this topic in class, it is at this stage that I can usually sense a feeling of unrest
among my students, with elbow nudging and whispering. ‘What about the average?’ I hear
them say. ‘We did that in primary school.’
So let’s have a quick look at average next.
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MEAN AND STANDARD DEVIATION
Measure of central location: the mean
First of all, the word ‘average’ is such a commonly used expression, covering such a
multitude of sins, that statisticians have coined a more specific word to be used in its place.
That word is mean. The mean, like the mode, is a measure of central location.
• The mean of a data set is the total of all the values of the data set divided by the number
of members of the data set.
When the credits roll at the end of an episode of Taggart on TV, the singer is enthusing about
Glasgow being ‘no mean city’, in the sense of ‘no average city’ or ‘no ordinary city’ (but
those words wouldn’t fit with the music).
The mean is the average which you did, in fact, learn about in primary school.
The mean of set A is (39 + 41 + ... + 60 + 63) ÷ 15 = 780 ÷ 15 = £52. That this is
exactly the same as the mode is coincidence, and as much a surprise to you as it was to me
when I worked it out. Coincidences do occur. Sometimes they mean something, at other
times they don’t. This one doesn’t.
Activity 1a
What is the mean of set B?
Note that all our measures have a unit attached. In this case it is £s. It could have been $,
people, cm, seconds, elephants or whatever. But there is always a unit attached, so try to
get into the habit of using units.
The measure of dispersion usually associated with the mean is called the standard
deviation, and we will be looking at that in detail later in this section.
Measure of central location: the median
The third commonly used measure of central location is the median, and this is actually
gaining in popularity, mainly thanks to an eminent American statistician called Tukey
(pronounced Chookie) who died as recently as 2000.
We’ll look at the median in more detail later but, briefly, the median is the middle number of
a data set when the data is arranged in order of size.
If we look back at set A, the middle number is
39 41 45 47 52 52 52 52 52 54 55 58 58 60 63
There are seven numbers to the left of the circled £52 and seven to the right of it, so the
median is £52. We have £52 again – and this is not so much of a coincidence this time. It
certainly looks as if £52 could represent set A quite happily.
OUTCOME 3: NUMERACY/HIGHER
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MEAN AND STANDARD DEVIATION
Activity 1b
What is the median of set B?
Are you having difficulty working this out? What’s the problem? What’s a good compromise?
So which measure is better, the mean or the median?
Let’s go back to our original example of the workers at the factory gate. Suppose you ask
one extra person what their earnings were for that day, and you just happen to pick the
Boss, whose Rolls Royce is quietly purring its 6-litre supercharged engine round the corner.
‘Actually, I reckon I earned £350 for today’s efforts.’
Let’s look at our two sets now:
Set A:
39 41 45 47 52 52 52 52 52 54 55 58 58 60 63 350
Set B:
25 28 35 38 42 47 52 52 52 59 64 67 350
Activity 1c
Calculate the mean and the median of the new Set A and the new Set B. Comment on the
changes between the values you have just obtained and the old ones.
You have just met one of the main disadvantages of the mean – that one very extreme value
can pull it out of its more natural, representative position.
Some statisticians get round this problem by using a so-called truncated mean, which is
the mean of all the numbers except a percentage of them from the top and bottom, i.e. it
allows you to ignore outliers (extreme values at the top or bottom of the range).
The measure of dispersion associated with the median is called the semi-interquartile
range (or quartile deviation in some older textbooks). We’ll study this more closely in
Section 2.
To end this introduction, here is a summary of the advantages and disadvantages of the
three measures of central location.
The mean
Advantages
• It is the best-known measure of average.
• It can be calculated easily.
• It uses all the data.
• It is used in more complicated mathematical formulae, probably its most important
feature.
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OUTCOME 3: NUMERACY/HIGHER
MEAN AND STANDARD DEVIATION
Disadvantages
• It is affected by extreme values (as in the example above).
• It can give impossible values (e.g. 1.8 children).
• It need not be a member of the data set.
The median
Advantages
• It is not influenced by extreme values.
• It can be obtained even if some values are missing.
• It is usually represented by an actual data value.
Disadvantages
• It does not reflect the full range of values.
• It might not be typical of the data set if the set is very small or irregularly distributed.
• In grouped distributions it can only be estimated.
• It cannot be used in further statistical formulae.
The mode
Advantages
• It is not affected by extreme values.
• It is very easily found from a table or graph.
• The whole distribution is not necessarily required.
Disadvantages
• In grouped distributions it cannot be found exactly.
• There may be more than one mode.
• It ignores dispersion around the modal value.
• It cannot be used in further statistical work.
In classical statistics, i.e. statistics as developed since the middle of the nineteenth century,
the mean is probably the single most important statistic which is calculated and, when used
in conjunction with the standard deviation, it forms the basis of much statistical analysis and
testing.
Since the 1970s, however, a modern approach to statistics known as exploratory data
analysis has emerged. It puts great emphasis on the median, and there is much research
currently being undertaken into its applications.
OUTCOME 3: NUMERACY/HIGHER
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MEAN AND STANDARD DEVIATION
Using the Mean and Standard Deviation
We now look in more detail at the calculations which are required to find the mean and the
standard deviation. You will fully understand what’s going on after you have read through
Example 1a on pages 11–12.
As we have seen, the mean of a set of data values is the arithmetic average of those values,
i.e. the total of the values divided by the number of values there are.
The standard deviation measures how much the data is spread about. It tells you how far
away, on average, each member of the data set is from the mean.
The mean and standard deviation, used on their own just for one data set, give us a limited
summary of what is going on. However, we often have several data sets which we want to
compare, and there are various recognised procedures for making comparisons. The means
and standard deviations of the data sets play a crucial part in the calculations.
Notation
If the data set is a sample, which more often than not it is, we use the following notation in
the Latin alphabet (x is usually read as ‘x bar’):
Mean: x
Standard Deviation: s
If the data is a population, i.e. if it is all the data which exists about a certain situation, then
we use the Greek alphabet. This is an internationally recognised convention:
Mean: µ
Standard Deviation: σ
µ is pronounced ‘myoo’ and is simply the Greek letter m; σ is pronounced ‘sigma’ and is the
lower case Greek letter s.
(The capital Greek s, Σ (also pronounced ‘sigma’), is used to mean ‘the sum of’ and you will
meet it very soon – almost immediately in fact.)
Suppose you are analysing the wages paid to building trades people in a particular company.
If you use the wages of everyone in the company in your calculation then you are dealing
with a population and you use the Greek letters. On the other hand if all the data is
unavailable and you use only the wages of the workers on one particular site, then you are
dealing with a sample and you use the Latin letters.
(There is actually more to it than that – some of the formulae change depending on whether
it’s a sample or population, and also on what kind of comparisons we want to make, but you
don’t need to worry about this yet.)
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OUTCOME 3: NUMERACY/HIGHER
MEAN AND STANDARD DEVIATION
However, I will use x and s for the mean and standard deviation in all the answers from now
on, for simplicity.
Data in the form of a short set of (individual) raw scores
The formulae we use are:
Σx
Mean:
n
Standard Deviation:
Σ(x – x )
2
n
The formula for the mean tells you to add up all the data values (that’s the top line) and
divide by n which is the letter that represents the total number of values that there are.
The standard deviation formula is a bit harder to explain. We start off by subtracting the
mean (x) from each member of the data set (x). But this will give us some positive numbers
and some negative which will then add up to give us zero. So to get round this problem we
square the results of all our subtractions (i.e. we square all these ‘deviations from the
mean’), add up all these now positive numbers, then find their average by dividing by n.
However, our units are all wrong! Suppose we are analysing wages. The wages are in
‘pounds’ and we are at the moment calculating an average of ‘square pounds’ (?). So we take
the square root of the answer to bring our units back to the original.
The alternative definition of standard deviation is the root mean square deviation from
the mean, which is precisely what the formula is calculating.
Example 1a
Find the mean and standard deviation of the numbers
3, 7, 2, 7, 4
Solution:
First find the mean:
x=
Σx
3+7+2+7+4
23
=
=
= 4.6
n
5
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OUTCOME 3: NUMERACY/HIGHER
11
MEAN AND STANDARD DEVIATION
Next, make up a table like this:
In this column we
square each number
in the previous
column (i.e. multiply
it by itself)
In this column we
are subtracting the
mean (4.6) from
each number in
the first column.
V
V
3
7
2
7
4
TOTAL
–1.6
2.4
–2.6
2.4
–0.6
0
(x – x) 2
2.56
5.76
6.76
5.76
0.36
21.20
V
x – x
= x – 4.6
x
Notice there are no
negative numbers in
column 3. Squaring
any number always
gives us a positive
answer.
Now perform the calculation:
s =
Σ(x – x )2
n
=
21.20
=
5
4.24 = 2.06 (to 2 d. p.)
This means that, on average, each number is 2.06 away from the mean of 4.6.
(The 0 total in the table just helps to check that the column entries are correct.)
Data in the form of a simple frequency table
We use a frequency table in situations where each member of the data set occurs more than
once.
The data of the previous example could have been arranged as
a frequency table like this, although for such a simple example
it would have been a bit of a waste of time.
The column of xs shows the values of the variable x.
x
2
3
4
7
Total
The frequency (f) column shows the number of times each
value occurs.
We have to amend our formulae slightly
x or µ =
12
Σfx
Σf
OUTCOME 3: NUMERACY/HIGHER
s or σ =
Σfx 2
 Σfx 
–

Σf
 Σf 
2
f
1
1
1
2
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MEAN AND STANDARD DEVIATION
Note that fx means f × x and that fx2 means either fx × x or f × x 2 but emphatically NOT
fx × fx.
You need some fairly complicated maths to see that the formula for the standard deviation
calculates the same thing as the first formula did on page 11, but please take my word for it
that it does.
Example 1b
The table below shows the number of strokes required by golfers to get round a particular
course during the first round of a competition. (It must have been an easy course!)
No. of strokes (x)
No. of golfers (f)
67
15
68
23
69
38
70
27
71
25
72
19
73
16
74
10
75
4
Find the mean and standard deviation.
Solution
How do you apply the formulae to this table?
Start by writing the data in a column, not a row, and adding a column for fx and another
column for fx2.
Thus the formula for the mean gives (total number of strokes) ÷ (total number of golfers)
giving the average number of strokes per golfer.
Before you do the calculations, look at the data and see if you have some sort of idea as to
the answers to expect.
You see there is a quick build-up of numbers to 69, then a long tail to 75. You could make a
rough guess that the mean will be about 70.
The standard deviation is also easy to estimate. Take the range of your data (= the largest
minus the smallest member of the data set, in this case 75 – 67 = 8). In many situations the
standard deviation is, very roughly, between one third and one quarter of the range. 1/ 3 of 8
is 2.7 and ¼ of 8 is 2. So we would expect the standard deviation to be somewhere
between 2 and 2.7. And, if not, then not too far away.
OUTCOME 3: NUMERACY/HIGHER
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MEAN AND STANDARD DEVIATION
Now for the actual calculations:
Here we multiply
each number of the x
column by its partner
in the f column. The
first line means 15
golfers each scored
67. This makes a
total of 15 × 67 =
1005 strokes.
V
f
15
23
38
27
25
19
16
10
4
177
V
This is the total
number of golfers
in the first round.
x=
Σfx
12,432
=
= 70.24
Σf
177
fx
1,005
1,564
2,622
1,890
1,775
1,368
1,168
740
300
12,432
fx 2
67,335
106,352
180,918
132,300
126,025
98,496
85,264
54,760
22,500
873,950
V
V
V
x
67
68
69
70
71
72
73
74
75
TOTALS
This is the total
number of
strokes played by
everyone in the
entire first round.
2
s=
The last column
required by the
formula. The
67335 we get by
multiplying 1005
by 67,
alternatively, 15 ×
67 × 67 or 15 ×
67 2 .
V
This column shows
how many people
score how many
strokes. E.g. 15
people each score
67, then 23 people
each score 68 etc.
V
This column shows
that the lowest
number of strokes
for a round in the
competition is 67
and the highest
number is 75.
Σfx 2  Σfx 
–
 =
Σf
 Σf 
This total doesn’t
actually mean very
much but you need
it for the formula.
2
873,950 12,432 
–
 = 2.07
177
 177 
As you see, both tie in with our initial estimates. Estimates won’t help if you make a small
error in the calculations, but they do tell you if you are miles out.
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OUTCOME 3: NUMERACY/HIGHER
MEAN AND STANDARD DEVIATION
? 1A
NOTE: It can help considerably if you are familiar with your calculator and are able to use
the memory keys efficiently. You can then add each of your multiplications into the memory
and get the final column total without having to key all the numbers back in again. Check
your manual.
1.
Find the mode and the range of each set of numbers. Next, have a guess at what you
think the mean and standard deviation of each set is. Finally, calculate the mean and
standard deviation as shown in Example 1a.
(a)
(b)
2.
2, 5, 6, 6, 7, 10
1, 3, 5, 7, 7, 8, 8, 8, 9, 11
Below is a table which gives the breaking stress of each of 9 test cubes of concrete,
the units of measurement being N/mm2. Calculate the mean breaking stress and the
standard deviation. (This is also done as Example 1a; it just looks a bit different.)
Cube
Breaking stress
3.
B
25.2
C
21.7
D
23.8
E
25.9
F
20.3
G
21
H
26.6
I
22.4
In absorption tests on 200 bricks the following figures were obtained. Calculate the
mean absorption rate and the standard deviation. (Follow the procedure of Example
1b.)
% absorption
Frequency
4.
A
19
6
1
7
5
8
13
9
31
10
51
11
47
12
33
13
14
14
4
15
1
The number of hours worked each week by each of 100 workers in a construction
company was recorded as follows. Calculate the mean number of hours worked and
the standard deviation.
Hours per worker
Number of workers
40
3
41
4
42
8
43
12
44
17
45
21
46
19
47
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OUTCOME 3: NUMERACY/HIGHER
48
5
15
MEAN AND STANDARD DEVIATION
Data in the form of a grouped frequency table
We use a grouped frequency table if the range of values in the data set is far too large to
have each number shown individually in the frequency table. We have to group the values in
bundles, or class intervals (to use the correct terminology).
Example 1c
Here we see the wages of a bunch of employees, grouped in intervals of £50:
Wages (£)
Frequency (f)
100-150 150-200 200-250 250-300 300-350 350-400 400-500
17
25
34
47
26
15
3
The first interval is defined as £100–150. The £100 is called the lower limit and the £150
is called the upper limit. But look at the first two intervals. The first is £100–£150 and the
second is £150-£200. Thus the lower limit of the second is exactly the same (£150) as the
upper limit of the first.
So suppose an employee earns exactly £150. Which interval is that person in? Is he or she
one of the 17 in the first group, or one of the 25 in the second group?
The convention in such a situation is that the first interval includes everyone earning from
£100.00 to £149.99 inclusive, and the second interval from £150.00 to £199.99 inclusive
and so on.
In other words, the upper limit of an interval is exclusive and the lower limit of an interval is
inclusive. This convention only applies where the intervals are defined in such a way that
each one starts on exactly the same number as the previous one finished on.
Consider the first interval again. The 17 employees could all be earning as much as
£149.99 each or as little as £100.00 each, we have no way of knowing. So, in order to
make any calculations at all, we have to make a fairly big assumption – that they all earn an
amount that is in the middle of the interval. The £149.99 is so close to £150 as makes no
odds, so pretend they all earn £(100+150) ÷ 2 = £125.
We make the same assumption for all the other intervals, and all these mid-values become
the x of our formula. The mean and standard deviation thus found aren’t absolutely exact,
but (unless we are very unlucky) they are close enough to the real values to be usable.
In statistics, ‘near enough is good enough’!
We can only do the best we can with the information we have.
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OUTCOME 3: NUMERACY/HIGHER
MEAN AND STANDARD DEVIATION
Note that the last interval is slightly different (actually twice the width of the others), and
our last mid-value will thus be slightly different in appearance from the rest as well.
Before launching into calculations step back a bit and really LOOK at the data. Try to get a
feel for it. What is it trying to tell you?
Roughly what do you think the mean is? Off the top of my head I’d say it was about £250.
What about the standard deviation? Again we’re going to be operating with the mid-values
of the intervals.
The mid-value of the first interval is half of (100 + 150) = 125. The mid-value of the last
interval is half of (400 + 500) = 450.
So the range of these is 450 – 125 = 325.
Divide this range by 3 (approx 108) and divide the range again by 4 (approx 80). So a
standard deviation should be somewhere around 80 to 108. Let’s see!
The x of our formula is
the mid-value of each
interval.
Add together the
lower and upper limit,
then halve the answer.
These three columns are exactly
as before.
17 × 125 = 2,125
then
2,125 × 125 or 17 × 1,252
(either of them) equal 265,625.
V
V
V
There is no point in
adding this column.
f
17
25
34
47
26
15
3
167
V
Mid-val x
125
175
225
275
325
375
450
V
Wages
100-150
150-200
200-250
250-300
300-350
350-400
400-500
TOTALS
fx
fx 2
2,125.0
265,625.00
4,375.0
765,625.00
7,650.0 1,721,250.00
12,925.0 3,554,375.00
8,450.0 2,746,250.00
5,625.0 2,109,375.00
1350.0
607,500.00
42,500.0 11,770,000.00
V
V
V
Find these column
totals as usual.
OUTCOME 3: NUMERACY/HIGHER
17
MEAN AND STANDARD DEVIATION
Now for the formulae:
x=
42,500
Σfx
=
= £254.49
167
Σf
2
2
11,770,000  42,500 
Σfx 2  Σfx 
–
–
 =
 = £75.59
167
f
Σf
Σ
 167 


s=
The mean we have calculated is close to our estimate. The standard deviation is lower than
we estimated but not miles away, so we are happy with the answers.
? 1B
Read the important note on page 20 before looking up the answers.
1.
The table below represents the weights of a sample of 100 male students at a
college. Find the mean weight and the standard deviation.
Weight (kg)
Frequency
2.
60-64
5
76-80
8
20-25
5
25-30
17
30-35
52
35-40
18
40-45
7
45-50
1
470-480 480-490 490-500 500-510 510-520 520-530 530-540
17
28
63
105
87
35
30
The table below shows the ages of passengers on a cruise ship. Find the mean age
and the standard deviation. Note: the last interval is ‘open-ended’ and theoretically
could go all the way up to 100 or more. But we use the convention that it is the
same width as the one next to it.
Age, x
Number, f
18
72-76
27
The daily output of components produced by a machine was logged over a period of
a year. The table below gives the results. Find the mean daily output and the
standard deviation.
Daily output, x
No. of days, f
4.
68-72
42
The following table gives the moisture content in 100 test samples of a particular
cement mix after a certain time interval. Calculate the mean moisture content and
the standard deviation:
% Moisture content
Number of samples
3.
64-68
18
18-25
46
25-35
157
OUTCOME 3: NUMERACY/HIGHER
35-50
218
50-65
308
65-75
146
75 & over
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MEAN AND STANDARD DEVIATION
The best uses of the mean and standard deviation are in an area of statistics called
inferential statistics or hypothesis testing, which is not covered in this section.
However, it is worth mentioning a few relevant points about this.
Most people intuitively understand what the mean is. Standard deviation is another matter.
The standard deviation of one frequency distribution is fairly meaningless on its own. The
meaning only becomes apparent if we have more than one distribution to look at.
Here are three distributions, each presented in the form of a histogram, with their means
and standard deviations calculated.
(A histogram is a special type of bar chart. The class limits (as you have studied them so
far) appear at the boundaries of the bars. There are no spaces between the bars. Ask your
tutor if you want to know more.)
Each histogram is for 176 people arranged by age in 5-year class intervals.
In this histogram the data is well
spread out, with almost equal
numbers in each interval.
70
60
50
40
Mean = 28.01 years
30
20
St. Dev. = 9.48 years
10
0
In this histogram, the data is more
‘normal’ with fewer people at the
ends and more people bunched in
the middle intervals.
Mean = 26.79 years
10
15
20
25
30
35
10
15
20
25
30
35
40
45
70
60
50
40
30
20
St. Dev. = 6.65 years
10
0
40
45
The means are approximately the same but the standard deviation of the second
distribution is smaller than that of the first one, showing less spread (or dispersion, to use
the technical jargon).
OUTCOME 3: NUMERACY/HIGHER
19
MEAN AND STANDARD DEVIATION
In this histogram the spread of the
data is very small, with well over
90% of the people in two class
intervals. The range of the data is
only 20 years (35 – 15 = 20).
90
80
70
60
50
40
30
Mean = 24.97 years
20
10
St. Dev. = 3.11 years
0
15
20
25
30
35
As you see, the standard deviation is considerably smaller now.
The larger the standard deviation, the more spread out the data set is, the further away
each one is, on average, from the mean.
The smaller the standard deviation, the less spread out the data is, the closer each number
is, on average, to the mean.
*
Important note
If you have already seen the calculation of standard deviation in, say, a Higher Mathematics
course you may well have seen a slightly different formula for it. This is because there is a
very slight technical problem which I will explain but which will not really concern you
unless your job happens to depend on your knowledge of statistics.
One important use of the science of statistics is to collect samples, which are small but
hopefully representative pieces of data about a much larger population, and then come to
some conclusion about the population from calculations made on the sample.
For instance, an opinion poll shows that 60% of a sample of 1,028 people in a constituency
plan to vote for party X in a local by-election, and so the conclusion is made that about 60%
of the entire electorate will vote that way too.
Various mathematical formulae are used in this branch of statistics called inferential
statistics. Many of them depend on using the standard deviation of a sample as an estimate
for the standard deviation of the population. But the standard deviation of the sample is
usually slightly smaller than the standard deviation of the population and this knocks the
sums out. And the smaller the sample, the worse it gets.
So the sample standard deviation is adjusted to make it slightly higher to alter the
calculations to make the predictions more reliable.
20
OUTCOME 3: NUMERACY/HIGHER
MEAN AND STANDARD DEVIATION
This can be done in two ways.
One way is to calculate the true standard deviation (which is what we have been doing) and
then adjust it if necessary by using a measure called Bessel’s correction factor. (You don’t
need to know what this is about.)
The other way is to adjust the standard deviation at source by using a slightly different
formula, the reasoning being that, since the majority of data are, in fact, samples, many of
which are going to be used for inferential purposes anyway, you might as well adjust the
standard deviation before you start.
This is not an argument which is relevant here. I have to point this different formula out,
though, in case you have come across it in other textbooks and are confused.
Here are the two sets of formulae for reference. Decide yourself which set looks easier to
use.
Σ(x - x )
2
True standard deviation:
n
Σ(x - x )
2
or
2
Adjusted version:
n –1
Σx 2  Σx 
–   or
n
 n 
1  2 ( Σx )
 Σx –
n –1
n
2
or
Σfx 2  Σfx 
–

Σf
 Σf 

 or


2
1  2 ( Σfx )
 Σfx –
Σf –1
Σf
2




All answers in this section have been calculated using the true
standard deviation formulae, which are the ones in the top row.
OUTCOME 3: NUMERACY/HIGHER
21
MEAN AND STANDARD DEVIATION
Answers
Activity 1a
Mean of set B = £(25 + 28 + 35 + ... + 67)/12 = £561/12 = £46.75
Activity 1b
Median of set B:
25
28
35
38
42
47
V
52
52
52
59
64
67
Because there are 12 numbers (i.e. an even number as opposed to odd) there is no middle
term. By convention we take the average of the two middle numbers, in this case (47 +
52)/2 = 99/2, so the median is £49.50.
Activity 1c
Mean of new set A = £1130/16 = £70.63
Median of the new set A is the average of the last two 52s, i.e. £(52 + 52)/2 = £52
Mean of new set B = £910/13 = £70
Median of new set B is the first of the 52s = £52
To summarise:
Mean
Median
Old A
£52
£52
New A
£70.63
£52
Old B
£46.75
£49.50
New B
£70
£52
As you see, the mean is considerably increased by the inclusion of an extra large number,
whereas the median only moves along ‘half a space’ and thus changes either only a little bit
or not at all.
There is more about the median in another section of the study pack.
22
OUTCOME 3: NUMERACY/HIGHER
MEAN AND STANDARD DEVIATION
?1A: Answers
1.
(a)
Mode = 6
Range = 10 – 2 = 8
Mean = 36/6 = 6
Σ(x – x )
2
s=
n
34
6
= 2.38
=
(b)
Mode = 8
Range = 11 – 1 = 10
Mean = 67/10 = 6.7
Σ(x – x )
2
s=
78
10
= 2.79
=
2.
205.9
9
= 22.9 N/mm2
x=
56.06
9
= 2.5 N/mm2
s=
n
x
2
5
6
6
7
10
Total
x–x=
x–6
–4
–1
0
0
1
4
check = 0
(x – x)2
16
1
0
0
1
16
34
x
1
3
5
7
7
8
8
8
9
11
Total
x–x=
x – 6.7
–5.7
–3.7
–1.7
0.3
0.3
1.3
1.3
1.3
2.3
4.3
check = 0
(x – x)2
32.49
13.69
2.89
0.09
0.09
1.69
1.69
1.69
5.29
18.49
78.10
x
19.0
25.2
21.7
23.8
25.9
20.3
21.0
26.6
22.4
205.9
x–x=
x – 22.9
–3.9
2.3
–1.2
0.9
3
–2.6
–1.9
3.7
–0.5
(x – x)2
15.21
5.29
1.44
0.81
9
6.76
3.61
13.69
0.25
56.06
OUTCOME 3: NUMERACY/HIGHER
23
MEAN AND STANDARD DEVIATION
3.
4.
x
6
7
8
9
10
11
12
13
14
15
Totals
f
fx
1
5
13
31
51
47
33
14
4
1
200
6
35
104
279
510
517
396
182
56
15
2,100
fx2
36
245
832
2,511
5,100
5,687
4,752
2,366
784
225
22,538
x=
2,100
= 10.5%
200
22,538  2,100 
–

200
 200 
= 1.56%
2
s=
∑f = 100;
∑fx = 4,460;
∑fx2 = 199,284
x = 44.6 hours; s = 1.92 hours
?1B: Answers
As you will have seen by now, the calculations can get tortuous, with numbers running to 6
or more digits.
The best advice for someone doing these on a hand-held calculator is to learn how to use
the memory function. This will allow you to find a column total without having to key in all
the numbers in the column all over again: well worth the effort.
1.
Using 62, 66, 70, 74, 78 as mid-values we get
∑f = 100;
x = 70.6 kg;
2.
∑fx2 = 499,952
Using 22.5, 27.5, 32.5, etc. as mid-values we get
∑f = 100;
x = 32.9%;
3.
∑fx = 7,060;
s = 3.89 kg
∑fx = 3,290;
s = 4.78%
∑fx2 = 110,525
Using 475, 485, 495, etc. as mid-values we get
∑f = 365;
∑fx = 185,095; ∑fx2 = 93,944,325
x = 507.11 components;
s = 14.89 components
4.
Using 21.5, 30.0, 42.5, 57.5, 70.0, and 80.0 as mid-values we get
∑f = 898;
∑fx = 44,734;
x = 49.82 years;
24
∑fx2 = 2,437,251
s = 15.25 years
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
SECTION 2
Another common measure of central tendency is the median.
The median is the middle member of a data set when the values are all arranged in order
of size, smallest to largest.
It is fairly obvious why the data must be arranged in order of size.
Suppose we have seven people of different ages: 23, 42, 35, 18, 50, 27, 62.
The age of the person in the middle is 18.
Suppose we now arrange them in a different way: 35, 50, 18, 62, 27, 23, 42. The
middle age is now 62, different from last time.
So we MUST put the ages in order of size first, before saying which is the middle one:
18
23
27
35
42
50
62
and then we isolate the median:
Median is 35 years
V
18
V
23
V
27
We have the same
number of values on
this side of the
median...
35
42
50
V
62
V
...as we have on
this side
Notice that the term median applies to the variable. Thus it’s not the person who is 35 that
is the median, it’s the age of 35 years that is the median.
If there is more than one value which is the same, line them all up as before. The median
may well not be an individual, but it’s a compromise we just have to make.
For instance, here is a list of the numbers of passengers in cars passing a traffic control
point: 1, 3, 2, 0, 2, 1, 2, 0, 1, 3, 4, 1, 0, 2, 0, 1, 1, 0, 3, 4, 1, 2, 1
OUTCOME 3: NUMERACY/HIGHER
25
MEDIAN AND QUARTILES
There are 23 numbers in that list, so line them up and the median will be in the 12th
position so that there are 11 on either side of it, even though some of the 11 may well be
the same as the median:
0
0
0
0
0
1
1
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
The median number of passengers is 1, or 1 passenger is the median number of passengers.
What happens if we have an even number of numbers? In the spirit of compromise we
average the two numbers on either side of the median position, like this:
Here someone has counted the number of matches in 10 boxes which are supposed to have
‘average contents 40’: 42, 40, 38, 39, 40, 42, 41, 39, 38, 40
First line them up in order: 38
38
39
39
40
40
40
41
42
42
Then find the median:
Median is the average of these two
½(40 + 40) = ½ × 80 = 40
so median is 40
V
38
V
38
39
39
V
40
40
40
41
V
The same number
on this side...
42
42
V
...as on this side
Some textbooks give you a formula which can use to find the position in line (or rank) of the
median:
The rank of the median is
1
2
( n +1)
Thus if there are 11 numbers in line, the median rank is
1
2
(11+1) = 12 ×12 = 6
i.e. the median is the sixth number in line (from either end, though we do usually count from
the bottom up).
And if there are 12 numbers in line, the median rank is
1
2
26
(12 +1) = 12 ×13 = 6.5
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
Now, you may well say there is no such number as the ‘6.5th’ number, but what this means
is ‘the average of the 6th and 7th numbers’.
In the same way, a rank of 8.5 means ‘the average of the 8th and 9th numbers’, and a rank of
19.5 means ‘the average of the 19th and 20th numbers’.
In the list above, the numbers on either side of the median position were both the same (i.e.
both 40) but if they were different you would still follow the procedure.
So if we have the numbers 17 19 12 24 22 26 30 18 as being the number of
cars passing a checkpoint on the road within a fixed interval of time, the median would be:
12
17
18
19
22
24
26
30
V
Median is the average
of these two numbers
and so the median is 20.5. Thus the median is not itself a member of the data set. This is
not a problem. The mean (arithmetic average) is almost never in the data set either.
Now we know that there is no such thing as 20.5 cars passing a point, but we can use this
number as a measure and compare it with the median derived from other checkpoints, or
from the same checkpoint at other times.
What it means is during half of the time intervals there were less than 20.5 cars passing the
checkpoint and during the other half of the intervals there were more than 20.5 cars
passing the checkpoint.
Why do we need a median anyway?
This has already been discussed in the mean and standard deviation section but a quick
second look will do no harm.
The mean (or arithmetic average) of a set of numbers is all very well if there are no ‘rogues’
which are abnormally different from the rest. But where this does happen, the mean gets
pulled out of its natural position and gives a false impression.
Suppose you have the set of numbers, written in order:
10
10
13
14
18.
The mean (average) is (10 + 10 + 13 + 14 + 18) ÷ 5 = 13.
The median is the middle one of these numbers: 13 is in the middle of the five of them. By
sheer coincidence it has turned out to be the same as the mean.
OUTCOME 3: NUMERACY/HIGHER
27
MEDIAN AND QUARTILES
Now let’s add another number which is much bigger than the others: 10
18 95.
10
13
14
The mean is now (10 + 10 + 13 + 14 + 18 + 95) ÷ 6 = 26.7. As you see, it has moved
quite dramatically upwards. Is it now a good representative of the data set? No.
What about the median? Well, we have an even number of values so we average out the two
middle ones: (13 + 14) ÷ 2 = 13.5. This is much closer to the spirit of the data set than
the mean’s 26.7, and so is a better measure of ‘average’.
2.1 Quartiles from a Simple Data Set
As with the mean, there is a measure of dispersion associated with the median, but before
we can find this we have to know about quartiles.
Quartiles are those members of the data set which split the data (when taken in order of
size) into quarters. Thus we have:
Smallest
number in
the data set
Lower
Quartile
Q1
Q2
25% of the data
Upper
Quartile
25% of the data
Q3
Largest
number in
the data set
25% of the data
}
}
}
}
25% of the data
Middle Quartile
= MEDIAN
The procedure is to find and isolate the median, and then to find the median of each half of
the remaining data. This ensures that the same number of data values lie within each
‘quarter’ of the data set.
A few examples will help. Each X represents a member of the data set. the positions of the
quartiles are marked. The median is, of course, the same as the middle, or second, quartile.
X
X
28
X
X
X
X
X
X
X
X
X
V
V
V
Q1
Q2
Q3
X
X
X
X
X
X
X
X
V
V
V
Q1
Q2
Q3
OUTCOME 3: NUMERACY/HIGHER
X
X
X
X
MEDIAN AND QUARTILES
X
X
X
X
X
X
X
X
X
X
X
V
V
V
Q1
Q2
Q3
X
X
Example 2.1a
Find the three quartiles of 1.82 1.57 1.64 1.85 1.70 1.75 1.83
where these numbers represent the heights in metres of nine students.
1.80
1.81
Solution:
Start by lining the numbers up in order. Since there are nine numbers, the median will be
the 5th number.
1.57
1.64
1.70
1.75
1.80
1.81
1.82
1.83
1.85
Q2
We now take the numbers on either side of the median, not including the median, and find
the middle of each.
Left side
1.64 1.70
1.57
1.75
1.81
Right side
1.82 1.83
1.85
V
V
Q1 = ½(1.64 + 1.70) = 1.66
Q3 = ½(1.82 + 1.83) = 1.825
The lower quartile is 1.66m, the middle quartile is 1.80m, the upper quartile is 1.825m.
Once we have found the actual quartiles we calculate the quartile deviation (or the semiinterquartile range: SIQR for short) using the formula
1
2
(Q3 – Q1)
This is because it shows how much, on average, each quartile differs from the median:
Distance of lower
quartile from median
V
V
Distance of upper
quartile from median
(Q3 – Q 2 ) + (Q2 – Q1 )
Q – Q 2 + Q 2 – Q1
Q – Q1
= 3
= 3
2 V
2
2
Add up and divide by
2 to find the average
OUTCOME 3: NUMERACY/HIGHER
29
MEDIAN AND QUARTILES
The semi interquartile range is
1
2
(1.825 – 1.66) =
1
2
× 0.165 = 0.0825m
This means that the upper and lower quartiles are, on average, 0.0825 metres away from
the median.
You will realise that
(Q3 – Q1 )
1
= is the same as (Q3 – Q1 )
2
2
Here’s another example.
Example 2.1b
Find the quartile deviation of the weights of this sample of parcels (units are kilograms).
18
25
22
17
21
19
21
19
20
13
17
25
20
19
Solution:
Line up in order of size, then find the positions of the quartiles (always find the median first).
13
17
17
18
19
19
19
V
20
20
21
21
V
22
25
25
}
}
}
}
V
V
Q1
V
V
Q2
V
Q3
As you see there are the same number of values (in this case
3) between each quartile position and also from the ends.
The median is mid-way between 19 and 20, so it is 19.5kg.
The lower quartile is 18kg and the upper quartile is 21kg.
The semi-interquartile range (or quartile deviation) is half of (21 – 18)kg, i.e. it is 1.5kg.
This means that the upper and lower quartiles are, on average, 1.5kg away from the
median.
30
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
? 2.1
Find the median and the semi-interquartile range of each set of numbers.
1.
These are hourly rates of pay in £s, to the nearest whole £, paid to employees of a
building firm.
16
2.
15
11
16
17
18
16
13
20
16
19
15
16
11
16
These are times taken in minutes for samples of paint to dry.
120 125 130 126 127 122 131 125 128 129 123 125
3.
These are lengths of bits of wire (in cm) cut by an erratic machine.
2.1 2.0 2.5 2.3 2.7 2.5 2.4 1.9 1.8 2.5 2.7 2.3 2.5
2.2 Quartiles from a Simple Frequency Table
If your data set is very small, it’s easy enough to line them up in order and count out the
positions of the quartiles. But what happens if there are 50 or 100 numbers and not just a
dozen or so?
Example 2.2a
Shoe size x
No. of pairs f
5
2
5½
5
6
8
6½
13
7
10
7½
4
8
1
The table above shows how many pairs of shoes of different sizes a shop has. Find the
median, quartiles and semi-interquartile range.
Solution:
There are 43 pairs altogether. The median is therefore the size of the 22nd pair when they
are all lined up in order of size. And the table does actually line them up like this.
There are 21 pairs of shoes on either side. The middle one of each is the 11th.
So we have the following positions for the quartiles:
1st 2nd ... 10th
11th
12th ... 21st
22nd
23rd ... 32nd
33rd
34th ... 43rd
TEN NUMBERS
Q1
TEN NUMBERS
Q2
TEN NUMBERS
Q3
TEN NUMBERS
To save us doing a lot of counting, we incorporate a cumulative frequency (c.f.) column in
our table (conventionally written as a column, but it could just as well be in a row). The
cumulative frequency is just a running total. It tells us how many numbers we have passed
as we count along the line.
OUTCOME 3: NUMERACY/HIGHER
31
MEDIAN AND QUARTILES
Shoe size
x
5
5½
6
6½
7
7½
8
Total
Pairs
f
2
5
8
13
10
4
1
∑f = 43
c.f.
2
7
15
28
38
42
43
2 + 5 = 7. This means 7 pairs were size 5.5 or less.
7 + 8 = 15. This means 15 pairs were size 6 or less.
15 + 13 = 28. This means 28 pairs were size 6.5 or less.
28 + 10 = 38. This means 38 pairs were size 7 or less.
38 + 4 = 42. This means 42 pairs were size 7.5 or less.
42 + 1 = 43. This means 43 pairs were size 8 or less.
Now look down the cumulative frequency (c.f.) column. Where a c.f. number is just larger
than a quartile position, the variable number (x) it refers to is the quartile.
Shoe size
x
5
5½
6
6½
7
7½
8
Total
Pairs
f
2
5
8
13
10
4
1
∑f = 43
The lower quartile is the 11th number. We
c.f.
2
7
15
28
38
42
43
pass 11 in this line. Thus Q1 = shoe size 6.
The middle quartile (median) is the 22nd
number. We pass 22 in this line. Thus Q 2 =
shoe size 6½.
The upper quartile is the 33rd number. We
pass 33 in this line. Thus Q3 = shoe size 7.
We can now find the quartile deviation (semi-interquartile range, SIQR for short)
SIQR =
1
2
(7 – 6) = 0.5
That perhaps took a long time to explain, but it’s actually done very quickly.
Finding the exact positions of the quartiles like this is a palaver. But there is a formula for it!
And you can use the formula to find the positions of pentiles (where the data set is split into
fifths) and deciles (split into tenths) and percentiles (split into hundredths) so it’s worth a
mention.
The exact position (or rank) of the pth q-tile of n observations is given by
 pn 
pn
INT   + 1 but this formula is ONLY used if
is NOT a whole number
q
q
 
where INT means the integral, or whole number part.
In this example we have 43 observations (so n = 43) and we want quartiles (so q = 4) and p
takes the values 1 (for Q1), 2 (for Q2) and 3 (for Q3) in turn.
32
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
1× 43 
Q1: = INT 
 = INT(10.75) = 10, so rank of Q1 is 10 + 1 = 11
 4 
 2 × 43 
Q2 : = INT 
 = INT(21.5) = 21, so rank of Q2 is 21 + 1 = 22
 4 
 3 × 43 
Q3 : = INT 
 = INT(32.25) = 32, so rank of Q3 is 32 + 1 = 33
 4 
and then the exact values of the quartiles can be found as before.
(If we want pentiles, q = 5; if we want deciles, q = 10 and so on.)
Example 2.2b
The table below gives the ages of members of a youth club. I’ve put the cumulative
frequency column in right away. Find the quartiles.
Age x
13
14
15
16
17
18
19
Total
f
23
26
28
22
10
9
2
∑f = 120
c.f.
23
49
77
99
109
118
120
Reminder:
23 + 26 = 49
49 + 28 = 77
77 + 22 = 99
and so on
The last number (120) should be the same
as the total of the f column.
Solution:
Calculate the ranks of the quartiles:
1×120
= 30 (a whole number) so rank of Q1 is average of 30th and 31st numbers, i.e. 30.5
4
2 ×120
= 60 (a whole number) so rank of Q2 is average of 60th and 61st numbers, i.e. 60.5
4
3 ×120
= 90 (a whole number) so rank of Q3 is average of 90th and 91st numbers, i.e. 90.5
4
Now check when we pass each of these rankings as we move down the c.f. column;
Age x
13
14
15
16
17
18
19
Total
f
23
26
28
22
10
9
2
∑f = 120
c.f.
23
49
77
99
109
118
120
Pass 30.5 here, so Q1 = 14 years
Pass 60.5 here, so Q2 = 15 years
Pass 90.5 here, so Q3 = 16 years
OUTCOME 3: NUMERACY/HIGHER
33
MEDIAN AND QUARTILES
The median age is thus 15 years. The SIQR is half of (16 – 14) i.e. 1 year. This means that,
on average, the upper and lower quartiles are 1 year away from the median.
One quarter of the members are below 14 years old; a quarter are between 14 and 15; a
quarter are between 15 and 16; and the oldest quarter are between 16 and 19.
It is just coincidence that the three quartiles all find themselves next to each other like this.
Two of them could easily be the same, it just depends on how the numbers are spread out.
Just in case you think I’m being too finicky about the ranks of the quartiles, here’s a
manufactured example to show you why this is necessary.
I’ll also stop using the expression ‘whole number’ and use the more correct ‘integer’. The
difference is imortant to mathematicians but not so much so to normal people (!). A whole
number is, technically, 0, 1, 2, 3, 4, and so on. Integers include all of these plus negative
whole numbers, i.e. -1, -2, -3 and so on.
Example 2.2c
The table below shows the final golf scores in a tournament of 214 players. Find the
quartiles and the semi-interquartile range.
Solution:
We can calculate the rankings of the quartiles before going to the table, given that we know
the value of n.
V
1× 214
= 53.5, not an integer, so rank of Q1 is 53 + 1 = 54
4
2 × 214
= 107, an integer, so rank of Q2 is 107.5
4
(rank Q2 is average of 107 and 108)
3 × 214
= 160.5, not an integer, so rank of Q3 is 160 + 1 = 161
4
Strokes
274
275
276
277
278
279
280
281
282
Total
f
1
9
15
32
50
25
20
47
15
214
c.f.
1
10
25
57
107
132
152
199
214
We pass 54 here, so Q1 = 277 strokes.
Aha! The rank of Q2 is 107.5 but the table
clearly shows that the 107th person is the
last one to score 278. So the 108th person
is the first to score 279. The median is the
average of 278 and 279, i.e. 278.5 strokes.
We pass 161 here, so Q3 is 281 strokes.
The semi-interquartile range is thus (281 – 277)/2 = 2 strokes.
34
OUTCOME 3: NUMERACY/HIGHER
pq
is exactly
n
an integer we
add 0.5 to it to
get the rank,
instead of 1.
If
MEDIAN AND QUARTILES
? 2.2
In each question find the median, the lower and upper quartiles, and the semi-interquartile
range.
1.
The number of faulty items made per day over a period of 30 days is catalogued as
follows.
Faults x
Frequency f
2.
2
3
3
7
4
4
7
3
9
2
10
1
11
3
0
10
1
21
2
15
3
13
4
8
5
5
6
2
7
0
8
1
For a statistics project, school pupils noted how many people there were in cars
passing the school gates during a certain time interval in the morning.
People in car x
No. of cars f
4.
1
2
During a 75 day period a Fire Brigade attended incidents per day as follows:
Incidents x
Frequency f
3.
0
5
1
28
2
35
3
20
4
8
5
5
6
1
This table shows the lengths of components produced by a machine,
Length of component x (mm)
Number of components f
73.0 73.1 73.2 73.3 73.4 73.5 73.6 73.7
25
34
67
100 178 213 104 35
We now look at grouped frequency tables and compare the SIQRs of various related
distributions.
2.3 Grouped Frequency Tables
If the range of data is so great that we cannot conveniently or sensibly make up a simple
frequency table, where each and every data value is represented, we need to group the data
to be able to illustrate and analyse it meaningfully.
This applies particularly in situations where the variable is a continuous variable, i.e. it can
take any value within a given range, to umpteen decimal places if necessary. In particular,
measurements fall into this category.
OUTCOME 3: NUMERACY/HIGHER
35
MEDIAN AND QUARTILES
Example 2.3a
Wages (£)
150-200
200-250
250-300
300-350
350-400
400-450
Total
No. of Employees
10
25
39
18
9
5
106
Here we have the wages of 106 people in a place of
work. Each of the 106 could be earning a different
amount. We can’t have a table with 106 entries! So
we group the data in class intervals.
You’ll notice that the number 200 is represented in
both the first and second intervals. By convention, a
wage of exactly 200 will be placed in the second
interval. So the first interval contains £150.00 to
£199.99 inclusive, the second interval contains
£200.00 to £249.99 inclusive and so on.
Find the quartiles and the semi-interquartile range.
(We have to make a rather large assumption: that the wages are spread evenly within each
interval. This assumption could be totally wrong, of course, but the numbers may be
squashed at the end of one interval in some places and spread out in others. It’s a matter of
swings and roundabouts. In the grand scheme of things everything evens out at the end.
Any answers we get will be close enough to be used in further analysis. They will only be
approximate, but they’ll be the best we can do. As the great (recently deceased) statistician
John W. Tukey said, ‘It is better to have the approximate answer to the right question than
the accurate answer to the wrong question.’)
Solution:
As usual, we start by making a column of cumulative frequencies, then calculate the ranks of
the quartiles to find where they lie. But since we know that any answers we get will be
approximate, we don’t bother with trying to pin down the quartiles exactly. An
approximate position will do.
We simply find a quarter, a half, and three quarters, of the total number 106.
Rank of Q 1 = ¼ of 106 = 26.5
Rank of Q2 = ½ of 106 = 53
Rank of Q 3 = ¾ of 106 = 79.5
36
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
Wages (£)
150-200
200-250
250-300
300-350
350-400
400-450
Total
No. of
Employees
10
25
39
18
9
5
106
Scanning down the cumulative frequency
column as before,
Cumulative
Frequency
10
35
74
92
101
106
We pass 26.5 here, so Q 1 is in the 200250 group
We pass 53 here so Q 2 is in the 250-300
group
We pass 79.5 here so Q3 is in the 300350 group
Let’s now find the approximate value of the lower quartile.
It lies in the 200-250 interval. The numbers 200 and 250 are called the boundaries of that
interval, 200 being the lower boundary and 250 being the upper boundary. The upper
boundary of the 150-200 group is the same as the lower boundary of the 200-250 group.
Imagine that all the employees are strung out in a line according to their wages and we walk
along the line. By the time we pass the last of the 150-200 group we have walked past 10
people. The lower quartile is the wage earned by the ‘26.5th’ person in the line so we have
16.5 people still to go before we reach that person.
There are 25 people in the 200-250 interval. So, assuming they are nicely spread out, we
have to walk past the fraction 16.5/25 of that interval. So we take the same fraction of the
width of that interval, which is £50.
But the interval’s lower boundary is £200, so we add the fraction onto this lower boundary.
Here’s the calculation:
16.5
of £50 = £33, then add this on to £200, so Q1 = £233
25
We repeat this procedure for the other two quartiles.
Below is a formula which allows you to slot numbers in and get an answer without much
thought, but if you compare the formula for the first quartile with what you’ve just read you
will see exactly where each number comes from.
First you must identify the interval in which the lower quartile lies. The formula for Q1 is
then
Lower Class Boundary of the class that Q1 is in
+
1
Total – cumulative frequency of the previous interval
4
× the class width
frequency of the class that Q1 is in
OUTCOME 3: NUMERACY/HIGHER
37
MEDIAN AND QUARTILES
This is quite a mouthful, so I’ll do it in shortcuts:
LCB Q1 +
1
4
Total – cfint before
× width
fQ1int
Substituting the data we get
£200 +
26.5 – 10
× £50 = £233 as before
25
(When using a calculator put brackets round the (26.5 – 10) otherwise you’ll get a different
(wrong) answer.)
The beauty of this formula is that, with a minor adjustment, you can calculate the values of
all the quartiles:
Lower quartile Q1 = LCB Q1 +
Median Q2 = LCBQ2 +
Upper quartile Q3 = LCB Q3 +
1
4
Total – cfbefore
× width
fQ1interval
1
2
Total – cfbefore
× width
fQ2interval
3
4
Total – cfbefore
× width
fQ3interval
So we calculate the values as follows:
26.5 –10
× £50 = £233
25
53 – 35
Median Q 2 = £250 +
× £50 = £273.08
39
79.5 – 74
Upper quartile Q3 = £300 +
× £50 = £315.28
18
Lower quartile Q1 = £200 +
and the quartile deviation, or semi-interquartile range, becomes
1
1
(Q3 – Q1 ) = (£315.28 – £233) = £41.14
2
2
This last number won’t mean much to you yet, but it will soon.
38
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
Example 2.3b
Two different machines are producing components, supposedly of the same size. Compare
the two using medians and quartiles.
Machine A
Component
size (mm)
1200-1205
1205-1210
1210-1215
1215-1220
1220-1225
1225-1240
Total
Machine B
No. of
components
143
217
83
21
17
2
483
Component
size (mm)
1200-1205
1205-1210
1210-1215
1215-1220
1220-1225
1225-1240
Total
No. of
components
25
174
193
32
15
0
439
Solution:
Add a cumulative frequency column to each table, then identify the interval in which each
required quartile lies. But find the quartile ranks first.
Machine A:
Q1 ranking 1/4 of 483 = 120.75
Q2 ranking 1/2 of 483 = 241.5
Q3 ranking 3/4 of 483 = 362.25
Q1 in this interval
Q2 in this interval
Q3 in this interval
Component
No. of
Cumulative
size (mm) components frequency
1200-1205
143
143
1205-1210
217
360
1210-1215
83
443
1215-1220
21
464
1220-1225
17
481
1225-1240
2
483
Total
483
Tracking down the
c.f. column
we pass 120.75 here
we pass 241.5 here
we pass 362.5 here
As you see, each quartile is in a different interval. This often happens, but it need not be so.
All the quartiles could be in the same interval, if the data was sufficiently squashed up.
OUTCOME 3: NUMERACY/HIGHER
39
MEDIAN AND QUARTILES
Now for the calculations:
120.75 – 0
× 5mm = 1204.2mm
143
241.5 –143
Median Q2 = 1205mm +
× 5mm = 1207.3mm
217
362.25 – 360
Upper quartile Q3 = 1210mm +
× 5mm = 1210.1mm
83
1
1
Quartile deviation (SIQR) = (Q3 –Q1 ) = (1210.1–1204.2) = 2.95mm
2
2
Lower quartile Q1 = 1200mm +
Machine B:
Q1 ranking 1/4 of 439 = 109.75
Q2 ranking 1/2 of 439 = 219.5
Q3 ranking 3/4 of 439 = 329.25
Q1 in here
Q2 in here
Q3 in here
Component
No. of
Cumulative
size (mm) components frequency
1200-1205
25
25
1205-1210
174
199
1210-1215
193
392
1215-1220
32
424
1220-1225
15
439
1225-1240
0
439
Total
483
Pass 109.75 here
Pass 219.5 here
Pass 329.25 here
As you see here, two of the quartiles are in the same interval.
Now for the calculations:
109.75 – 25
× 5mm = 1207.4mm
174
219.5 –199
Median Q2 = 1210mm +
× 5mm = 1210.5mm
193
329.25 –199
Upper quartile Q3 = 1210mm +
× 5mm = 1213.4mm
193
1
1
Quartile deviation (SIQR) = (Q3 –Q1 ) = (1213.4–1207.4) = 3mm
2
2
Lower quartile Q1 = 1205mm +
As you see, the quartile deviations (or semi-interquartile ranges) are virtually the same (B’s
being only marginally larger), showing that the two distributions have a similar spread. But
A’s median is lower than B’s median.
Another point to note is that the two sets of figures are of different sizes.
40
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
Box plots
A box plot (or box-whisker chart) is an extremely good way to show median and quartiles
for comparison purposes. Many graphic calculators include box plots in their library of
statistical charts.
A box plot has two components. The first is a line showing the scale of numbers required:
1200
1205
1210
1215
1220
1225
1230
The second is the ‘box’ with its two ‘whiskers’.
¼ of the data
values lie in
here
V
V
V
V
V
¼ of the data
values lie in
here
V
V
V
¼ of the data ¼ of the data
values lie in values lie in
here
here
V
V
V
This point
marks the
position of
the smallest
number
This point
marks the
position of
the lower
quartile Q1
V
V
This point
marks the
position of
the upper
quartile Q3
This point
marks the
position of
the middle
quartile, Q2 ,
the median
This point
marks the
position of
the largest
number
Now you marry the two together, lining up the various parts of the boxes with the scale.
As an example, here is the data for Example 2.3b.
1200
1205
1210
1215
1220
1225
1230
Machine A
Machine B
We have to take the two whiskers out as far as they theoretically can go; if you know the
actual maximum and minimum values of the data set you would stop the whiskers at those
points. (Note that sometimes box plots are drawn vertically instead of horizontally.)
OUTCOME 3: NUMERACY/HIGHER
41
MEDIAN AND QUARTILES
The actual range of A (1230 – 1200 = 30 mm) is larger than the range of B (1225 – 1200 =
25mm) but the SIQR of B is, as we saw earlier, marginally larger than that of A. Those very
few values of A which stretch out to 1230 aren’t enough to counteract the dispersion of B in
the most vital area, that of the middle 50%.
We can see that B has larger components, with more bunching for the third 25% (between
Q2 and Q3) than in A. The smallest 25% of B are dispersed more than the corresponding
values of A, since the leading whisker of B is longer.
If we added two more machines C and D, without actually showing the class intervals and
so on, you can tell even more:
1200
1205
1210
1215
1220
1225
1230
Machine A
Machine B
Machine C
Machine D
Which machine do you think is performing best? Which is the worst?
Note that there are conventions about what constitutes an outlier (i.e. a rogue value which is
substantially higher or lower than all the other values) and how the presence of an outlier
can affect the construction of the box plot, but this topic has been deliberately omitted
here.
Here is a final example about boundaries before you do the last exercise of this section.
42
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
Example 2.3c
This table shows the life spans of Christmas tree bulbs. Find the quartiles.
Life (hrs)
200-249
250-299
300-349
350-399
400-449
450-499
500-549
550-599
No. of bulbs
4
15
27
45
73
51
19
6
c.f.
4
19
46
91
164
215
234
240
Rank of Q1 is 240/4 = 60.
We pass 60 here.
Rank of Q2 is 240/2 = 120.
We pass 120 here.
Rank of Q3 is 3 × 240/4 = 180.
We pass 180 here.
You’ll notice that the second interval starts at 250, but the first interval ended at 249.
This is not what you’ve been seeing so far. But it is very common.
The upper limit of the first interval is 249.
The lower limit of the second interval is 250.
The boundary between them is 249.5 (you split the difference between the limits).
The second class has lower boundary 249.5 and upper bounday 299.5.
The third class has lower boundary 299.5 and upper boundary 349.5 and so on.
This complication only arises if the intervals are so defined that each one starts at a slightly
higher number (usually 1 unit more) than the one on which the previous one ended.
The quartile formulae use the lower class boundary as the kick-off.
Note also that the width of each interval is 50 hours and NOT 49 hours as you might think.
The width of an interval is defined as the difference between upper and lower boundaries.
(60 – 46)
Q1 = 349.5 +
× 50 hours = 365.06 hours
45
(120 – 91)
Q2 = 399.5 +
× 50 hours = 419.36 hours
73
(180 –164)
Q3 = 449.5 +
× 50 hours = 465.19 hours
51
I’ve given more complete
answers here so you can check
them on your calculator. But
remember these are just
approximate. So 2 decimal
places aren’t really justified.
The nearest whole number may
well be appropriate.
OUTCOME 3: NUMERACY/HIGHER
43
MEDIAN AND QUARTILES
? 2.3
For each question find the median, the quartiles, and the SIQR. You should also practise
drawing box plots.
1.
The table represents the weights of a sample of 100 male students at a college.
Weight (kg)
Frequency
2.
60-62
5
64-66
42
66-68
27
68-70
8
The following table gives the moisture content in 100 test samples of a particular
cement mix after a certain time interval.
% moisture content
Number of samples
3.
62-64
18
20-25
5
25-30
17
30-35
52
35-40
18
40-45
7
45-50
1
This table gives the diameters (in mm) of a sample of washers produced by a
machine.
Diameter x
10.0-10.4 10.5-10.9 11.0-11.4 11.5-11.9 12.0-12.4 12.5-12.9
No. of items f
4
15
27
56
75
53
4.
Here is the data for the daily output of three machines over a period of one year.
Illustrate each with a box plot and make some relevant comments about the
machines’ performance. Each machine is supposed to produce 500 components per
day.
Machine A
Daily output
(units)
480-490
490-500
500-510
510-520
520-530
Total days
44
No. of days
(f)
15
28
142
97
83
365
Machine B
Daily output
(units)
470-480
480-490
490-500
500-510
510-520
520-530
530-540
Total days
OUTCOME 3: NUMERACY/HIGHER
No. of days
(f)
17
28
63
105
87
35
30
365
Machine C
Daily output
(units)
490-495
495-500
500-505
505-510
Total days
No. of days
(f)
30
203
126
6
365
MEDIAN AND QUARTILES
Answers
?2.1: Answers
1.
11 11 13 15 15 16 16 16 16 16 16 17 18 19 20
V
V
V
Q1 = £15 per hour, Q2 = £16 per hour, Q 3 = £17 per hour, SIQR = £1 per hour
2.
120 122 123 125 125 125 126 127 128 129 130 131
V
V
V
Q1 = 124 minutes, Q 2 = 125.5 minutes, Q3 = 128.5 minutes, SIQR = 2.25 minutes
3.
1.8 1.9 2.0 2.1 2.3 2.3 2.4 2.5 2.5 2.5 2.5 2.7 2.7
V
V
V
Q1 = 2.05 cm, Q2 = 2.4 cm , Q3 = 2.5 cm , SIQR = 0.225 cm
?2.2: Answers
1.
1× 30
= 7.5, Rank = 7 + 1 = 8
4
2 × 30
Q2 :
= 15, Rank is 15.5
4
3 × 30
Q3 :
= 22.5, Rank = 22 + 1 = 23
4
Ranks: Q1:
x
0
1
2
3
4
7
9
10
11
f
5
2
3
7
4
3
2
1
3
c.f.
5
7
10
17
21
24
26
27
30
pass 8 here
pass 15.5 here
pass 23 here
Q1 = 2 faulty items, Q 2 = 3 faulty, Q3 = 7 faulty, SIQR = 2.5 faulty items
OUTCOME 3: NUMERACY/HIGHER
45
MEDIAN AND QUARTILES
2.
Ranks are Q1: 19, Q2: 38, Q3: 57
Q1 = 1 incident, Q2 = 2 incidents, Q3 = 3 incidents, SIQR = 1 incident
3.
Ranks are Q1: 25, Q 2: 49, Q3: 73
Q1 = 1 person in car, Q2 = 2 people, Q3 = 3 people, SIQR = 1 person
4.
Ranks are Q1: 189.5, Q2: 378.5, Q3: 567.5
Q1 = 73.3 mm, Q 2 = 73.4 mm, Q 3 = 73.5 mm, SIQR = 0.1 mm
(Note: In question 4, all the ‘point 5s’ in the ranks made no difference, but they might have.)
?2.3: Answers
1.
1×100
= 25
4
2 ×100
Q2 :
= 50
4
3 ×100
Q3 :
= 75
4
Q1:
Q1
Q2
Q3
x
60-62
62-64
64-66
66-68
68-70
f
5
18
42
27
8
c.f.
5
23
65
92
100
pass 25 here
pass 50 here
pass 75 here
25 – 23
× 2 = 64.10kg
42
50 – 23
Q2 = 64 +
× 2 = 65.29kg = median
42
75 – 65
Q3 = 66 +
× 2 = 66.74kg
27
1
SIQR = (66.74 – 64.10) = 1.32kg
2
Q1 = 64 +
60
61
62
63
64
65
66
67
68
69
70
The whiskers have to go out to 60 and 70 (strictly 69.99999) because, for all you
know, 60kg could be the lowest weight and 70kg could be the highest. If you
actually know what the lowest and highest weights are, you only take the whiskers
out that far.
46
OUTCOME 3: NUMERACY/HIGHER
MEDIAN AND QUARTILES
2.
3.
4.
25 – 22
× 5 = 30.3%
52
50 – 22
× 5 = 32.7%
Q 2 = 30 +
52
75 – 74
× 5 = 35.3%
Q3 = 35 +
18
1
SIQR = (35.3 – 30.3) = 2.5%
2
Q1 = 30 +
57.5 – 46
× 0.5 = 11.55mm
56
115 –102
Q2 = 11.95 +
× 0.5 = 12.04mm
75
172.5 –102
Q3 = 11.95 +
× 0.5 = 12.42mm
75
1
SIQR = (12.42 – 11.55) = 0.44mm
2
Q1 = 11.45 +
Machine A: Q1 = 503.4 units, Q2 = 509.8 units, Q3 = 519.1 units
Machine B: Q1 = 498.9 units, Q2 = 508.0 units, Q3 = 517.0 units
Machine C: Q1 = 496.5 units, Q2 = 498.8 units, Q3 = 501.6 units
the numbers don’t mean much until you see the box plot:
470
480
490
500
510
520
530
540
A
B
C
OUTCOME 3: NUMERACY/HIGHER
47
48
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
SECTION 3
A formula is a rule, usually written using mathematical symbols, which shows how two or
more variables are related.
Some formulae are explicit, i.e. they spell the relation out with no ambiguity. For example,
A = lb
is the formula for the area of a rectangle. It tells you clearly that if you know the length (l)
and the breadth (b) you have to multiply them together to get the area.
Other formulae are implicit, i.e. they imply the relationship without spelling it out. For
example
NA
n
= B
NB
nA
is a formula which relates the rotational speed (N) and the number of teeth (n) for two
interlocked gear wheels A and B.
When we transpose (or change the subject of) a formula we rearrange it so that a
different variable is on its own on one side of the formula. That variable becomes the
subject of the formula.
So if we know the area and the length of a rectangle, the formula which gives us the breadth
is
b=
A
l
and if we want to know the number of teeth for gear wheel B, given that we know
everything else, the formula for that is
nA =
NB nB
NA
Transposition as a technique is one of the most important skills used in mathematics (and
also physics, electronics and engineering and …). It may look complicated, but is helped by
the fact that, in mathematics, the rules are always consistent.
Two things to remember:
• You can frequently think of the letters in the formula as numbers. What is normally true
for ordinary numbers will be true for letters as well.
• When you are transposing a formula you follow the rules of equations: whatever you
do to one side of the equation you do at the same time to the other side.
OUTCOME 3: NUMERACY/HIGHER
49
TRANSPOSITION OF FORMULAE
3.1 Writing Formulae Using Numbers
Multiplication and division are inverse operations. This means that each one is the
opposite of the other. Whatever one does, the other undoes.
This means that any multiplication can be rewritten as a division and vice versa.
Thus the multiplication 5 × 2 = 10 can be rewritten as 5 =
10
10
and also as 2 =
2
5
Similarly, the multiplication 3 × 4 = 12 can be rearranged as 3 =
12
12
or 4 =
.
4
3
48
6
can be rewritten as the product (multiplication) 6 × 8 = 48. We don’t bother offering an
alternative 8 × 6 = 48 because everyone knows that it doesn’t matter which number goes
first with multiplication , the answers will be the same.
If we start off with a division, we can rearrange it to form a multiplication. Thus 8 =
In the language of mathematics we say that multiplication is commutative. Do you think
division is commutative? (Answer is on page 80.)
In the same way, addition and subtraction are inverse operations. Each one can be
rearranged as the other. Whatever one does, the other does the opposite.
Thus we can rearrange the sum 5 + 2 = 7 as the difference 7 – 2 = 5 or as 7 – 5 = 2.
Similarly the difference 10 – 6 = 4 can be rewritten as the sum 10 = 6 + 4. Again, we
wouldn’t offer 4 + 6 = 10 as an alternative because addition is also commutative, i.e. it
doesn’t matter in what order you add two numbers, the answer will be the same.
Do you think subtraction is commutative?
? 3.1
Rewrite each product as a quotient (division) and vice versa.
Rewrite each sum as a difference and vice versa.
1.
4.
7.
2+3=5
15
3
8 = 12 – 4
5=
2.
5 × 7 = 35
3.
5.
12 = 10 + 2
6.
8.
18 = 3 × 6
9.
8–7=1
21
=3
7
5=7–2
It should be clear that adding and subtracting form one pair of inverse operations, and
multiplying and dividing form another pair. You can’t jump between pairs. For example,
subtraction won’t undo a division, and multiplication is not the opposite of addition.
50
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
This is most important in what follows.
3.2 Relating Letters to Numbers – Adding and Subtracting
Here is a true statement, 2 + 5 = 7, which we can rewrite either as 5 = 7 – 2 or as
2 = 7 – 5.
Now we do the same with letters: a + b = c which we can rewrite either as b = c – a or as
a = c – b.
Can you see how the equations with the letters are exactly the same as the ones written
above with numbers, if you replace 2 by a, 5 by b and 7 by c each time?
Here we have a similar situation, first with numbers and then with letters replacing the
numbers:
7–3=4
7=4+3
with the alternative
p–q=r
where p replaces 7, q replaces 3 and r replaces 4.
p=r+q
? 3.2
Fill in the blanks in these equations, using the example on the left to help you.
1.
8 + 2 = 10
k+t=s
(a)
8 = 10 – 2
k = .... – ....
(b)
2 = 10 – 8
.... = .... – ....
2.
4–3=1
m–r=u
(a)
4 = .... + ....
.... = .... + ....
(b)
3 = .... – ....
.... = .... – ....
3.
8–5=3
t–k=w
(a)
8 = ............
t = ............
(b)
5 = ............
k = ............
See if you can do the next questions using only letters or a mixture of numbers and letters.
(Having a mixture does not affect what you do in any way.) By all means imagine numbers
to be there if you need them, but don’t write any numbers down.
4.
(a)
t+u=h
t = .... – ....
(b)
m–v=k
m = .... + ....
(c)
f–x=w
f = .... + ....
5.
(a)
q–r=p
q = .... + ....
(b)
s+t=8
t = .... – ....
(c)
x+4=y
x = .... – ....
OUTCOME 3: NUMERACY/HIGHER
51
TRANSPOSITION OF FORMULAE
3.3 Solving Equations Involving only Addition and Subtraction
We will now apply what we have learned so far to solve some very simple equations, ones
in which you can spot the answer. We will use special techniques which will seem
unnecessary for such simple problems. However, you have to learn the techniques, because
you won’t be able to spot the answer to the more difficult equations you’ll meet later.
Suppose we want to solve the equation x + 2 = 5. This means we want to find the value of
x which makes the statement true. Obviously the solution is x = 3 because when you
replace the letter x by the number 3 you get 3 + 2 = 5, which is true.
Now we will use a procedure to do this.
The aim in the solution of any equation (or the transposition of any formula) is to isolate the
variable in which you are interested. As far as our equation is concerned, we have to get x on
its own on one side of the equation. This means getting rid of the 2 on the left.
But the 2 is being added to the x, so we get rid of it by using the inverse operation to
addition. We subtract 2 from the left-hand side (LHS) of the equation. But for the equals
sign to mean what it says (‘is the same as’) we have to do the same to the other side, the
right-hand side (RHS) of the equation. This is to satisfy the rule: whatever you do to one
side of the equation you must do to the other.
The working looks like this:
x+2 = 5
x+2–2 = 5–2
x = 3
V
x is left on its
own because
+2–2=0
Now let’s have a look at another simple example: x – 4 = 8, the solution to which is x = 12
because 12 is the number which, when it replaces x in the equation, makes it true:
12 – 4 = 8.
Again, we want to isolate x, which involves getting rid of the 4 on the LHS. Since the 4 is
being subtracted, we eliminate it by adding 4 to the LHS, which means we have to add 4 to
the RHS.
The working looks like this:
52
x–4 =8
x–4+4 =8+4
x = 12
V
x is left on its
own because –
–4 + 4 = 0
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
Let’s try one more easy example where you can spot the answer: 12 – a = 7 to which the
answer is a = 5 because 12 – 5 = 7.
There is a slight problem here because of the ‘minus’ sign in front of the a. Although this
may seem complicated to you, the easiest way (in the long run) to deal with this is to add a
to each side, and subtract 7 from each side (do these two operations in the same line of the
solution).
Watch what happens:
–a + a
gives 0
=7
=7+a–7
=a
=a
V
V
12 – a
12 – a + a – 7
12 – 7
5
7–7
gives 0
the last line of which can be reversed to read
a=5
Remember that the aim is to get the variable on its own, on one side of the equation (and it
doesn’t matter which side), and positive.
? 3.3
Try to solve these equations in the same way.
1.
4.
x+7=8
9–k=4
2.
5.
y –4=9
7 + t = 10
3.
6.
z+2=6
m – 12 = 17
3.4 The ‘Change Side, Change Sign’ Rule
This adding or subtracting from each side is all very well, but it can be tedious if there is a
lot of it to be done. So let’s see if there is a short cut which we can apply.
Look at one of the previous examples:
x+2 =5
x+2–2 =5–2
x =3
If we take away the bit which gives us zero we are left with
x+2 =5
x =5–2
x =3
It looks as if the +2 has
disappeared from the
LHS and reappeared as
a –2 on the RHS.
OUTCOME 3: NUMERACY/HIGHER
53
TRANSPOSITION OF FORMULAE
Let’s look at another one from before
x–4 =8
x–4+4 =8+4
x = 12
Again, remove the bit which gives zero:
x–4 =8
x =8+4
x = 12
It looks as if the –4 has
moved across from the
LHS and appeared as a
+4 on the RHS.
This rule, whereby it appears that if you move something from one side to the other you
change its sign as you go, is incredibly useful and we will apply it from now on.
It works only because we are adding equal quantities to, or subtracting equal quantities
from, each side of the equation, as you have already seen. It’s not magic.
More important, it applies ONLY to + and – signs.
Here’s a more complicated example:
3x – 5 = 2x + 4
To achieve our aim, we subtract 2x from each side and add 5
to each side:
3x – 5 – 2x + 5 = 2x + 4 – 2x + 5
Remember the Aim
To get the variable x on
one side of the equation,
and positive;
and to get the numbers
on the other side.
If we now do away with all the stuff that becomes zero on each side of the equation we get
3x – 2x = 4 + 5
and this finally resolves itself
x=9
Here’s a fully worked example:
54
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
Example 3.4
Solve the equation
5x – 7 = 6x – 3
Solution:
We notice first that if we move the 6x to the left we will have 5x – 6x which will be
negative.
– 7 + 3 = 6x – 5x
V
This 7 has a – sign glued to
the front of it.
V
V
V
So move the x terms to the right, and the numbers to the left:
It is staying on the LHS.
So the – sign stays with it!
This +3 has
flown in from the
RHS.
So its sign has
changed from –3
on the right to
+3 on the left.
This 6x was on
the RHS and is
staying on the
RHS so its sign
does not change.
This 5x had been at the
start of the LHS, so it was
assumed positive.
It has flown over to the
RHS, so has changed its sign
and appeared as a –5x
We end up with
– 4 = x or, reversed, x = – 4
You could not have guessed that one from the start!
For practice, you can try some straightforward numerical examples.
For instance, here is a true statement:
5 + 2 – 7 + 3 = 6 – 8 + 9 – 4 (each side is worth 3)
Rearrange so that the even numbers are on the left and the odd numbers are on the right:
2 – 6 + 8 + 4 = 9 + 7 – 5 – 3 (each side is now worth 8, so the equals sign still works!)
And the same thing can be done with letters:
a – x + y – b = c + z – d – w. Rearrange this so that the a, b, c, d are on the left and the w,
x, y, z are on the right.
A possible answer is a – b – c + d = z – w – y + x.
There are dozens of variations, such as a + d – c – b = x – w + z – y for instance.
OUTCOME 3: NUMERACY/HIGHER
55
TRANSPOSITION OF FORMULAE
This will happen throughout this section. So long as each variable or number has its
correct symbol stuck in front of it, it can take up any position in the line-up and still be
correct. So please expect legitimate variations in the answers at the back.
? 3.4
1.
Rearrange these as required:
(a)
(b)
(c)
(d)
2.
Solve these equations.
(a)
(c)
(e)
3.
2 + 3 – 4 = 7 – 12 + 6 so that the even numbers are on the left, odd on
the right
2 + 6 – 8 = – 4 + 7 – 3 so the odds are to the left, evens to the right
A – b + c = P + Q – r so that the capital letters are to the left, lower case
to the right
– X + y – z = W – T + t as for (c) above
5x + 4 = 4x + 7
7k + 2 = 6k – 4
7 – 3m = 2 – 4m
(b)
(d)
(f)
3y – 7 = 2y + 4
9 + 2t = t – 7
9w + 3 = 8w –11
Rearrange each of these so that the x is positive on one side of the equation, and
everything else, numbers or letters – it doesn’t matter, is on the other side.
(a)
(c)
(e)
x+a=5–p
w=y–2–x
2x – a = x + b
(b)
(c)
(d)
x+3–a=w+p
y = 10 + x
k=2+x–a
CONGRATULATIONS! You’ve just transposed your first formulae without even realising it!
56
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
3.5 Transposing Formulae with only + and – Signs
In a simple formula such as a = b – c we say that a is the subject of the formula, because it
is isolated (on its own) on one side of the equation, with everything else on the other side.
To change the subject of the formula, or transpose the formula, we rearrange the
equation so that a different letter is on its own.
Suppose we want to change the subject of the formula above to b. This means we want b on
its own.
Since b is positive on the RHS we keep it there and move the c over to the left. Like this:
The b stays on the RHS
because it is positive here.
V
a=b–c
a+c=b
b=a+c
V
The –c has moved
from RHS to LHS so
it has changed its sign
to +.
V
V
This line is really optional – it tidies the
solution up so that the new subject is
on the left. When you completely swap
sides like this there are no sign changes.
Signs only change when some things
move and other things stay.
Notice how the equals signs are all lined up neatly one
under the other. This is very important. It is much easier
to clearly see which side is the left and which side is the
right.
Suppose now we want to take the original formula and transpose it to c. This means making
c the new subject. Here is the solution:
V
When c was on the RHS it had a – sign
in front of it.
Moving it to the LHS changes its sign
to a +.
Because it is at the start of the line we
don’t write the + sign in, we understand it to be there.
V
a=b–c
c=b–a
The a was positive on the LHS. There
was no + sign in front of it but we
understood it to be there.
We have to move it right so that c can
be isolated on the left.
Moving a to the RHS changed its sign to
a –.
Here are three more examples. Try to follow them without any explanations:
p = q + r, change to r
p–q =r
r =p–q
s = u – v, change to u
s+v =u
u =s+v
w = a – k, change to k
k=a–w
OUTCOME 3: NUMERACY/HIGHER
57
TRANSPOSITION OF FORMULAE
? 3.5
Transpose each of these formulae as required:
1.
3.
5.
p=q+r
m=q–t
s=x–r–8
to q
to q
to r
2.
4.
6.
p=q+r
m=q–t
s=x–r–8
to r
to t
to x
3.6 Relating Letters to Numbers – Multiplying and Dividing
We will now repeat for × and ÷ what we did for + and – on page 51.
Here is a true statement, 5 × 2 = 10, which can be rewritten as a division 5 =
10
10
or 2 =
.
2
5
c
c
or b = .
b
a
As you see, the pattern is the same, where a replaces 5, b replaces 2 and c replaces 10.
Using letters: a × b = c, which can be rewritten as a division a =
Here’s another example. We start with a division, which can be rewritten as a
multiplication and then a different division:
12
12
which rearranges to 12 = 4 × 3 and then 3 =
.
3
4
Replacing 4 by p, 12 by r and 3 by s we get
4=
p=
r
r
which rearranges to r = p × s and then s = .
s
p
? 3.6
Fill in the blanks in these equations, using the example to help you.
1.
5 × 3 = 15
(a)
a×b=c
2.
6 × 2 = 12
(a)
r×w=k
3.
58
7=
21
3
h=
x
t
(a)
5=
15
3
a=
⋅⋅⋅⋅
⋅⋅⋅⋅
6=
12
2
r=
⋅⋅⋅⋅
⋅⋅⋅⋅
21 = 7 × 3
x = .... × ....
OUTCOME 3: NUMERACY/HIGHER
(b)
(b)
(b)
3=
15
5
b=
⋅⋅⋅⋅
⋅⋅⋅⋅
2=
12
6
w=
⋅⋅⋅⋅
⋅⋅⋅⋅
3=
21
7
t=
⋅⋅⋅⋅
⋅⋅⋅⋅
TRANSPOSITION OF FORMULAE
See if you can do these without any help.
4.
(a)
p×q=b
(b)
p=
5.
j=
(a)
t=
m
f
(c)
m=
t
8
(b)
t=
b=r×q
q=
g=y×a
a=
(c)
7=
r
x
r=
3.7 Solving Equations Involving only Multiplication and Division
In the previous section you saw how subtracting undid an addition and vice versa, because
adding and subtracting are inverse operations.
This next section involves only multiplication and division. I hope you remember that each
is the inverse operation of the other. So to undo a multiplication we will be dividing. And
vice versa.
First of all, some algebraic conventions.
Convention 1:
Unless it will lead to confusion, all multiplication signs will be omitted.
Thus ab means a × b and pqr means p × q × r.
Convention 2:
Since multiplication is commutative, the order of variables does not matter,
although alphabetical order is preferable.
Thus bca is the same as cab but abc is best of all.
Convention 3:
If a number is involved in multiplying, it always comes first.
Thus you write 4x and not x4 to mean ‘4 times x’.
Convention 4:
In addition/subtraction, you can only simplify like terms, or terms which
look alike.
For instance, 3x + 2x = 5x because 3x and 2x are both the same sort of
term, but terms like 3x + 2y or 3x – 2 cannot be simplified any further.
With multiplying, however, you can simplify virtually anything. Thus
3x × 2y = 6xy and 3x × 2 × 5z = 30xz and so on.
Now let’s solve some equations.
2x = 8
The x is being multiplied by 2. We want to isolate x, i.e. make it on its own. To do this we
have to separate the 2 from the × x. The only way to do this is to use the inverse operation,
i.e. division.
OUTCOME 3: NUMERACY/HIGHER
59
TRANSPOSITION OF FORMULAE
So we divide each side by 2:
2x
8
=
2
2
On the LHS the 2s cancel out top and bottom, leaving x.
On the RHS we divide 8 by 2 to give the answer 4.
When we say that we CANCEL a
number top and bottom, we are
dividing the numerator and the
denominator BOTH by the same
number (in this case by 2) so that
the expression looks different but
is still worth the same.
x=4
In practice we do the LHS of the middle line in our heads, and the solution as written out
looks like this:
2x = 8
8
2
x=4
x=
Quite easy, really: easier than moving terms from left to right and so on.
Here’s another example:
3a = 7
To separate the 3 from the a we have to use the inverse operation which will unstick the
glue which holds them together. They are stuck together with multiplication, to which the
inverse operation is division.
So divide each side by 3. This will isolate the a, which is what we’re after.
3a
7
=
3
3
60
OUTCOME 3: NUMERACY/HIGHER
7
3
V
a=
V
V
The 3s cancel out top
and bottom, leaving you
with the a only.
In practice we do this line
mentally, and go straight
on to the last line.
We could turn this into a
decimal if we want, but
mathematical convention
is to leave it as an exact
fraction.
TRANSPOSITION OF FORMULAE
Example 3.7a
Transpose the formula A = BC to B.
Numerical equivalent
Solution:
We could start by reversing the entire formula so that the
BC is on the left.
BC = A
12 = 3 × 4
3 × 4 = 12
4=
We want to isolate B. B is being multiplied by C. So divide
each side by C:
B=
12
3
A
C
Example 3.7b
Transpose the formula P = 3QRS to R.
Solution:
Don’t be put off by the extra number and letters. The method is exactly the same as before.
We want to isolate Q. Q is being multiplied by 3RS (it doesn’t matter that the Q is in the
middle. It could have been at the start or at the end).
So divide each side by the 3RS. We can reverse the equation either at the start or at the
end.
P = 3QRS
P = 3QRS
3QRS = P
R=
or
P
3QS
Numerical equivalent
P
=R
3QS
R=
120 = 3 × 2 × 4 × 5
120
4=
3 × 2 ×5
P
3QS
? 3.7
Transpose these formulae as required:
1.
4.
s = 7k
t = km
to k
to m
2.
5.
m = 9t
B = 4hg
to t
to g
3.
6.
C = πd
A = 5bmx
to d
to m
OUTCOME 3: NUMERACY/HIGHER
61
TRANSPOSITION OF FORMULAE
3.8 Equations with Fractions
The first, and only, rule about fractions is get rid of them. Fast.
This is so that your equation is a ‘one liner’, i.e. all written on one line, because it is then
more easily dealt with.
Let’s go back to a numerical example
5=
20
4
Although this means ‘5 equals 20 divided by 4’ it can also mean ‘a fraction equivalent of 5
is 20 over 4’.
Since the 20 and the 4 are glued together by division, we can unglue them by using the
inverse operation of division. We can multiply both sides by the denominator of the fraction
(the number on the bottom).
4 × 5=4 ×
20
4
On the right-hand side of this the 4s cancel out top and bottom, and we are left with
4 × 5 = 20
which I’m sure you knew anyway (as far back as page 50). But it’s the technique that’s
important.
Example 3.8a
Transpose the formula P =
m
to m.
8
Solution:
You multiply both sides by 8.
62
V
This reversal is optional,
but it does make the
solution look a lot better!
V
V
m
8 × P=8 ×
8
V
8P = m
m = 8P
OUTCOME 3: NUMERACY/HIGHER
In practice you can do this
line mentally, without
actually writing it down,
although it might help to
write it for now.
8 cancels out top and
bottom on the RHS.
TRANSPOSITION OF FORMULAE
Example 3.8b
From here on, read comments in
boxes in their numerical order.
3y
Transpose the formula X =
to y.
2wz
Solution:
This looks difficult, but it is only one step longer than the last example.
2. The 2wz cancels top and
bottom on the RHS.
V
2wzX = 3y
3y = 2wzx
2wzX
y=
3
V
3. Reverse equation (i.e.
swap sides about) to
make it look good.
3y
2wz
2wz × X = 2wz ×
V
V
4. Divide each side by 3
to isolate the y.
Transposition complete.
V
1. Multiply each side by
the denominator.
For the last example in this section we’ll take the same formula as before, but change the
new subject.
Example 3.8c
Transpose the formula X =
3y
to w.
2wz
Solution:
The first step is to get rid of the fraction as before. In (almost) every formula which has
fractions in it, the very first step is always to get rid of the fraction. It doesn’t matter what
the subject of the formula is to be.
2wz × X = 2wz ×
w=
V
2wzX = 3y
3y
2wz
3y
2Xz
V
1. The first two lines are
EXACTLY the same
as before.
It doesn’t matter if we write 2Xz or 2zX on the bottom line.
2. Note that we don’t need to
reverse the equation because
the required w is on the left
already.
3. To isolate w we divide each
side by whatever is
multiplying w. We divide
each side by 2zX.
It does matter, however, if you write a lower case x instead of the capital X which is in the
formula. In a particular context, lower case and capital letters may represent different
variables, so you have to write the letters as they are.
? 3.8
Transpose these as required:
1.
w=
P
4
to p
2.
X=
2r
S
to r
3.
m=
ab
cd
4.
T=
2π
to t
5t
5.
k=
3abc
4pqr
to a
6.
k=
3abc
4pqr to q
to a
OUTCOME 3: NUMERACY/HIGHER
63
TRANSPOSITION OF FORMULAE
3.9 Cross Multiplication
Cross multiplication is an incredibly useful component of your transposition toolkit, used in
situations where the formula or equation is in the form
FRACTION = FRACTION
In fact, we have been sort of cross multiplying in the last section. How does it work?
Let’s first look at a numerical example.
If you multiply both the numerator AND the denominator of a fraction by the SAME number,
you get an equivalent fraction; the two fractions look different but they are both worth
the same.
For example, if you multiply the numerator and denominator of the fraction ½ by 3, you get
1
3
=
2
6
Now watch this:
1×6=2×3
which is perfectly true: both sides are equal to 6.
Here are a few more examples for you to think about:
3
6
=
4
8
3 × 8=6 × 4
5 20
=
4 16
5 × 16 = 20 × 4
2
20
=
7
70
70 × 2 = 7 × 20
If an equation is written in the form
A
C
=
B
D
then we can cross multiply and arrive at the one liner
A×D=B×C
Of course we would write this without the times signs, AD=BC, and it could be written in
many other ways, depending on convenience: DA = CB or BC = DA and a few others.
Cross multiplication is just a convenient shortcut. You could transpose this formula by
multiplying both sides; it would just take longer:
64
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
V
LHS: B cancels out
top and bottom.
A
C
=
B
D
A
C
= BD ×
BD ×
B
D
DA = BC
V
Multiply both sides by the product of the denominators, i.e. by B times D
RHS: D cancels out
top and bottom.
Any of the four variables can now become the new subject with one quick division.
Example 3.9a
Solve the equation
Have you figured out why it’s
called ‘cross multiplication’?
5
2
=
x
3
V
x=
Solve for x:
V
15
= 7 12 or 7.5
2
V
2x = 5 × 3
V
Solution:
Cross multiply:
A
C
=
B
D
A
C
=
B
D
A × D=B × C
Example 3.9b
Transpose the formula
p
2π
=
to make t the subject.
q
t
Solution:
Cross multiply:
pt = 2πq
t =
Solve for t:
2πq
p
Example 3.9c
Transpose the formula X =
3Y
to make Z the subject.
5wZ
Solution:
Put the X ‘over 1’ to make the LHS look like a fraction. You can do this with any letter or
3
12
a
5t
; 12 = ; a = ; 5t = ; and so on.
1
1
1
1
V
You would do this first
step no matter what
transposition you
intended to make.
X
3Y
=
1
5wZ
5wZX = 3Y × 1
Z=
3Y
5wX
V
number: 3 =
You wouldn’t actually write
3Y × 1 because you know
that the answer will be 3Y,
since anything times 1 is
itself.
I’ve written it here so you
can see what’s happening.
OUTCOME 3: NUMERACY/HIGHER
65
TRANSPOSITION OF FORMULAE
? 3.9
Transpose these formulae as required.
aC
360
1.
A=
4.
NA
n
= B
NB
nA
to a
2.
A
7
=
w
p
to w
3.
A
7
=
w
p
to p
to nA
5.
A=
BC
D
to B
6.
A=
q
pr
to r
3.10 Formulae Involving all Operations
This is crunch time: time to put everything together. If you’ve managed up until now, one
more small step and you’re there.
It’s sometimes difficult to give rules which will cover every eventuality, but as a rule of
thumb, for the formulae you can cope with so far:
1.
Get rid of fractions either by multiplying both sides by the denominator (if there is
only one), or by multiplying both sides by the lowest common multiple of the
denominators, often the product (if there is more than one).
2.
Move terms around from side to side so as to isolate on one side the term involving
the variable you want (where it is positive) and have all the other terms on the other
side. This involves changing side/sign if there are + and – signs involved.
3.
Finally, isolate the variable you want, probably by division.
Example 3.10a
Transpose a = bc – d to d.
Solution:
V
3. d was negative on the RHS,
has moved left, so has
become positive.
a = bc – d
1. The proposed subject, d, is negative
on the RHS, so move it to the
other, changing sign as you go.
d = bc – a
4. a was positive on the LHS,
has moved right, so has
become negative.
V
2. If d comes over from the left, it will
join the a, but you want d isolated.
So move the a over to the right,
changing sign as you go.
V
5. bc has stayed on the RHS,
it has NOT moved, so
does NOT change.
66
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
The transposition is complete.
Without all the comments, the problem looks like this:
a = bc – d
d = bc – a
Example 3.10b
Transpose a = bc + d to d.
Solution:
2. But if you leave d on the RHS it will
not be isolated, so move the bc to
the left, changing its sign as you do
so.
a = bc + d
1. The proposed new subject, d, is
positive on the RHS so LEAVE IT
HERE, do not move it. Leaving it
thus will NOT change its sign. (Signs
only change when things move.)
a – bc = d
Finally, reverse the formula:
d = a – bc
The completed problem, without comments, but with all the = signs neatly one under the
other:
a = bc + d
a – bc = d
d = a – bc
Example 3.10c
Transpose a = bc + d to b.
Solution:
V
2. If you have decided to isolated
bc on the RHS, then the d must
come over to the LHS. Because
it has moved, it has changed
sign.
a = bc + d
V
V
3. a has not moved, so its
sign does not change.
a – d = bc
V
a–d
=b
c
V
5. Finally, reverse the fomula
to make it look good.
1. You want b, which is glued
together with c by
multiplication. You MUST unglue
+ and – BEFORE ungluing x.
Since bc is positive on the RHS
leave it here, without changing
its sign.
You are thus isolating the term
which involves the variable you
want.
b=
a–d
c
4. Having isolated bc, you now
isolate b. Unglue it from c by
dividing both sides by c.
Transposition complete.
OUTCOME 3: NUMERACY/HIGHER
67
TRANSPOSITION OF FORMULAE
Example 3.10d
Transpose a = b – cd to d.
Solution:
a = b – cd
○
V
○
V
○
1. As usual, isolate the term
involving the variable you want.
This means get the cd term on
its own to the side which makes
it positive.
Since it is negative on the RHS
move it left to make it positive.
V
○
○
V
○
○
cd = b – a
○
2. As cd moves left it joins a which
is already there. But you don’t
want it there because you want
the cd to be alone.
So move the a to the RHS,
changing its sign from + to – as
you go.
d=
b–a
c
V
3. Finally, divide both sides
by c to isolate d. The c’s
cancel out top and bottom
on the LHS so don’t write
them there. But do write
c on the RHS.
Transposition complete.
? 3.10
Transpose the following as required.
1.
4.
p = q – rs
s = ap – 3nq
to q
to a
2.
5.
k = 3t + ah
T = 4r – abcd
to h
to c
3.
6.
x = 2y + 3bc
K = π – pqr
to b
to q
3.11 Coping with Brackets
A bracket is yet another layer of ‘glue’ to be unstuck when solving an equation or
transposing a formula.
The general rule is:
• get rid of any fractions outside the brackets first
then
• multiply out the brackets
then
• get rid of any fractions which are inside the brackets
then
• do any moving terms about from one side to the other.
If various terms are inside a
bracket, you cannot get at
them individually until after you
have multiplied the bracket out.
Using another analogy:
Stuff inside a bracket is like
money in your pocket. You
can’t see how much you’ve got
until it’s out of the pocket and
on the table.
When multiplying out brackets, remember that everything inside the bracket gets multiplied
by whatever is immediately outside. If it is a negative number that’s outside, then the signs
inside will change as you go.
68
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
Reminders:
Watch the sign changes in the
last two examples.
2(x + 3)
3(a – 2b + 5)
x(x + y)
3m(2m + 5n – 3)
–4(2x – 3z + 1)
–3a(1 – 2a + 3a2 – 4b)
=
=
=
=
=
=
2x + 6
3a – 6b + 15
x2 + xy
6m 2 + 15mn –9m
–8x + 12z – 4
–3a + 6a2 – 9a3 + 12ab
Multiplying by a negative
number always changes the
sign, because
Negative × Positive = Negative
and
Negative × Negative = Positive
Example 3.11a
Transpose x = y(a + b) to y.
Solution:
Just to confuse you, we’ll start off this section without multiplying out brackets!
Since we want y isolated, and y is being multiplied by the single term (a + b), simply divide
both sides by the bracket (a + b) like this:
x
y(a + b)
=
and the (a + b) will cancel out top and bottom on the RHS leaving you
(a + b)
(a + b)
with
x
x
= y which can be reversed to give you y =
(a + b)
(a + b)
Example 3.11b
Transpose x = y(a + b) to a.
The a we want is inside the bracket.
A rule of thumb (which you can break when you become proficient at all of this, but not yet)
is that you must multiply out any bracket to be able to access individually the terms inside it.
x = y (a + b)
x – yb = ya
V
3. Reverse
V
x = ya + yb
2. Move yb to the LHS to
isolate the positive ya
on the right.
V
V
1. Multiply out the
bracket.
4. Divide both sides by y
to isolate a as required.
ya = x – yb
a=
x – yb
y
OUTCOME 3: NUMERACY/HIGHER
69
TRANSPOSITION OF FORMULAE
Example 3.11c
Transpose a = b(3x – 5y) to y.
Solution:
See if you can follow this without any further explanation.
a = b(3x – 5y )
a = 3bx – 5by
5by = 3bx – a
3bx – a
y=
5b
Example 3.11d
Transpose a =
x
to b.
a–b
Solution:
a
x
=
1
(a – b)
a(a – b) = x
1. Note that I have added a
pair of brackets round the
a – b. This is MOST
IMPORTANT and will avoid
mistakes later.
2. Cross multiply as usual.
Remember that x times 1
equals x.
a2 – x = ab
b=
a2 – x
a
V
a2 – ab = x
4. Isolate the term involving b.
V
V
3. Multiply out the bracket.
Having put the bracket in
ensures that this step is
done correctly. (A very
common mistake is to get
the a 2 correct but then
write –b, having forgotton
to multiply it by a as well.)
V
a
to achieve ‘fraction = fraction’ format
1
V
The easiest way is to rewrite the LHS as
5. Continue as usual. I have
combined the ‘isolating’ step
and the ‘reversing’ step into
one line.
Example 3.11e
Change the subject of the formula x =
p+q
to p.
p–q
Solution:
Start as before: put brackets round the p + q and also the p – q , write the x as ‘x over 1’,
then cross multiply:
x
(p + q )
=
1
(p – q )
x (p – q ) = 1(p + q )
70
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
Since the required p is inside a bracket and, what’s worse, on both sides of the equation as
well, we have no choice but to multiply out both brackets:
1. Move the terms involving
the required p to one side
of the equation...
2. ...and the other terms
to the other side
V
xp – p = q + xq
V
xp – xq = p + q
We now have two terms involving the required p but we only want to have one term.
If x was a number it would have been easy: 5p – p = 4p; 9p – p = 8p and so on. But we don’t
have that, we have xp – p.
We cope with the problem by taking out p as a common factor:
q(1 + x )
p=
(x – 1)
p=
V
q + xq
(x – 1)
q(x + 1)
(x – 1)
V
V
V
4. To isolate p, divide both
sides by whatever is
multiplying it, in this case
the entire (x – 1) bracket.
p=
V
p(x – 1) = q + xq
3. Remember that px is
exactly the same as xp.
5. This line is actually OK as an
answer...
6. ...but it looks better if you
notice that taking out q as a
factor on the top line is a
possibility...
7. ...and changing the order of
the top bracket to make it
look similar to the bottom
one is the icing on the cake!
To be fair, that was quite a stinker of a transposition. Here is another one but, if you
followed the last one without difficulty, omit this next one and move on.
Example 3.11f
Make r the subject of the formula E =
R + nr
.
R – nr
2. Put the numerator and the
denominator of the fraction
each into a bracket as a
safety precaution.
E(R – nr ) = 1(R + nr )
ER – Enr = R + nr
V
ER – R = nr + Enr
ER – R = r (n + En)
V
V
4. Now a slight problem. If
you put the terms involving
the required r to the left,
you get too many minus
signs, which make handling
the equation difficult...
E
(R + nr )
=
1
(R – nr )
5. ...so move the terms
involving r to the RHS
instead...
V
V
3. Cross multiply, then multiply
out the brackets as before.
V
1. Write ‘E over 1’ to put the
equation into ‘fraction =
fraction’ form to make cross
multiplication possible.
V V
Solution:
6. ...then take out r as a
common factor.
OUTCOME 3: NUMERACY/HIGHER
71
TRANSPOSITION OF FORMULAE
Now divide both sides by the bracket which is multiplying r to get r on its own, and reverse
the entire equation:
r=
r=
(ER – R )
(n + En)
R(E – 1)
n(E + 1)
8. And a final cosmetic
embellishment to make it
look as if you really know
what you’re doing!
V
V
7. Put a bracket round the
numerator. This stops you
making silly mistakes like
cancelling out the Es, for
example.
But what happens if you can’t just put one side ‘over 1’ and cross multiply?
Well, you just go back to multiplying each side by something which will get rid of all the
fractions at one go:
Example 3.11g
Rearrange the formula R =
k
3k
+
to make k the subject.
2x
x +1
Solution:
The trick is to multiply both sides by the lowest common denominator of all the fractions.
In this case, it’s 2x times (x + 1).
To avoid possible mistakes, put a bracket round the entire RHS.
3k 
 k
2x ( x + 1) R = 2x ( x + 1) 
+

2x 
x +1
Now multiply out the RHS. You can leave the LHS as it is.
2x ( x + 1) R = 2x ( x + 1)
V
k
3k
+ 2x ( x + 1)
2x
x +1
V
V
V
Cancel out the entire (x + 1)
bracket top and bottom.
Cancel out the 2x top and
bottom.
and you are left with
V
V
2x ( x + 1) R = 2xk + ( x + 1) 3k
2x (x + 1)R = k (2x + 3x + 3)
V
2x (x + 1)R = k (5x + 3)
72
OUTCOME 3: NUMERACY/HIGHER
2x (x + 1)R
(5x + 3)
V
k=
1. Take out k as a common
factor. You get
k(2x + (x + 1)3)
= k(2x + 3x + 3)
2. Divide each side by
(5x + 3). Keeping it in a
bracket means you avoid
mistakes such as
cancelling out x top and
bottom, for instance.
TRANSPOSITION OF FORMULAE
And now, an exercise to practise on.
? 3.11
A quick word about cancelling. You can only
cancel entire factors. The only reason that
the fraction 6/8 cancels down to be ¾ is that
you are dividing the top and the bottom by a
factor of 2:
Change the subject of each formula as required:
1.
2.
T = 3(x –2y)
P = t(3 + 5m)
to (a) x
to (a) t
3.
a
Y=
+ bc
2
4.
t
2m
X= –
3
5
5.
2w + 3
K=
7
6.
H=
3(x – y )
x
to (a) y
(b) x
7.
P=
a + 2b
a–b
to (a) a
(b) b
8.
T=
3(x – 2y )
2(x + 3y ) to (a) x
(b) y
to (a) a
to (a) t
(b) y
(b) m
6
8
(b) b
=
2×3
2×4
3
=
4
If we write the fraction as
(b) m
6
8
=
can’t cancel out the 5s and get
to w
5 +1
5+3
1
3
you
or anything
else.
Similarly, if you have a fraction like
x + 3y
3
the 3 of the denominator is dividing into the
WHOLE of the top line, not just a selected
bit of it. So you can’t cancel out the 3s and
get x + y or anything else.
2x + 6
can be simplified
2
ONLY because the top can be written as a
product of two factors, one of which is 2,
and we can then cancel out the 2’s like this:
A fraction like
We’re nearly at the end now: one last topic
to look at.
2x + 6
2
=
2 × (x + 3)
2
=
x +3
1
=x +3
3.12 Powers and Roots
First of all, a power and the appropriate root are inverses of each other. This means that
we can use one to unravel the other.
For example, 3 to the power 2 (or 3 squared) is equal to 9. This means that 3 times 3 is 9.
The power tells you how many times a number multiplies itself.
The inverse of this is that the square root of 9 is 3.
Written using symbols, we have
32 = 9 so 9 = 3
In the olden days, the square root was written like this: 2 9 = 3 but since the square root
is the one most often used, we have become lazy and the 2 has been dropped.
Here are some more examples:
OUTCOME 3: NUMERACY/HIGHER
73
TRANSPOSITION OF FORMULAE
52 = 25 so 25 = 5
7 2 = 49 so
100 = 10 because 10 2 = 100
49 = 7
2.25 = 1.5 because 1.52 = 2.25
and we can do the same with symbols
a 2 = b so a = b
x 2 = y so x =
t = w so t = w 2
y
bc = d so bc = d 2
Believe it or not, we’re actually ‘doing the same to each side of the equation’ in these
examples; in this case we are squaring each side.
x= a
Square each side
(x)
2
=
( a)
2
x2 = a
If you ‘square a square root’ like this, you effectively get access to whatever is under the
root sign. Thus:
( t) = t
( 5b ) = 5b
2
2
(
3x 2 y
)
2
= 3x 2 y
and so on.
There’s actually more to it than this, but that’s all you need to know here.
The same applies to higher powers, for example:
23 = 8 (meaning 2 × 2 × 2 = 8) so
3
8 = 2 (read ‘the cube root of 8 is 2’)
54 = 625 (meaning 5 × 5 × 5 × 5 = 625) so
4
625 = 5 (read ‘the fourth root of 625 is 5’)
But we will be sticking mainly to squares and square roots, with the occasional cube and
cube root thrown in.
Now for some more worked examples.
74
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
Example 3.12a
Transpose a =
x
to t.
3t
Solution:
Start by squaring both sides. This will gain you access to everything under the root sign:
 x 
( a ) =  
 3t 
x
a2 =
3t
2
2
From now on it’s plain sailing, I hope:
a2
x
=
1
3t
3ta 2 = x
t=
x
3a 2
Example 3.12b
Transpose p = 2π x – 3y to y.
Solution:
There’s often more than one way to solve this kind of problem, but I'll just stick to the one.
As a rule of thumb, you should start by isolating the entire rooted term:
V
p
=
2π
x – 3y
2. Square both sides to be
able to access what’s under
the root sign.
 p 
  =
 2π 
V
3. Note the bottom line here:
2π × 2π = 4π 2 and NOT
2π 2.
(
x – 3y
)
2
V
2
V
1. Divide both sides by 2π.
p2
= x – 3y
4π 2
3y = x –
y=
V
5. To get one y divide both
sides by 3. This would look
cumbersome, but
multiplying by one third
does the same job.
p
4π 2
V
2
4. Since the 3y term is clear of
interference, move it to the
LHS to make it positive.
This involves moving other
stuff to the RHS.
1
p2 
x –

3
4π2 
OUTCOME 3: NUMERACY/HIGHER
75
TRANSPOSITION OF FORMULAE
Example 3.12c
Rewrite the formula
a
=
b
3w – π
to make w the subject.
w
Solution:
See if you can follow this without a commentary.
2
 3w – π 
a

  = 
w 
 b

(3w – π )
a2
=
2
b
w
2
2
a w = b (3w – π )
2
a 2 w = 3b 2 w – b 2 π
b2 π = 3b2 w – a 2 w (*)
2
2
2
πb2
(3b – a 2 )
w=
2
V
b π = w (3b – a )
Since π is a number
(3.14159 etc) it technically
should go in front, just as
when we write 3x instead
of x3.
Notice how in the line marked (*) I've moved the w terms to the right. This is because the
alternative, moving them left, would give us
a 2 w – 3b 2 w = – b 2 π
w (a 2 – 3b 2 ) = – b 2 π
w=
–b 2 π
(a 2 – 3b 2 )
Technically speaking this is OK, but it looks cumbersome because of the negative sign on the
top line. We can counter this by multiplying numerator and denominator both by –1. This
has the effect of changing signs but nothing else, so the result would be
w=
b2 π
– a 2 + 3b2
and we would then rewrite the bottom line with the positive term first
w=
b2 π
3b 2 – a 2
So it’s a lot easier doing it the first way from scratch.
76
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
Example 3.12d
Transpose the formula y = 3x2 + 5 to make x the subject.
Solution:
x=
y –5
3
V
y –5
= x2
3
1. Do whatever you have to
do to isolate the term
involving x2.
V
y – 5 = 3x 2
2. Do whatever you have to
do to isolate the x2 itself.
V
y = 3x 2 + 5
3. Now that you have x 2
isolated, you square root
each side to get x.
Notice that it is important to write the square root sign so that it encompasses the entire
relevant part of the equation. Some people are careless, and would write the answer with a
smaller sign, making it look like this:
x=
y –5
3
This is completely wrong, as it has a totally different meaning. In the first (correct)
expression, you subtract 5 from y, divide the answer by 3, then take the square root of the
final answer. In the second (wrong) expression, you are subtracting 5 from y, square rooting
the answer, then dividing that answer by 3.
Example 3.12e
Change the subject of the formula 3t =
5
to X.
X –2
3
Solution:
Start as usual
3t
5
=
3
1
(X – 2)
3t(X 3 – 2) = 5
3tX 3 – 6t = 5
OUTCOME 3: NUMERACY/HIGHER
77
TRANSPOSITION OF FORMULAE
Now treat the cube in the same way as you would a square – isolate first the term involving
it, then the X3 itself:
NB You canNOT cancel the 6
with the 3, or the t with the t.
To avoid this mistake, put
brackets round the top line
(effectively gluing it together).
3tX 3 = 5 + 6t
(5 + 6t )
X3 =
3t
Now, instead of taking a square root of each side, you take a cube root.
X=
3
5 + 6t
3t
Now for an exercise on roots and powers.
? 3.12
A few more things you need to know about
roots:
Transpose these formulae as required:
36 = 6
1.
Q = 3p2
2.
y=
3.
k =3 t
4.
m=
5.
y = 3x2 + ab
to
6.
z=
x +y
to y
7.
t=
x
x +y
to
8.
m = 5x 2 +
25
x2
to p
but we can also write
36 =
to x
2
5 x
4×9 =
4 ×
9 = 2×3 = 6
This illustrates the rule that
to t
ab =
a ×
b
We can similarly show that
to x
(a) a
(b) x
a
b
=
a
b
This does NOT work for adding and subtracting!
(a) y
(b) x
2y to y
a + b is NOT the same as
a +
b
A simple illustration:
9 + 16 =
25 = 5 we know is true
BUT if we try two separate ones
And that’s all there is to it (!)
If you have managed to survive this long,
you have actually achieved a great deal.
9 + 16 = 3 + 4 = 7 is different.
The same applies to subtraction.
In my mind, transposition of formulae is probably the single most difficult technique to
master, and we’ve done it all with the exception of exponentials and logarithms, which is a
complete topic all to itself.
Here is a final exercise, a complete mixture of everything you’ve covered so far, using ‘real’
formulae, in other words, formulae which are actually used in the real world.
78
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
? 3.13
Hope it’s lucky for you.
Transpose these formulae as required. In many cases you are asked for more than one
transposition, but remember that in most cases your initial steps are always the same, you
don’t have to go back to the very beginning each time.
Take care to copy down formulae correctly using capital letters and lower case letters as
shown, especially in questions 13 and 14.
1.
2.
C = πd
V = πr2h
to d
to
(a)
h
(b)
r
3.
F=
1
2
mv2
to
(a)
m
(b)
v
4.
V=
3
4
πr3
to r
5.
f =
(b)
m1
(b)
d
6.
T = 2π
7.
8.
v = u + at to
v2 = u2 + 2as
to
(a)
(a)
u
s
(b)
(b)
a
u
(u + v)t
to
(a)
t
(b)
u
1
2
Gm1m2
d2
1
2
to
l
g
to l
9.
s=
10.
s = ut +
at2
to
(a)
u
(b)
a
11.
to
(a)
b
(b)
c
12.
a = b2 – c2
A = 2πr(r + h)
13.
I=
iR
R +r
to
(a)
i
(b)
r
(c)
R
14.
I=
nE
R + nr
to
(a)
E
(b)
R
(c)
r
to h
(d)
n
OUTCOME 3: NUMERACY/HIGHER
79
TRANSPOSITION OF FORMULAE
Answers
Page 50
Division is not commutative because the order of division does matter. For example, 6 ÷ 3
is not the same as 3 ÷ 6. The answer to the first one is 2, the answer to the second is ½.
Similarly, subtraction is not commutative either. For instance, 5 – 2 = 3 but 2 – 5 = –3
which is a totally different answer.
?3.1: Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
2 = 5 – 3 or 3 = 5 – 2
35
35
or 7 =
7
5
8 = 7 + 1 (or 1 + 7 but this is the last time a variation like this will be given here)
15 = 5 x 3 (or 3 × 5 but this is the last time a variation like this will be given here)
10 = 12 – 2 or 2 = 12 – 10
21 = 7 × 3
12 = 8 + 4
5=
18
18
or 3 =
3
6
7=5+2
6=
?3.2: Answers
1.
2.
(a)
(a)
3.
(a)
4.
5.
(a)
(a)
k=s–t
4=1+3
m=u+r
8=3+5
t=w+k
t=h–u
q=p+r
(b)
(b)
(b)
(b)
(b)
t=s–k
3=4–1
r=m–u
5=8–3
k=t–w
m=k+v
t=8–s
(c)
(c)
f=w+x
x=y–4
?3.3: Answers
1.
4.
80
x=1
k=5
2.
5.
y = 13
t=3
3.
6.
OUTCOME 3: NUMERACY/HIGHER
z=4
m = 29
TRANSPOSITION OF FORMULAE
?3.4: Answers
1.
(a)
(b)
(c)
(d)
2 – 4 + 12 – 6 = 7 – 3 (each side equals 4)
– 7 + 3 = – 4 + 8 – 6 – 2 (each side equals -4)
A–P–Q = b–c–r
T – X – W = t + z – y (try to start each line or side with a positive, unless
you have no choice)
(2)
(a)
5x – 4x = 7 – 4
x=3
(b)
3y – 2y = 4 + 7
y = 11
(c)
7k – 6k = –4 – 2
k = –6
(d)
2t – t = – 7 – 9
t = –16
(e)
–3m + 4m = 2 – 7
m = –5
(f)
9w – 8w = –11 – 3
w = – 14
(a)
(c)
(e)
x=5–p–a
x=y–2–w
x=a+b
(b)
(d)
(f)
x=w+p–3+a
y – 10 = x
k–2+a=x
3.
?3.5: Answers
1.
3.
5.
q=p–r
q=m+t
r=x–8–s
2.
4.
6.
r=p–q
t=q–m
x=s+r+8
?3.6: Answers
1.
(a)
a=
c
b
(b)
b=
c
a
2.
(a)
r=
k
w
(b)
w=
k
r
3.
(a)
x=h×t
(b)
t=
4.
(a)
p=
(b)
m=t×f
5.
(a)
t=8×j
(b)
a=
b
q
x
h
g
y
b
r
(c)
q=
(c)
r=7×x
OUTCOME 3: NUMERACY/HIGHER
81
TRANSPOSITION OF FORMULAE
?3.7: Answers
1.
4.
s
7
t
m=
k
k=
2.
5.
m
9
B
g=
4h
t=
3.
d=
6.
m=
3.
a=
C
π
A
5bx
?3.8: Answers
1.
p = 4w
2.
4.
t=
2π
5T
5.
XS
2
4kpqr
a=
3bc
r=
6.
cdm
b
3abc
q=
4kpr
?3.9: Answers
1.
4.
360A
C
NB nB
nA =
NA
a=
2.
5.
Ap
7
AD
B=
C
w=
3.
6.
7w
A
q
r=
Ap
p=
?3.10: Answers
1.
q = p + rs
4.
a=
s + 3nq
p
2.
5.
k – 3t
a
4r – T
c=
abd
h=
3.
6.
x – 2y
3c
π–K
q=
pr
b=
?3.11: Answers
1.
x=
T + 6y
3x – T
; y=
and you can’t cancel the 6 and the 3!
3
6
2.
t=
P
P – 3t
; m=
3 + 5m
5t
3.
Multiply each side by 2 first, giving 2Y = a + 2bc after which
a = 2Y – 2bc or 2(Y – bc); b =
4.
Multiply each side by 15 (= 3 × 5) first, giving 15X = 5t – 6m after which
t=
82
2Y – a
and you can’t cancel the 2s.
2c
15X + 6m
5t – 15X
; m=
.
5
6
OUTCOME 3: NUMERACY/HIGHER
TRANSPOSITION OF FORMULAE
5.
6.
7k – 3
but remember it could be written ½(7k – 3). This applies to lots of
2
other answers as well.
w=
Whether it’s x or y you’re trying for, the first steps for both are
Hx = 3x – 3y
3y = 3x – Hx
after which, obtaining y requires one compulsory and one optional move:
3x – Hx
(compulsory)
3
(3 – H )x
(optional)
y=
3
y=
and to get x we must do this:
3y = (3 – H )x
3y
x=
(3 – H )
7.
Put P over 1 and cross multiply, multiply out the bracket you get, then shift the a
terms to one side and the b terms to the other. You do all that whether it’s a you
want or b.
P(a – b) = a + 2b
Pa – Pb = a + 2b
Pa – a = Pb + 2b
a(P – 1) = b(P + 2)
Now we can get either a or b:
a=
8.
b(P + 2)
a(P – 1)
; b=
(P – 1)
(P + 2)
Same idea as in Question 7
x=
6y (t + 1)
(3 – 2t )
; y=
(3 – 2t )
6(t + 1)
In both cases, the brackets round the 3 – 2t are optional, although the brackets
round the t + 1 obviously aren’t.
OUTCOME 3: NUMERACY/HIGHER
83
TRANSPOSITION OF FORMULAE
?3.12: Answers
Q
3
1.
P=
3.
k2
k
t =   or
9
3
5.
a=
25
5
or
y
y
2.
x=
4.
4
 2 
x=
 or
25m
 5m 
2
2
6.
y – 3x 2
; x=
b
y = z2 – x
7.
y=
8.
Start with m – 5x 2 =
y – ab
3
x (1 – t 2 )
t2 y
;
x
=
t2
1 – t2
2y to isolate the rooted term, then square both sides:
(m – 5x2)2 = 2y and finally isolate y and reverse: y = ½(m – 5x2)2 or
prefer.
?3.13: Answers
1.
d=
C
π
2.
h=
V
; r=
πr 2
3.
Multiply both sides by 2 first
2F
; v=
v2
m=
V
πh
2F
m
3V
4π
4.
r=
5.
m=
6.
Remember to isolate the rooted term first: l =
84
3
fd 2
; d=
Gm2
Gm1m 2
f
OUTCOME 3: NUMERACY/HIGHER
gT 2
4π2
(m – 5x 2 )2
if you
2
TRANSPOSITION OF FORMULAE
v –u
t
7.
u = v – at; a =
8.
s=
v 2 – u2
; u = v 2 – 2as
2a
9.
t=
2s
2s – vt
2s
or u =
or
–v
(u + v )
t
t
10.
u=
2s – at2
2s – 2ut
2(s – ut)
or a =
or
2t
t2
t2
11.
b = a2 + c2 ; c = b2 – a2
12.
h=
A
A – 2πr 2
–r
or
2πr
2πr
13.
i=
R(i – I )
I (R + r )
Ir
iR
iR – I R
; r=
– R or
or
; R=
I
I
I
R
i–I
14.
E=
I (R + nr )
nE
nE – nr I
n(E – r I )
; R=
– nr or
or
I
I
I
n
r=
I  nE
IR
E R
nE – I R

– R  or
–
or
; n=

I
In
E – Ir
n I
n

OUTCOME 3: NUMERACY/HIGHER
85
86
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
SECTION 4
Introduction – Bivariate Data
This title does look forbidding, but believe me it’s not really that difficult. Bivariate data
simply means data where the information comes in pairs of mumbers, each from a different
source, in other words, each number is a different variable. It could be that one number is
the amount of money spent on advertising a particular product and the other is the value of
the sales made as a result.
As usual, a first step to analysing this sort of information is often to draw a graph. The kind
of graph most usually associated with the analysis of bivariate data is the scattergraph or
scatter diagram.
4.1 Scattergraph or Scatter Diagram
A scatter diagram is a method of plotting the values of two variables, using axes and plotting
points just as you did for line graphs, but without trying to join any of the points together
(this is because usually there are too many dots for any obvious line to be drawn). The
resulting distribution of dots or crosses can then give us an idea of whether the relationship
between the two variables is weak or strong. The technical word for this relationship is
correlation, and the calculations and study of correlation form a significant part of many
maths and statistics courses. You’ll learn more about this very soon.
Here is a very simple scatter diagram which plots four
models of aeroplane. One axis shows the number of
passengers the plane can carry, the other axis shows the
hourly cost of operating the plane.
Carrying Capacity
Example 4.1a
•B
•D
•C
A shows a plane which costs little to operate and carries
•A
few passengers.
O
Operating Costs
Operating
Costs
B shows a plane which costs little to operate (but a bit
more than A) and carries many passengers.
C shows a plane which costs a lot to operate and carries few passengers (but a few more
than A).
The diagram does not show any relationship whatever between the two variables (i.e.
between operating costs and carrying capacity).
OUTCOME 3: NUMERACY/HIGHER
87
CORRELATION AND REGRESSION
Activity 4.1a
What do you think the scatter diagram shows about aeroplane D?
Example 4.1b
Each dot represents an aeroplane in an airline’s fleet.
The diagram shows that there is a fairly high
correlation (i.e. a strong relationship) between the
number of passengers a plane can carry and the profit
which that plane generates for the airline’s coffers.
The correlation is positive, i.e. the higher the value of
one variable, the higher the value of the other.
Profit
Here is another scatter diagram.
•
•
•
•
•
•
• •
•
• •
•
•
•A
Carrying
CarryingCapacity
Capacity
0
Activity 4.1b
What does the diagram tell you about the aeroplane marked A? Can you suggest any reasons for
this?
Here, a factory manager has plotted the age of each
machine in the factory against the efficiency of that
machine. Again, as with the aeroplanes in Example
4.1b, the correlation is high.
In fact it is higher here than in Example 4.1b. How do
we know? Because the dots lie closer to a straight line
here. In fact, if the dots actually all lie on a straight line
then we have perfect correlation: you can’t get better
than that.
Efficiency Index
Example 4.1c
•
• •
• •
•A
0
•
• • •
•
•
•
Age
Ageofof Machine
Machine
In this case our correlation is negative. The higher the age of the machine, the lower the
efficiency index. (In other words, the older it is, the worse it works.)
Activity 4.1c
The machine marked A sticks out like a sore thumb on the diagram, whereas it might have been
buried in a page of figures. What does the diagram say about machine A? Suggest a reason.
88
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
Example 4.1d
Here, someone has plotted the relationship between
the wages paid to employees of a company and the
physical height of those employees.
•
•
•
•
Clearly, we would expect there to be no relationship
at all, and the scatter diagram shows this with a
‘plum pudding’ effect.
•
•
0
•
•
•
•
• •
•
•
•
Wage paid by Firm
There is no correlation at all, either positive or
negative. We say that the correlation is zero.
If the dots had happened to lie reasonably close to a straight line, we would hope that the
correlation was spurious, i.e. completely coincidental, and that the firm did not have any
policy of paying by height!
OUTCOME 3: NUMERACY/HIGHER
89
CORRELATION AND REGRESSION
? 4.1
For each diagram say whether it displays positive correlation, negative correlation, or zero
correlation. Write a sentence or two about each one. Try to put the strengths of the
correlations in order of size.
(2)
Volume of Production
Company Profits (£)
(1)
0
Company Sales (£)
Examination Marks
0
Rainfall (mm)
0
Age of Student
(6)
Traffic Speed (mph)
Production Costs (£)
(5)
0
90
Absenteeism by Employees
(4)
Crop Yield (tonnes)
(3)
0
Production Level (units)
OUTCOME 3: NUMERACY/HIGHER
0
Volume of Traffic (000s)
CORRELATION AND REGRESSION
Important Note
It is important to remember, and this will be repeated several times, that the existence of
correlation does not, repeat NOT, imply that one effect is CAUSED by the other. It simply
shows that the two variables are related. Both of them could, of course, be caused by a
third variable you don’t even know about.
So when we talk about graphing the independent variable horizontally and the dependent
variable vertically, these are expressions we use to denote a notional dependence – ‘if there
is a dependence, this is how it would be’ – but the existence of correlation does NOT imply
causation.
4.2 Drawing Scatter Diagrams
When drawing a scatter diagram, plot the variables along a pair of axes in the same way as
you would for a line graph, the only difference being that you don’t (in fact can’t) join them
up with a line. As usual, the variable which, it is suspected, is dependent on the other is
plotted vertically and the independent variable is plotted horizontally.
Example 4.2a
This table shows the diameters (in inches)
of a variety of cylindrical components
produced on a lathe and the time taken (in
seconds) by the machinist to make them.
Draw a scatter diagram to illustrate this.
Solution:
If anything, the time taken to make the
component will depend on the diameter of
the component (although they might both
depend on something else), so we plot time
vertically and diameter horizontally.
Component
A
B
C
D
E
F
G
H
I
J
K
L
Diameter
3.006
3.012
3.001
2.998
3.015
3.009
3.013
3.000
2.997
3.005
3.010
3.016
Time
195
205
200
185
210
215
200
190
195
205
200
220
The lowest diameter is 2.997 and the highest is 3.016, so a horizontal scale from 2.990 to
3.020 will suffice.
The lowest time is 185 and the highest is 220 so a vertical scale from 180 to 220 will do.
Each point is plotted in its correct position relative to the axes. I have also typed the
reference letter beside each one, though this is not usually done. As you see, joining the
points up with a line is just nonsense.
OUTCOME 3: NUMERACY/HIGHER
91
CORRELATION AND REGRESSION
Scatter Diagram showing Diameters and Production Times
220
L
215
F
210
E
Time
205
B
J
200
C
195
K
190
G
A
I
H
185
D
180
2.990
2.995
3.000
3.005
3.010
Diameter (inches)
3.015
3.020
The scattergraph shows a reasonably strong positive (sloping up from left to right) linear
(i.e. straight line) relationship between the two variables. In general, the larger the
diameter of the component, the longer it takes to make.
Activity 4.2a
On the diagram above, plot the following:
Component M, diameter 3.007 inches, taking 208 seconds
Component N, diameter 3.003 inches, taking 210 seconds
and make a comment about each one.
You can try to draw a best-fitting straight line (or line of best fit) through these points
which would help you to decide, for example, how long it would take to produce a
component of diameter 3.008 inches (i.e. within the known range of data) or of diameter
3.03 inches (outwith the known range of data). Such a line is called the line of regression
of y on x. Drawing one by eye is very difficult and not at all reliable, as the graph below
shows. You may well disagree with the one shown.
Better ways of doing this will follow.
Scatter Diagram showing Diameters and Production Times
220
L
215
F
210
E
Time
205
200
C
195
G
H
185
92
K
A
I
190
180
2.990
B
J
D
2.995
3.000
3.005
3.010
Diameter (inches)
OUTCOME 3: NUMERACY/HIGHER
3.015
3.020
CORRELATION AND REGRESSION
Example 4.2b
School
This table shows the pupil:staff ratio at
fourteen secondary schools and also the
percentage of pupils who passed 5 or more
Highers.
Draw a scatter diagram to illustrate the
data.
Percentage
A
B
C
D
E
F
G
H
I
J
K
L
M
Pupils per
teacher
17
19
12
10
16
13
18
20
22
13
19
8
19
N
17
20
10
12
16
15
10
14
9
8
10
17
7
23
19
Solution:
The resulting scatter diagram is shown below.
Scatter Diagram of Teaching Ratio and Pass Rate
25
Percentage
20
15
10
5
0
0
5
10
15
Pupil : Teacher Ratio
20
25
Here we see a relatively strong negative (because the general slope of the ‘line’ is down, left
to right) correlation between the two variables. In general, the more pupils there are per
teacher, the fewer of the pupils do well in their exams. The strength of the relationship is
clearly not as high as it was in Example 4.2a, so drawing the line of regression by eye is even
more problematic.
OUTCOME 3: NUMERACY/HIGHER
93
CORRELATION AND REGRESSION
Drawing a line of best fit by eye
Drawing a line of best fit is not actually as hit and miss as you might think. There is a point
through which the best line will go.
Find the average of the values of each variable (separately) and you will have a point on
which to anchor your ruler. How the line will slope is then a matter of conjecture.
In the example above,
the average of the pupil:teacher ratio column is (17 + 19 + ... + 19 + 17)/14 = 15.9
the average of the percentage column is (10 + 12 + ... + 19 + 20)/14 = 13.6
Plot that point on the graph, then use your skill and judgment to draw the line. A possible
example is:
Scatter Diagram of Teaching Ratio and Pass Rate
Point (15.9 , 13.6)
25
20
15
10
5
0
0
5
10
15
Pupil : Teacher Ratio
In some textbooks this point is called the mean centre.
94
OUTCOME 3: NUMERACY/HIGHER
20
25
CORRELATION AND REGRESSION
? 4.2
Draw a scatter diagram to illustrate each of the following tables.
Once you have drawn the graph, draw on it what you think is the best fitting straight line.
You are looking for a straight line which passes through as many points as possible, with as
many points above it as below it, but also passing through the mean centre as calculated on
the previous page.
Keep your answers; you’ll need them later.
1.
An advertising executive is investigating
the relationship between the size of the
budget for certain advertising campaigns
(horizontal scale) and the resulting sales
(vertical scale).
Use your line to predict the sales for an
advertising expenditure of £7 million.
2.
A town planner is looking at the
population density of certain areas of the
town (horizontal scale) and their
distance from the town centre (vertical
scale).
Use your line to predict how far away
from the town centre you would find a
population density of 40 persons per
hectare.
3.
An economist from an energy company is
looking at the demand from a power
station (vertical scale) and how it is
affected by the average daily
temperature (horizontal scale).
Use your line to predict the demand if
the temperature is 5.5°C.
Advertising
Expenditure (£m)
3.4
4.0
5.2
6.5
7.9
8.3
9.2
10.0
Sales (£m)
Persons per
Hectare
65
21
78
48
56
53
31
34
Distance from
Centre (km)
0.8
3.7
1.6
2.7
1.5
2.6
3.4
1.9
Temperature (°C)
-2.0
1.0
2.3
4.1
4.9
8.3
6.5
3.8
12.0
15.8
19.7
19.2
26.0
24.7
29.2
29.7
Demand (MWh)
60.5
58.9
56.4
51.3
48.0
42.6
43.2
53.1
OUTCOME 3: NUMERACY/HIGHER
95
CORRELATION AND REGRESSION
4.3 Correlation and Regression
We can have different types of correlation and we try to draw a line of best fit, if we can,
through the points plotted on the scatter diagram, which will allow us to make point
estimates of values of y for any given x.
The line we draw is called the line of regression of y on x. It allows us to predict a value
of y for a given value of x, but not the other way round.
We will be looking at linear regression, where the line of best fit is a straight line. But
other forms of regression exist, and many scientific calculators with statistical functions can
apply more than one form of regression to given data so that you can choose the best one
(note – so that you can choose, the calculator won’t make the choice for you!)
Here we have positive linear correlation.
y
The line of points is best approximated by a straight line
whose form is
y = a + bx
x
where b will have a positive value because the line is sloping
up from left to right.
Here we have negative linear correlation.
y
The line of points is best approximated by a straight line
whose form is
y = a + bx
x
where b will have a negative value because the line is
sloping down from left to right.
Here we have quadratic correlation.
y
The shape of the points is best approximated by a
parabola, whose equation is of the form
y = ax2 + bx + c
96
OUTCOME 3: NUMERACY/HIGHER
x
CORRELATION AND REGRESSION
Here we have exponential correlation.
y
The shape of the points is best approximated by a curved
line whose equation is of the form
y = ab x or y = axb
x
Here we have logarithmic correlation.
y
The shape of the points is best approximated by a curve
whose equation is of the form
y = a + b logex
x
We will be looking only at linear correlation.
Associated with the regression line is the coefficient of correlation. There are several of
these measures used; the one we will be looking at was invented by someone called
Pearson, and its full title is Pearson’s Product Moment Correlation Coefficient.
Its calculation involves means and standard deviations, and a few other things besides, but
the formula he devised essentially boils down to this:
The closer the points all lie to the line of regression, the nearer the coefficient is to either
+1 (if the the line has a positive slope) or –1 (if the line has a negative slope).
The further away from the line that the points lie, the closer the coefficient (which is given
the letter r as an identifier) is to 0.
All values of r thus lie between +1 and –1, which both show perfect correlation, i.e. the
strongest possible relationship between the two variables. The value of 0 for r shows no
correlation at all, i.e. no relationship whatever between the two variables.
OUTCOME 3: NUMERACY/HIGHER
97
CORRELATION AND REGRESSION
For example:
y
r = 0.82
x
y
r = -0.95
x
y
r = -0.15
x
I will now give you the formulae required. But first, a little disagreement with our American
partners. For reasons I won’t go into here, American textbooks often give the regression
equation in the form
y = ax + b
whereas British textbooks give the equation as
y = a + bx
There is thus confusion between the values of a and b as shown in various formulae.
(Thankfully, the formula for r is the same no matter which convention you use.)
So make sure you know what’s what!
Even calculators are not consistent. At college I tend to use two calculators both made by
the same manufacturer and both able to calculate correlation. One model uses ax + b and
the other uses a + bx.
And to confuse things just a little bit more, mathematics textbooks use y = mx + c!
98
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
Here is the convention and the formulae I will be using in this section:
For the line of regression of y on x of the form y = a + bx:
b=
r=
nΣxy – ( Σx )(Σy )
nΣx – (Σx )
2
2
; a=
Σy – bΣx
n
nΣxy – (Σx )(Σy )
 nΣx 2 – (Σx )2   nΣy 2 – (Σy )2 



These look complex, but the formulae actually tell you what columns of figures you have to
add up. We need an x column, a y column, an xy column, an x 2 column and a y2 column.
The n of the formula is the number of points we have as data.
In case you haven’t met it before, the Σ sign is a Greek symbol (read ‘sigma’) which means
‘the total of’.
Where do these formulae come from?
There is a method known as least squares. It takes the distance each point is from the
regression line, squares it (to get rid of the problem of some of these distances being
positive and some negative, due to their position above or below the line) and then uses
calculus to minimise the total of these squared distances.
The best fit line is thus sometimes called the least squares line.
Don’t worry, the vast majority of people who use these formulae don’t know (or care)
where they come from, they just use the results. You won’t at any point be expected to
know calculus!
Example 4.3a
We’ll start with an example where we can find the answer just by looking.
It costs £40 to set up a computer to print a logo on a T-shirt, then it costs a further £5 per
shirt.
So there is a start-up cost of £40, then you have to multiply the number of shirts (= x) by £5
to get the cost of the actual shirts, and add it on to the 40.
So the formula will be y = 40 + 5x where y is the total cost. If you’ve done any
accountancy or economics you’ll recognise the format as ‘fixed cost + variable cost’.
OUTCOME 3: NUMERACY/HIGHER
99
CORRELATION AND REGRESSION
Make up a table of values (the x and y will be given to you in a normal situation) then add
columns for x2, y2 and xy.
Multiply each
number of the x
column by itself.
0 ×0 = 0
3 ×3 = 9
5 × 5 = 25
and so on
V
V
0
3
5
8
10
26
y
40
55
65
80
90
330
x2
0
9
25
64
100
198
y2
1,600
3,025
4,225
6,400
8,100
23,350
V
V
V
The numbers in
this row are the
totals of each
column.
x
xy
V
The x and y values
are normally given
to you to start with
Multiply each x
by its partner y.
Thus:
0 × 40 = 0
3 × 55 = 165
5 × 65 = 325
and so on
Multiply each
number of the y
column by itself.
40 × 40 = 1,600
55 × 55 = 3,025
65 × 65 = 4,225
and so on
0
165
325
640
900
2,030
The calculations are now as follows:
5 × 2,030 – 26 × 330
1,570
=
=5
2
5 × 198 – 26
314
330 – 5 × 26
200
a=
=
= 40
5
5
1,570 (= same numerator as for b)
r=
=
[314 = same denominator as for b]× 5 × 23,350 – 3302 
b=
1,570
314 × 7,850
=
1,570
1,570
Here is the graph with points plotted and the line of regression drawn on.
90
80
70
60
yy
50
40
30
20
10
0
0
1
2
3
4
5
xx
6
7
8
9
10
The values of b = 5 and a = 40 confirm what we already knew, that y = a + bx is, in
fact, y = 40 + 5x.
Furthermore, we get a value of 1 for our correlation coefficient, which comes as no surprise
since the points all exactly fit the formula.
100
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
Just to confirm what I said earlier about drawing by eye:
the average of the xs is (0 + 3 + 5 + 8 + 10)/5 = 5.2
the average of the ys is (40 + 55 + 65 + 80 + 90)/5 = 66
Looking at the graph you can vaguely see that the line passes through that point. But, more
importantly, if we take x as being 5.2 and substitute into the formula 40 + 5x =
40 + 5 × 5.2, we get 40 + 26 = 66 as expected.
Example 4.3b
Now let’s take three of our points and change them a bit, and see how that affects the graph
and the calculations.
x
y
30
55
40
80
80
285
0
3
5
8
10
26
Changing the y values of three of the
points changes some of the table entries
and also some of the column totals.
x2
0
9
25
64
100
198
y2
900
3,025
1,600
6,400
6,400
18,325
xy
0
165
200
640
800
1,805
The calculations change as well:
5 × 1,805 – 26 × 285
1,615
=
= 5.143
5 × 198 – 262
314
285 – 5.143 × 26
151.282
a=
=
= 30.256
5
5
1,615 (= same numerator as for b )
r=
[314 = same denominator as for b]× 5 × 18,325 – 2852 
1615
1615
=
= 0.894
=
1807.097
314 × 10400
b=
90
80
70
60
yy
50
40
30
20
10
0
0
1
2
3
4
5
6
7
8
9
10
xx
OUTCOME 3: NUMERACY/HIGHER
101
CORRELATION AND REGRESSION
The best fitting line no longer goes through all the points, so we expect the value of r to be
less than 1. In fact, r is now 0.894, still showing a strong positive relationship between the
two variables but not as strong as before.
The equation of the line of regression is now y = 30.256 + 5.143x thus reflecting its
altered position on the diagram.
How do we draw the line?
The technique is to pick two values of x, as far apart as is practical, and substitute them in
the formula to calculate y.
Choose, say, x = 0. Then y = 30.256 + 5.153 × 0 = 30.256 so we could plot (0, 30) on
the diagram because any greater degree of precision isn’t possible.
Now pick another x, say 8. Then y = 30.256 + 5.153 × 8 = 71.480 so plot the point
(8, 71). Then draw the line connecting these two points.
What do we use the line for?
The line of regression will now give us a point estimate of the cost for any number of
shirts. In reality, we would have to qualify our answer with a ‘plus or minus ...’ but the
calculation of this is extremely complicated (it involves things called confidence intervals) so
I’ll give you a simplified version.
What is the cost for 7 shirts? Looking at the graph, you see that the cost will be round
about £68.
If we replace x by 7 in the equation, we get y = 30.256 + 5.143 × 7 = £66.257. But
remember that this is just an estimate, so ‘round about £66’ will be a good answer.
This answer is fairly reliable, because the value of r is close to 1. In addition, our estimate
was for a value within the known range of data (the values of x we were given lay between 0
and 10) thus adding to its reliability.
What is the cost of 20 shirts? This can be calculated as 30.256 + 5.143 × 20 = £133.116.
But here we are forecasting beyond the known range of data (i.e. outside the limits of the
graph as we can see it) and quite far beyond. We would expect the answer of ‘around £133’
to be a bit less reliable than our previous answer.
The further we go beyond the range of data that we know about, the less reliable any point
estimates will be, since there may be factors ‘out there’ affecting results in a manner that we
don’t know. Further study of the range of values that a point estimate can represent is not
covered in this study pack.
102
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
Example 4.3c
Here is now a full worked example, from start to finish.
A factory produces batches of items. The table shows the number of items produced (in
hundreds) and the corresponding cost of producing that batch (in thousands of £s).
No. of items produced (00) 10
24
30
33
Cost of production (£000) 380 400 510 500
45
45
485 500
54
58
60 69
590 523 600 618
Draw a scatter diagram, calculate the coefficient of correlation and the equation of the
regression line of y on x, and draw the line on your diagram.
Then estimate the cost of a batch of (a) 4,320 items and (b) 7,540 items and comment on
the reliability of these estimates.
Solution:
The first and most important matter to sort out is which variable is x and which is y?
The clue is in the last line of the question ‘...estimate the cost of a batch of 4,320 items...’
The regression line is the line of regression of y on x and allows us to estimate y for a given x.
We are given 4,320 items. It follows that Number of items is x and Production cost is y.
We have to write the data in two columns, and also draw the graph. Here’s the graph.
y = cost (£000)
700
600
500
400
300
0
10
20
30
40
50
60
70
= no.
xx=
no.ofofitems
items(00)
(00)
Next, put in the extra columns and evaluate the formulae:
OUTCOME 3: NUMERACY/HIGHER
103
CORRELATION AND REGRESSION
x (00 Items)
b=
x2
y (£000 cost)
y2
Reminder:
xy
10
380
100
144,400
3,800
24
400
576
160,000
9,600
30
510
900
260,100
15,300
33
500
1,089
250,000
16,500
45
485
2,025
235,225
21,825
45
500
2,025
250,000
22,500
54
590
2,916
348,100
31,860
58
523
3,364
273,529
30,334
60
600
3,600
360,000
36,000
69
618
4,761
381,924
42,642
428
5,106
21,356
2,663,278
230,361
nΣxy – (Σx )(Σy )
nΣx – ( Σx )
2
2
=
First Line:
100 comes from
102
144,400 comes
from 380 2
3,800 comes
from 10 × 380
10 × 230,261 – 428 × 5,106
118,242
=
= 3.893
2
10 × 21,356 – 428
30,376
Σy – bΣx
5,106 – 3.893 × 428
=
= 344.0
10
n
118,242
118,242
= 0.905
=
r=
2
30,376 × 561,544
[30,376]× 10 × 2,663, 278 – 5,106 
a=
The line of regression of y on x has equation y = 344.0 +3.893x
This shows two things:
(a)
it costs 344.0 (£000), in other words £344,000, to produce no items at all. We get
this by setting x equal to 0 in the equation. Note the units: each unit of y is actually
£1,000.
(b)
3.893 is the gradient of the line (compare with y = mx + c). In other words it is the
rate of y for each x. Each x costs 3.893 (y) to produce. But since x is in hundreds of
items and y is in £000, this means that every 100 items cost £3,893 to produce.
We can easily draw the line of regression on the graph – simply choose two easy values of x
quite far apart, calculate y, plot the points then join them up:
x = 0 , y = 344.0 + 3.893 × 0 = 344 (we already knew that)
x = 70, y = 344.0 + 3.893 × 70 = 616.51
The line is plotted on the next page.
Since the value of the correlation coefficient is very high (0.905 is very close to 1) this
means that there is a very strong relationship between the number of items produced (in
hundreds) and the production cost (in £000). The points on the graph all lie close to the
line of regression.
104
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
y = cost (£000)
700
600
500
y=
400
.0
344
.89
+3
3x
300
0
10
20
30
40
50
60
70
= no.
(00)
xx =
no.ofofitems
items
(00)
Now for the forecasts:
(a)
For 4,320 items, we have to put x = 43.20 into the regression equation (which has
x in hundreds of items) and we get y = 344.0 + 3.893 × 43.20 = 512.178.
This means a cost of £512,178 although it would make more sense to round it off to,
say, £512,000 since it is a point estimate after all. This is roughly the answer which
the graph would give you.
We would expect this estimate to be quite reliable because
(i)
the value of r is close to 1
(ii)
our value of x is within the range of known data (i.e. 43.20 is between 10 and
69).
(b)
For 7,540 items, we have to put x = 75.40 into the regression equation and we get
y = 344.0 + 3.893 × 75.40 = 637.532 meaning £637,532 which can be rounded off
to perhaps £638,000.
This estimate is not as reliable as the previous one because, despite the high value of r, we
are forecasting beyond the known range of data (75.40 is beyond 69).
OUTCOME 3: NUMERACY/HIGHER
105
CORRELATION AND REGRESSION
? 4.3
1.
Dunedin Chemicals Ltd plan to use data from the previous six months to obtain a
formula for estimating the total power costs for a given output. The relevant data are
shown.
Output (units)
Power costs (£000)
12
6.2
18
8
19
8.6
20
24
30
10.4 10.2 12.4
Draw a scatter diagram, calculate the equation of the line of regression and the value
of the correlation coefficient, then draw the line of regression on your diagram.
Estimate the total power costs for an output of (a) 22 units (b) 35 units and
comment on the reliability of your estimates.
2.
A hire purchase company predicts the level of business it expects by considering
current rates of interest. Graph the given data on a scatter diagram, calculate the
coefficient of correlation, and comment on the strength of the relationship between
the two variables.
Value of contracts (£million) 5.0 5.5 4.8 4.0
Rates of interest (%)
10.0 10.5 11.0 11.0
6.0
9.0
5.5 6.2
10.5 9.5
7.0
8.5
Calculate the equation of the line of regression and draw it on the diagram. Use it to
estimate the value of contracts if the interest rate is 8%. Comment on the reliability
of your estimate.
3.
A consumers’ organisation awarded marks out of 20 for each of nine washing
machines tested, and these marks are shown below together with the recommended
retail price (RRP) of each machine.
Machine
Mark/20
RRP (£)
A
16
300
B
10
220
C
13
250
D
7
175
E
19
295
F
8
210
G
13
245
H
11
255
I
14
260
Calculate the correlation coefficient and also the equation of the line of regression.
How much would you expect to pay for a machine which scored full marks in the
survey ?
106
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
4.
The following table shows the labour cost (£) and the corresponding output (tonnes)
of a manufacturing plant.
Output
Labour costs
7,600
8,400
8,800
8,000
9,100
10,000 9,700
60,000 63,000 69,000 62,000 74,000 95,000 87,000
As you see, the numbers are all on the large side, so you may wish to alter their size
before proceeding further. Draw the scatter diagram, estimate what the labour cost
would be for an output of 12,400 tonnes, and say how reliable you think this estimate
will be.
5.
6.
A company has a new management accountant who is attempting to install a new
budgetary control system into one of the company’s operating plants. The following
historical data has been established for plant overheads for the last 12 months. Help
him decide whether the overhead costs vary more with the output or the overhead
costs vary more with the number of employees, and find the equation of the
appropriate regression line.
Month
(1998)
No. of
Employees
January
February
March
April
May
June
July
August
September
October
November
December
200
180
220
200
250
220
190
150
160
240
240
180
Output
(standard
hours)
16,000
14,400
18,200
18,000
20,100
18,600
17,500
12,500
16,600
20,400
19,700
14,000
Overhead
Costs (£)
52,400
48,500
56,400
55,700
60,300
57,000
55,000
46,500
53,000
61,000
59,500
47,900
Look back at the three questions in Exercise 4.2. For each one calculate the
correlation coefficient and the equation of the line of regression.
Then calculate the same estimate as you did last time by eye. How close were you?
OUTCOME 3: NUMERACY/HIGHER
107
CORRELATION AND REGRESSION
Answers
Activity 4.1a
D’s operating costs are a bit lower than C’s but considerably higher than A’s or B’s.
D’s carrying capacity is a bit lower than B’s but considerably higher than A’s or C’s.
Activity 4.1b
Its carrying capacity is the highest of all the planes but the profits generated by it are almost
the worst. Perhaps the route is not popular and so the plane flies aound half empty, a
smaller plane should be assigned to that route.
Note: if you are asked for a reason, it’s bound to be one which requires no specialist
knowledge but just a bit of common sense.
Activity 4.1c
It’s the third youngest machine but has almost the worst efficiency. Perhaps there are
unexpected teething troubles, or the operators have not been properly trained.
?4.1: Answers
1.
2.
3.
4.
5.
6.
Positive
Negative
Negative
None
Positive
Negative
Strongest to weakest: 5, 1, 6, 2, 3, 4
The stronger the relationship between the two variables, the closer the dots lie to a straight
line.
You will have noticed that in question 6 a curve fits the dots better than a straight line. Not
all correlation is linear; more about this later.
108
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
Activity 4.2a
Scatter Diagram showing Diameters and Production Times
220
215
N
210
Time
T 205
i 200
m
e 195
190
. .
M
L
F
E
B
J
C
K
G
A
I
H
185
180
2.990
D
2.995
3.000
3.005
3.010
Diameter (inches)
3.015
3.020
?4.2: Answers
1.
Your regession line will go through the point (Advertising 6.8, Sales 22.0).
Relation of Sales to Adv ertising Expenditure
30.0
Sales (£m)
25.0
20.0
15.0
10.0
3.0
4.0
5.0
6.0
7.0
8.0
Adv ertising Expenditure (£m)
9.0
10.0
Estimated sales will be around £23–24 million.
Your regression line will go through the point (Density 48.2, Distance 2.3).
Relationship of Population Density and Distance f rom Town Centre
Kilometres from Town Centre
2.
4
3.5
3
2.5
2
1.5
1
0.5
0
20
30
40
50
60
Persons per Hectare
70
80
Estimated distance is about 2.6km.
OUTCOME 3: NUMERACY/HIGHER
109
CORRELATION AND REGRESSION
3.
Your regression line will go through the point (Temperature 3.6, Demand 51.8).
Relationship of Power Demand to Temperature
65.0
60.0
55.0
50.0
45.0
40.0
-2.0
0.0
2.0
4.0
Temperature (
6.0
8.0
10.0
° C)
Estimated demand is approximately 48 MWh.
?4.3: Answers
1.
x is Output (horizontal axis), y is Power Costs (vertical axis). They MUST be that
way round.
x
12
18
19
20
24
30
123
y
x2
6.2 144
8
324
8.6 361
10.4 400
10.2 576
12.4 900
55.8 2,705
y2
xy
38.44
74.4
64
144
73.96
163.4
108.16
208
104.04
244.8
153.76
372
542.36 1,206.6
6 × 1,206.6 – 123 × 55.8
= 0.341689
6 × 2,705 – 1232
55.8 – 0.341689 × 123
a=
= 2.29537
6
6 × 1,206.6 – 123 × 55.8
r=
 6 × 2,705 – 1232  × 6 × 542.36 – 55.8 2 
= 0.956436
b=
Correlation coefficient is 0.96 (rounded off to 2 d. p.) This is very close to 1 and thus
shows a very strong relationship between the two variables.
The equation of the regression line is y = 2.30 + 0.34x.
To draw the line on your diagram, pick any two points reasonably far apart. (The
further apart these are, the less significant any errors you make will be, by having too
thick a pencil, say, or by missing your point altogether.)
Say x = 10, then y = 2.30 + 0.34 × 10 = 5.7.
Say x = 22, then y = 2.30 + 0.34 × 22 = 9.78.
(As you see, once you get to the regression line you can drop all the decimal places
your calculator gave you, and just stick to 2.)
Join up the two points just found and there’s your line.
110
OUTCOME 3: NUMERACY/HIGHER
CORRELATION AND REGRESSION
Required estimates:
(a)
For 22 units we get y = 9.78 but remember that this is in £000 so the
estimate is £9,780. This should be reliable because (i) r is very high and (ii)
we are estimating within the known range of data.
(b)
For 35 units we get y = 14.2 meaning £14,200. The 35, however, is beyond
the range of known data (remember x only went from 12 to 30) so the
estimate won’t be as reliable as for (a) although the high value of r would still
make it reasonable.
NB if you plot your x and y the wrong way round you get the regression line of x on y
and you would end up estimating the output for a given power cost. This is a
complication you do not need. So get them the right way round to start with!
2.
The interest rate is x (plotted horizontally) and the contract value is y (plotted
vertically).
The column totals are:
Σx = 80; Σy = 44; Σx2 = 806; Σy2 = 247.98; Σxy = 434.7
b= –0.88333 and a = 14.3333 so the regression line has equation y = 14.33 –
0.88x.
r = –0.88481 ..., showing a very high correlation. The line will slope down from left
to right.
For an interest rate of 8% we get y = 7.269 i.e. the value of the contracts will be
approximately £7.3 million, or £7,269,000.
The points I picked when drawing the graph are (8, 7.3) and (11, 4.7).
3.
x is the mark out of 20, y is the RRP.
The column totals are
Σx = 111; Σy = 2,210; Σx 2 = 1,485; Σy2 = 555,300; Σxy = 28,390
The equation is y = 125.06 + 9.77x.
The value of r is 0.937, showing a very strong relationship.
Points which could be used to draw the line are (0,125) and (20, 320).
Full marks is 20 out of 20, so x = 20 and y = 320.46, meaning £320.46, but ‘about
£320’ is much better. It is a less precise answer but shows that you understand that
an estimate is not expected to be accurate.
OUTCOME 3: NUMERACY/HIGHER
111
CORRELATION AND REGRESSION
This figure of £320 should be reliable, as the value of r is high and we are going only
a little way beyond the range of known data.
4.
It is easier on the calculator finger if you change the scale of your figures.
Make your x column (output) in thousands (7.6, 8.4, 8.8, etc.) and make your y column
(labour cost) also in thousands (60, 63, 63, etc.).
It is coincidental that the scale change is the same for each variable. You could make
the labour cost in hundreds (600, 630, 690, etc.) but it’s not as convenient.
Σx = 61.6; Σy = 510; Σx2 = 546.66; Σy2 = 38,244; Σxy = 4,555
y = –57.22 + 14.78x is the equation, r = 0.960 is very high.
For an output of 12,400 tonnes you have to make x = 12.4 (because the scale was
changed) and substitute this into the equation, giving y = 126.052. But this is in
thousands as well, so the estimate is £126,052 which looks more like an exact figure.
Therefore quote the estimate as ‘around £126,000’.
However, this estimate is being made for a value of x well beyond the data range, so
despite the high value of r and the strong relationship this implies, I wouldn’t bet my
life on it!
5.
This is really two questions in one.
(a)
Output = x (changed to thousands), overhead costs = y (thousands)
Σx = 206; Σy = 653.2; Σx2 = 3,606.88; Σy2 = 35,818.46; Σxy = 11,348.81
r = 0.9958
(b)
Employees = x (changed to hundreds), overhead costs = y (thousands)
Σx = 24.3; Σy = 653.2; Σx2 = 50.35; Σy2 = 35,818.46; Σxy = 1,338.2
r = 0.8931
Since the value of r is higher in (a) than it is in (b) the relationship is stronger there
and so we’ll find the regression equation for part (a): y = 21.45 + 1.92x.
6.
y = 4.60 + 2.56x; r = 0.98; estimate is £22.52 million.
7.
y = 4.26 – 0.041x; r = –0.78; estimate is 2.62 km. This would not be as reliable as
the answer to question 6 because r is not so high.
8.
y = 59.06 – 2.02x; r = -0.96; estimate is 47.95 MWh.
112
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
SECTION 5
A sequence is a list of numbers which are related to each other. Each item of the sequence
is called a term.
Invariably, this being Mathematics, there is a formula which generates each term of the
sequence.
Notation
We use the notation Un to denote the nth term of a sequence. The subscript of the letter U
tells you which term you are finding.
Thus U1 is the first term, U 2 is the second term, U 3 is the third and so on.
Just as the term before U8 is U7 , and the term after U8 is U9,
so the term before Un is Un–1 and the term after Un is Un+1.
5.1 Generating by Formula
Example 5.1a
Suppose we have a formula Un = 3n – 1. We can find any term of the sequence by replacing
n by the number of the required term.
So we can get the first three terms thus:
U1 = 3 × 1 – 1 = 2
U2 = 3 × 2 – 1 = 5
U3 = 3 × 3 – 1 = 8
Note that only the n
changes value as required,
all the other bits of the
formula stay as they are.
But the beauty of this is that, suppose we want to find, say, the 15th term. We don’t have to
find each and every term preceding the 15th. We can home in on it:
U15 = 3 × 15 – 1 = 44
Example 5.1b
Find the 8th term of each of these sequences:
(a)
Un = 3(n –1) 2
Solution:
U8 = 3 × (8 –1)2 = 3 × 72 = 3 × 49 = 147
OUTCOME 3: NUMERACY/HIGHER
113
RECURRENCE RELATIONS
(b)
Un = (–1)n+1n
Solution:
U 8 = (–1) 8+1 × 8 = (–1)9 × 8 = – 8
(c)
Un =
2n
n+5
Solution:
U8 =
2×8
16
=
8+5
13
? 5.1
1.
Find the first three terms of each of these sequences:
(a)
2.
(b)
Un = n2 + 1
(c)
U n = 5 – 2n
Find the required term in each of these:
(a)
(b)
(c)
(d)
(e)
3.
Un = 2n + 7
Un = 5n2; U4
Un = n2 + 3n – 2; U5
Un = (–1)n × 3n; U 8
Un = (–1)n+1 × (2n + 1); U8
Un = (n – 3)2; U7
Sometimes we can jump to premature conclusions. Calculate the first four terms of
each of these sequences, then make a rather obvious comment.
(a)
(b)
Un = 2n
Un = n2 – n + 2
5.2 Recurrence Relations
Sometimes it is extremely difficult to define a sequence by using a formula as above.
Sometimes we can only define it by relating each term to the previous one(s).
This means we also require to know what the first term is so that we have something to
start with. In these cases, U1 is the first term as before, but it is often easier to make U0 the
term which you are actually given, and U1 then becomes the first term which you have
actually calculated (i.e. the second term of the sequence).
114
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
I know that this seems complicated, but it can make life much simpler as you will see later
on. At the end of the day, it doesn’t actually matter which way you do it, so long as you
clearly state what your U0 and U 1 are, and get the correct answer.
The name given to the type of formula which relates a particular term to whatever went
before is recurrence relation.
Example 5.2a
A sequence is generated by the recurrence relation Un = Un–1 + 2 where U 1 = 5. Write
down the next four terms.
V
Solution:
The relation means ‘each term is equal to the previous term plus 2’
U1
=5
V
U2 = U1 + 2 = 5 + 2 = 7
V
U3 = U2 + 2 = 7 + 2 = 9
V
U4 = U3 + 2 = 9 + 2 = 11
U5 = U4 + 2 = 11 + 2 = 13
In the above example, we made U1 = 5 because making the subscript equal to 0 would
cause problems with the part of the formula which reads Un–1, because making n = 0 would
give us U–1 which, by convention, we cannot have (there is no such thing as a (–1)th term).
Here is another example where we define the relation slightly differently.
Example 5.2b
Write down the next four terms of the sequence generated by the recurrence relation
Un+1 = 3U n – 5 where U0 = 2. Each term is ‘3 times the previous term, then subtract 5’.
Solution:
U0
V
=2
V
U1 = 3 × U0 – 5 = 3 × 2 – 5 = 1
V
U2 = 3 × U1 – 5 = 3 × 1 – 5 = –2
V
U3 = 3 × U2 – 5 = 3 × (–2) – 5 = –11
U4 = 3 × U3 – 5 = 3 × (–11) – 5 = –38
You can do the calculations either mentally or using a calculator.
OUTCOME 3: NUMERACY/HIGHER
115
RECURRENCE RELATIONS
If you had made U1 = 2, the calculations would have been exactly the same, but the final
answer of –38 would have been called U5 and not U4.
Example 5.2c
Calculate the next four terms of this sequence: Un+1 = 0.3U n + 5 where U1 = 20.
Solution:
U n+1 = 0.3Un + 5
= 20
V
U1
V
U 2 = 0.3 × U1 + 5 = 0.3 × 20 + 5 = 11
V
U 3 = 0.3 × U2 + 5 = 0.3 × 11 + 5 = 8.3
V
U 4 = 0.3 × U3 + 5 = 0.3 × 8.3 + 5 = 7.49
U 5 = 0.3 × U 4 + 5 = 0.3 × 7.49 + 5 = 7.247
If you have a modern calculator with an ANS key, the calculations can be simplified
dramatically as follows:
Type in the first term, i.e. 20, then press the = (or EXE) key.
Next, type the whole formula, but press ANS instead of U, thus: 0.3 × ANS + 5 = and the
calculator will reward you with 11.
Now the good bit: JUST KEEP PRESSING THE = KEY. The calculator sequence loops
round, putting each successive answer back into the formula and generating the entire
sequence of numbers. You don’t have to worry about rounding off or anything.
116
OUTCOME 3: NUMERACY/HIGHER
V
V
By this line the third decimal place
looks like it has settled on 3...
V
By this line the first decimal
place has settled on 1.
20
11
8.3
7.49
7.247
7.1741
7.15223
7.145669
7.1437007
7.14311021
7.142933063
7.142879919
V
Try it. You should end up with this sequence of numbers:
By this line the second decimal
place has settled on 4.
...but in actual fact the
third place is 2.
RECURRENCE RELATIONS
The sequence is heading for a number called a limit. Such a sequence is called convergent.
If I keep pressing the = key my calculator display eventually stops changing at
7.142857143, but a calculator with a longer display will just keep on going.
Theoretically you cannot get an ‘exact’ limit. The limit is really the number which the
sequence would reach if you kept pressing that key forever, which is obviously impossible.
But we don’t split hairs about that so we simply say that the limit is 5, or 12.4, or whatever
it is.
In the last example we can round off the limit, and say it is 7.14 to two decimal places, or
7.1429 to four decimal places, whatever is required (or sensible).
? 5.2
1.
Calculate the first four terms of each sequence.
(a)
(b)
(c)
(d)
2.
Un = 2U n–1 + 3
Un = 0.5Un–1
Un+1 = 3U n
Un+1 = ¼U n – 2
where
where
where
where
U1 = 4
U1 = 100
U1 = 1
U0 = 60
Sometimes a recurrence relation depends on two (or more) previous terms, in which
case we are given two numbers with which to start. The famous Fibonacci Sequence
is one of these.
Find the first eight terms of this sequence: Un+2 = Un+1 + Un where U 1 = 1and U2 = 1.
(This means ‘add the previous two terms to get the next one’, so U3 = U2 + U1, and
then U 4 = U3 + U2 and so on.)
3.
Use the ANS key as shown above to determine quickly what the following sequences
do.
If they converge, i.e. tend to a limit, say what this limit is (round off to three decimal
places if you have to).
A sequence which does not converge is said to diverge or be divergent. When you
try to find the limit you get numbers which go off the scale or which wobble about
and can’t decide what they want to do. Divergent sequences do not have a limit.
(a)
(c)
(e)
Un = 2Un–1; U 1 = 1
Un+1 = 1.5Un + 2; U0 = 1
Un+1 = 0.75U n – 4; U0 = 6
(b)
(d)
(f)
Un+1 = 0.8U n; U 1 = 10
Un = 2Un–1 – 4; U1 = 3
Un+1 = –0.5Un + 3; U 1 = 8
OUTCOME 3: NUMERACY/HIGHER
117
RECURRENCE RELATIONS
I hope you weren’t put off by my apparently almost random use of U0 and U 1 as a starting off
number. I’ve done so deliberately to get you used to the idea that, in practice, it doesn’t
really matter which you use, so long as you can’t get a negative subscript.
So when do you know if there is a limit or not? You might have got an idea from the
previous exercise. We’ll come back to this later.
5.3 Growth and Decay Factors
Suppose you want to increase or decrease a quantity by a percentage. At school, you may
well have done it as in the next example.
Example 5.3a
Jimmy used to get £4.50 a week pocket money. After his birthday he was promised a 10%
rise, if he could work the new amount out correctly. What should his new weekly
allowance be?
Solution:
Jimmy’s increase is 10% of £4.50 which is £0.45.
New allowance
= Old amount + increase
= £4.50 + £0.45
= £4.95
Another way of calculating this is as follows.
If you think of his old amount (£4.50) as 100%, an increase of 10% will now give Jimmy
110%.
110
= 1.10.
100
Multiplying the old amount £4.50 by 1.10 automatically gives you £4.95, the new, increased
quantity.
Now, any percentage means ‘over 100’, so 110% =
We call the number 1.10 a growth factor.
An obvious disadvantage is that if you specifically want the £0.45 increase for some reason,
you have to subtract £4.50 from £4.95, but usually it’s the end product you want, not what
went on in between.
In the same way, if you want to increase a quantity by, say, 4%, you would multiply by 1.04;
to increase a quantity by 8.5%, multiply by 1.085 and so on.
118
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
? 5.3A
Complete the blank spaces in this table:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
To increase by
12%
72%
5%
1%
7.5%
96.5%
2%
0.5%
0.1%
100%
Multiply by
Example 5.3b
When Jimmy’s dad bought a new car it cost him £12,500 but, when he tried to sell it, he
found it had lost 35% of its value. What was it now worth?
Solution:
The depreciation is 35% of £12,500 which comes to £4,375.
The new value is therefore £12,500 – £4,375 = £8,125.
To do this another way, if we call the original £12,500 100%, and we have lost 35%, then
we have 65% of the original value left.
65% of £12,500 is, would you believe, £8,125. Again, we have homed in on the final value
without calculating the intervening loss.
Now 65% as a decimal is 0.65 and we call this the decay factor. Multiplying the original
value £12,500 by 0.65 automatically gives you £8,125.
Similarly, if you lose 20% of the value, you are left with 80%, giving you a decay factor of
0.80 and if you lose 5% you are left with 95%, a decay factor of 0.95.
OUTCOME 3: NUMERACY/HIGHER
119
RECURRENCE RELATIONS
? 5.3B
Fill in the blanks in this table to find the relevant decay factors:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
To decrease by
12%
72%
5%
1%
7.5%
96.5%
2%
0.5%
0.1%
100%
Multiply by
We will now use growth and decay factors in our study of recurrence relations.
5.4 Recurrence Relations of the form Un+1 = aUn
Suppose you invested £500 in an account which paid 8% per annum (each year, shortened
to p.a.), the interest being credited to the account at the end of each year.
The amount is being compounded annually at a rate of 8%.
What we have here is, in fact, compound interest.
Since the £500 is increasing by 8%, this means that we can apply a growth factor of 1.08 to
the £500.
At the end of year 1 the account is worth £500 × 1.08 = £540.
At the end of year 2 it is no longer £500 which is being increased, but £540.
The account is now worth £540 × 1.08 = £583.20.
At the end of year 3 we have £583.20 × 1.08 = £629.856 and so on.
Each year we multiply what we had the year before by 1.08.
A perfect recurrence relation!
120
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
U0 = 500
U1 = 1.08U0 = 1.08 × 500 = 540
U2 = 1.08U 1 = 1.08 × 540 = 583.20
U3 = 1.08U 2 = 1.08 × 583.20 = 629.856
U0 is what we start off with.
U 1 is the first calculated figure. It is the
amount at the end of year 1.
It is very important to be very clear about
the exact meaning or status of U1.
and so on.
The recurrence relation itself is U n+1 = 1.08Un with U 0 = 500.
If you want to use the ANS key on your calculator follow the instructions
in the box but make sure you keep count of how many times you pressed
equals. The calculator will keep umpteen decimal places for you but you
will need to round off to the nearest penny.
500 =
then
1.08 × ANS
then
=
=
=
So if the question is ‘How much is in the account after the 7th year?’ you press the key 7
times and round off the display to £856.91. Try it and see.
If you want to know how many years it will take for the investment to reach, say £1,000,
you have to keep track of how many times you press the = key.
The 8th press gives £925.465 . . .
The 9th press gives £999.50 . . . almost there!
The 10th press takes you beyond the £1,000 to £1,079.46 so technically it takes 10 years
to get beyond the £1,000 mark.
Example 5.4
A piece of machinery is bought new for £350,000 and depreciates at 15% each year.
(a)
(b)
How much is it worth after four years?
It is to be scrapped when it reaches 25% of its original value. If it was bought at the
start of 1996, in what year will it be scrapped?
Solution:
The machinery loses 15% a year, so the decay factor is 100% – 15% = 85% or 0.85.
Each year it is worth 0.85 times what it was worth the year before.
So the recurrence relation is
Un+1 = 0.85Un where U0 = 350,000 and U1 represents the value at the end of year 1,
which in this case is the end of 1996.
OUTCOME 3: NUMERACY/HIGHER
121
RECURRENCE RELATIONS
V V
V
1996
1997
1998
1999
2000
2001
2002
2003
2004
You’ll notice that I’ve ignored the pennies
here for the purposes of clarity. The
calculator keeps them all, though, so the
answer should be offered complete with
pennies, though I can’t see you being marked
wrong for rounding to the nearest £ since we
are dealing with thousands of them.
(a) ‘after four years’ means ‘end of
1999’, so the answer is £182,702.19.
V
£350,000
£297,500
£252,875
£214,943
£182,702
£155,296
£132,002
£112,201
£95,371
£81,065
V
Following the usual key sequence you will get the following results – they are typed out
here, you only need to keep track of where you are. Amounts refer to ‘end of year’.
(b) 25% of £350,000 is £87,500, and the
calculator sequence of numbers dips below
this here. So the machine would be scrapped
towards the end of 2004.
? 5.4
In each question you must write down the recurrence relation and clearly define what U 0
and U1 are, as well as answering the specific question(s) asked.
1.
£1,000 is invested in an account which has interest compounded at the end of each
year at a rate of 7%.
(a)
(b)
2.
An antique painting is bought at the beginning of July for £1.2 million. Each month it
appreciates (i.e. increases) in value by 5%.
(a)
(b)
3.
How much is it worth at the end of December?
In what month does its value pass £1.75 million?
A microscope slide has 35,000 microbes on it (give or take a few) at 6 a.m. Every
hour, the number increases by 12%. Assuming there’s room for them all,
(a)
(b)
(c)
122
How much is in the account at the end of the 6th year?
How many years pass before the account has doubled in value?
how many microbes are there by 9 a.m.
how long does it take for the number to double
when will the number increase to 100,000?
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
4.
A toxic element is accidentally introduced into a water supply and the concentration
is 1.4 parts per million (ppm). Measures taken to reduce this concentration are
successful in making it drop by 3% per hour. The safe level of this toxic element is
0.25 ppm.
(a)
(b)
How many hours does it take to reduce the concentration to 1 ppm?
How long does it take to make the supply safe?
5.
A car radiator has a leak through which 6% of the remaining water escapes every
day. The radiator starts full with a capacity of 5 litres of water in it, and the engine
will seize up when this drops to half. How many days does the driver have in which
to notice the fault before the engine seizes up?
6.
A pet shop sells hamsters, of which it initially has 8, which breed at a rate that
increases the number of hamsters in the shop by 15% per week. Normally,
schoolchildren buy the hamsters and thus manage to keep the numbers in the shop
under control. Unfortunately, the seven-week school holidays have just started, the
weather is super, children aren’t interested in the hamsters, and so the hamsters just
keep on breeding. The shop has enough space for 25 hamsters. Will it run out of
space by the time the holidays end?
5.5 An Alternative Formula
If you look back at the example of compound interest on page 120 you might realise that
there is available to us a slightly different method of solution.
U0 = 500
U1 = 1.08 × U 0
U2 = 1.08 × U1 but U1 = 1.08 × U0 so U2 = 1.08 × 1.08 × U0
In other words we can write U2 = 1.082U 0
and similarly
U3 = 1.083U 0
and finally the formula
Un = 1.08nU 0
So after seven years the amount is 1.087 × 500 = £856.91 as before.
OUTCOME 3: NUMERACY/HIGHER
123
RECURRENCE RELATIONS
Annual percentage rate (APR)
Suppose you borrow (or invest; it works the same way) a sum of money (call it U0) and the
monthly interest rate is 2%.
After 1 month the value is 1.02 × U0 which is the same as 1.02 1 × U0
After 2 months the value is 1.02 2 × U0
After 3 months the value is 1.02 3 × U0
.
.
.
After 12 months the value is 1.02 12 × U0
The value of 1.02 12 × U0 is 1.26824..., which I will round off to 1.269, which is the same as
126.9
or 126.9%.
100
100% represents the original amount of money, whatever it was, so the 26.9% represents
the increase over a period of one year.
This 26.9% is the annual percentage rate (APR).
The general procedure is as follows.
If the monthly rate is r%, we express it as a decimal, then calculate (1+r)12 and then
subtract 1. What’s left is the decimal form of the APR.
? 5.5
Find the APR if the monthly interest rate is
1.
2.
3.
4.
124
1%
1.5%
1.75%
2.38%
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
5.6 Recurrence Relations of the Form Un+1 = aUn + b
If you look back to Exercise 5.2 you will see that some of the sequences there were
convergent, i.e. they tended to a limit; others were divergent, and did not have a limit.
In this section we will look first at when a recurrence relation is convergent, then we’ll
discuss some real-life problems and see what relevance this limit has.
Activity 5.6a
1.
Remind yourself of the special key sequence which generates the list of terms.
2.
Check that this list of sequences is convergent, and find the limit of each.
(a)
(b)
(c)
(d)
3.
Check that this list of sequences is divergent.
(a)
(b)
(c)
(d)
4.
Un+1 = 0.7Un – 3; U 0 = 6
Un+1 = 0.3Un + 5; U0 = 4
Un+1 = –0.25Un +12; U0 = 10
Un+1 = –0.8Un – 1; U0 = –4
Un+1 = 1.5Un + 5; U0 = 4
Un+1 = 4Un – 3; U0 = 6
Un+1 = –2Un +12; U0 = 10
Un+1 = –2.8Un – 1; U0 = –4
See if you can come up with a conjecture (a theory, an intelligent guess, call it what you
will) about what feature of the sequences above makes those for question 2 convergent
and those for question 3 divergent.
The numerical answers to questions 1 and 2 are on page 136.
The answer to question 4 is below, but try to avoid looking at it without a bit of thought on
the subject first.
The answer is that a relation of the form Un+1 = aUn + b converges to a limit if the number
a, i.e. the number multiplying Un in the formula, is between –1 and 1.
A quick illustration:
Un+1 = 0.75Un + 3; U0 = 5
OUTCOME 3: NUMERACY/HIGHER
125
RECURRENCE RELATIONS
The sequence is
V
Difference between these two numbers is 1.3125
V
V
Difference between these two numbers is 1.75
Difference between these two numbers is 0.9843
V VV
5
6.75
8.0625
9.0468 ...
9.7851 ...
10.3388 ...
10.7541 ...
The difference between
each pair of sequence
terms is getting smaller,
showing that the sequence
is convergent.
(20 = presses later)
11.9960 ...
11.9970 ...
11.9977 ...
(12 more = presses later)
11.9999296
As you can see, the limit appears to be 12, but this sequence is taking its time getting there.
This is because the 0.75 is quite big, i.e. relatively close to 1.
If you repeat the whole process but replace the 0.75 with 0.25, the limit will change to 4,
but you can see this within 6 presses of the = key. Convergence is much faster.
Activity 5.6b
Now let’s investigate how the limit, when there is one, is affected by the initial value U 0.
Take the relation Un+1 = 0.4U n + 5 and find its limit with these different starting values
1.
2.
3.
4.
U0 = 10
U0 = 100
U0 = 2
U0 = –50
The answer is on the next page, but don’t look too soon.
126
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
The answer which I hope you got was that the limit is the same (8.3333333) no matter
what the starting value is. The limit of the relation Un+1 = aUn + b depends on the values of
a and b, but is independent of the value of U 0. You can start the sequence anywhere you like,
the limit will (eventually) be the same.
Needless to say, there is a formula for this. If you don’t understand the maths of it, don’t
worry, because you can simply apply the formula.
5.7 Formula for the Limit
Once we start getting close to the limit, successive terms get closer and closer to each
other, and the difference between two consecutive terms becomes so small that it gradually
disappears. When we ‘reach the limit’ we could thus say that two consecutive terms are
both the same.
In other words, both Un+1 and Un are equal to each other, and so are both equal to the limit
which we can call L. So, at the limit, the equation Un+1 = aUn + b can be written instead as
L = aL + b. This can be solved for L as follows:
V
V
V
2. Reduce the two terms
involving L to just one
term, by taking out L as a
common factor.
L = aL + b
L – aL = b
L(1 – a ) = b
b
L=
1–a
1. Move both terms involving L
to one side of the equation.
3. Divide both sides by the
bracket (1–a) to get L on
its own.
The last line is an easy one to remember.
Example 5.7
Find the limit of the sequence generated by U n+1 = 0.36Un + 72.
b
1–a
72
=
1 – 0.36
72
=
0.64
= 112.5
L=
It is most important to establish
this fact.
V
Solution:
A limit L exists because 0.36 lies between –1 and 1.
So the value of this limit is
Just because a formula exists does
not mean to say that it makes
sense.
If the 0.36 was actually, say, 1.36
the formula would give the answer
–200 which is nonsense in the
context of the sequence.
You can always check this out with your calculator, using any number you like as a starter.
OUTCOME 3: NUMERACY/HIGHER
127
RECURRENCE RELATIONS
? 5.7
Practise using the formula for the limit on the following sequences to find the limit where it
exists. Where there is no limit, say so.
1.
2.
3.
4.
5.
6.
7.
8.
Un+1 = 0.5Un + 6
Un+1 = 0.25Un +4
Un+1 = 1.6Un +7
Un+1 = 0.92Un – 8
Un+1 = 3Un + 6
Un+1 = –0.75Un +3
Un+1 = 0.43Un +2.5
Un+1 = –0.2Un – 10
5.8 Problems involving Recurrence Relations
And finally, we look at the sorts of questions which will turn up in an assessment.
Example 5.8a
Each evening, the school caretaker and his team manage to pick up 60% of the litter in the
playground, but the next day the little recalcitrants deposit another 3 tonnes of litter. On
Monday after school there are 10 tonnes of litter in the playground.
(a)
How much litter is there on
(i)
Wednesday afternoon immediately after school
(ii)
Friday afternoon immediately after school?
(b)
How much litter is likely to be in the playground long term (before and after cleanup)?
Solution:
The decay factor is 0.40, because if 60% of the litter is picked up there is still 40% left.
If we start with U0 = 10 tonnes, then on the Tuesday immediately after school there is
0.40U0 + 3, i.e. the 40% that’s left plus the new deposit of 3 tonnes.
We can thus set up the recurrence relation: Un+1 = 0.40U0 + 3 with U 0 = 10.
It is important to remember that each successive value of U is the amount of litter after the
pupils have dropped their day’s offering but before the caretaker cleans up.
128
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
Values on the calculator display are thus
Mon
Tue
Wed
Thu
Fri
10
7
5.8 = answer (a,i)
5.32
5.128 = answer (a,ii)
The difference between each
successive pair of numbers is
decreasing, thus suggesting
convergence.
The sequence is clearly converging, and this is shown by the fact that the decay factor 0.40
is between –1 and 1.
To get the answer to part (b), as we approach the limit L we get
L = 0.4L + 3
L – 0.40L = 3
0.60L = 3
3
0.60
=5
L=
The following graph should show you what is
happening:
Start with 10 tonnes
60% = 6 tonnes is picked up, leaving 4 tonnes
Day
3 tonnes is added during the day, making 7 tonnes
60% = 4.2 tonnes is picked up, leaving 2.8 tonnes
3 tonnes is added during the day, making 5.8 tonnes
60% = 3.48 tonnes is picked up, leaving 2.32 tonnes
and so on.
0
1
2
3
4
5
6
7
8
9
10
Amount before
clean-up
10
7
5.8
5.32
5.128
5.0512
5.02048
5.008192
5.0032768
5.0013107
Amount after
clean-up
4
2.8
2.32
2.128
2.0512
2.02048
2.008192
2.0032768
2.0013107
2.0005243
5.0005243
2.0002097
10
The top line is the limit
of the recurrence
relation, showing how
much there is after a
deposit but before the
clean-up. As you see, it
levels off at 5 tonnes.
9
No. of tonnes
8
7
6
5
4
3
2
1
0
1
3
5
7
9
11
Day
13
15
17
19
21
The bottom line shows
the amount after the
clean-up but before the
next deposit. It levels off
at 2 tonnes.
OUTCOME 3: NUMERACY/HIGHER
129
RECURRENCE RELATIONS
What actually happens is that the decrease of 60% is larger, at the start, than the increase
of 3 tonnes, but as the level of litter gets smaller, so does the amount of decrease (because it
is a fixed 60% of an ever smaller number), until it is eventually exactly balanced by the 3
tonne increase.
The next worked example will show how a sequence increases towards a limit.
Example 5.8b
A harbour has 4 metres of silt in it. A dredger is employed to dredge out a channel, and
succeeds in removing 25% of the silt each day. However, heavy rains bring down a further
1.5 metres each day. For ships to use the harbour there must be no more than 5 metres of
silt in it. Will the harbour be usable?
Solution:
25% of the silt is removed so 75% is still there. The decay factor is therefore 0.75, which
is just 75% expressed as a decimal.
Now we set up the recurrence relation:
U0 = 4
U 1 = 0.75U 0 + 1.5
where U 1 is thus the level of silt after the first day’s dredging and also after the first addition
of silt by heavy rain.
The recurrence relation is thus Un+1 = 0.75Un + 1.5 and we know a limit will exist because
0.75 is between –1 and 1.
L = 0.75L + 1.5
L – 0.75L = 1.5
0.25L = 1.5
1.5
0.25
=6
L=
Now this 6 metre limit is after the addition of 1½ metres of more silt, so the depth will vary
between 4.5 and 6 metres. In other words, there will be times when there are less than 5
metres of silt in the harbour, but also times when there are more.
So the harbour will not be safe for use.
130
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
6.5
2. Level before dredging is
above the 5m line. This
is the 6m limit of the
recurrence relation.
6
Depth of silt
5.5
5
4.5
1. Level after dredging is
below the 5m line, which
looks safe.
4
3.5
3
2.5
2
1
4
7
10 13 16 19 22 25 28 31 34 37 40
Day
Sometimes we have to form one recurrence relation within another.
Example 5.8c
A patient is being given a new experimental drug. He is initially given a 50mg dose which is
to be repeated every 4 hours. The body flushes out 10% of the drug every hour.
To be effective, there must be at least 100mg of the drug continually in the body, but it is
considered unsafe to give more than 50mg at a time, hence the problem.
How many doses are required before there are 100mg of the drug continually in the body?
Solution:
The first relation: 10% lost per hour, so decay factor 0.90
V
See page 123 ‘An
Alternative Formula’ if
unsure about this step.
V
V
U0 = 50
U1 = 0.90 × U 0
and so on
U4 = (0.90)4 × U0
= (0.90)4 × 50
= 32.805
Initial dose.
Amount remaining
after 1 hour.
Amount remaining
after 4 hours.
32.805
× 100 = 65.61% left from
50
each 50mg dose. In other words, the decay factor for each 4-hourly dose is 0.6561.
So there are 32.805mg left out of 50, which represents
So now we look at the recurrence relation which affects the patient
every 4 hours:
U0 = 50
U1 = 0.6561U0 + 50
NB The decay
factor shows how
much is left, not how
much has gone.
where U1 is the amount in the body after the second dose. (This is a situation where it
might have been marginally easier to call the first dose U1 instead of U 0.)
OUTCOME 3: NUMERACY/HIGHER
131
RECURRENCE RELATIONS
U n+1 = 0.6561U0 + 50
is the recurrence relation, and a quick check will show that this has a limit of 145.39mg.
On the face of it, this is above the required 100mg level so looks OK, but remember that it
is the level after the addition of 50mg.
So the amount in the body will vary between 95.39mg and 145.39mg, i.e. there are times when
there is less than the required 100mg in the body, so the drug will not be wholly effective.
Had the required effective level been, say, 90mg, then we would have been OK. And so
would the patient.
I hope this example demonstrates that you mustn’t just look at the value of the limit. You
must also look at the value immediately before, in other words at the whole range of values
which the limit actually represents.
? 5.8
1.
A pond has an absolutely massive supply of frog spawn. Every night another 400
tadpoles are born, and every day predators eat 70% of the tadpoles in the pond at
that time. If we start off with 500 tadpoles on the Monday morning, how many are
there on Tuesday morning? On Tuesday night? On Wednesday morning? What is the
long-term position? (In other words, is there a limit, and if there is, what is it?)
2.
A large herd of deer is causing a nuisance on a Highland estate. Every spring
approximately 150 new calves are born, so the owner allows a cull of 20% of the
deer population each autumn. In the summer of 1995 there were 1,000 deer on the
estate. How many will there be by the summer of 1996? Of 1997? Of 1998? Will
the number of deer level off?
3.
A fish farm stocks its ponds with 5,000 new fish. Every year the fish population
increases by approximately 15%, and the owner allows 1,000 fish to be caught. If
you assume the increase comes before the fish are taken out, calculate the first few
terms of the sequence and decide if the sequence is convergent or divergent. Is the
owner allowing too many or too few fish to be caught?
4.
A forestry company has a forest with an area of approximately 750 hectares, of
which only 500 hectares is actually planted. During the summer, approximately
18% of the planted area is cut, and another 75 hectares is planted out in the autumn.
Does the company have spare capacity for the foreseeable future, or will it have to
buy more land to keep up this rate of activity?
5.
A city manager finds that every day the population dumps 15 tonnes of litter in the
city centre, but squads of cleaners, who operate during the night, only manage to lift
132
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
60% of it. On the evening of 17 May there are 25 tonnes of litter in the city centre.
How many tonnes are there
(a)
on the evening of 20 May
(b)
on the evening of 22 May?
What’s going on?
6.
In the same city centre as in question 5, there is a new manager. She pays her squads
more and they lift 80% of the litter. How long does it take to get the amount of litter
permanently below 18 tonnes, starting at 25 tonnes again?
7.
The same situation as in question 5, but two years and a massive public education
campaign later. The public now drop only 10 tonnes of litter per day, and squads
pick up 95% of the total at night. How much litter is on the streets every morning?
8.
A captured spy is given 150 units of a truth drug every hour to make him talk. Every
hour the body gets rid of 30% of the drug. The drug is only effective if there are at
least 300 units continuously in the body. If the spy is given the first dose at 8 a.m., at
what time can questioning begin?
9.
Another spy is given a different drug. This drug is given every 6 hours in 100mg
doses. The body gets rid of 5% of the drug every hour.
(a)
(b)
10.
How many doses are needed to keep the level continuously above 250mg?
In the long term, what will be the range of values between which the amount
of drug in the body will lie, and how will this affect the spy if a dose of over
360mg is fatal?
In order to safeguard depleted fishing stocks, the appropriate ministry orders a cull
of seals, of which there are 200,000. Each spring this number increases by 15%.
(a)
(b)
How many seals must be culled each summer to keep the number of seals at
its present level?
In fact, 40,000 seals are killed each summer. How many culls must be made
to bring the number of seals down to 100,000?
11.
A car engine holds 5 litres of oil, but a leak is causing it to lose 22% of its oil each
month. At the end of each month the owner adds 600ml of oil. The engine will seize
up if the oil level drops at any time to less than half the capacity. Is the owner's
replacement policy a safe one?
12.
It is winter, and a plant is 80cm tall. Each spring it is pruned by 20%. It has been
genetically modified to grow exactly 35 cm each year. The target height is a
maximum of 1.8m. Will this pruning policy achieve this?
OUTCOME 3: NUMERACY/HIGHER
133
RECURRENCE RELATIONS
Answers
?5.1: Answers
1.
(a)
U1 = 2 × 1 + 7 = 9
U2 = 2 × 2 + 7 = 11
U3 = 2 × 3 + 7 = 13
(b)
U1 = 12 + 1 = 2
U2 = 22 + 1 = 5
U3 = 32 + 1 = 10
2.
(a)
(b)
(c)
(d)
(e)
U4 = 5 × 42 = 80
U5 = 52 + 3 × 5 – 2 = 38
U8 = (–1)8 × 3 × 8 = 24
U8 = (–1)9 × (2 × 8 + 1) = –1 × 17 = –17
U7 = (7 – 3)2 = 16
3.
(a)
(b)
2, 4, 8, 16
2, 4, 8, 14
(c)
U 1 = 5 – 21 = 3
U 2 = 5 – 22 = 1
U 3 = 5 – 23 = –3
First three terms are the same, the fourth (and subsequent) terms are different. You
frequently need more than just three terms to know what’s what.
?5.2: Answers
1.
(a)
(b)
(c)
(d)
2.
1, 1, 2, 3, 5, 8, 13, 21, 33, 54
3.
(a)
(b)
(c)
(d)
(e)
(f)
134
4, 11, 25, 53, 109
100, 50, 25, 12.5, 6.25
1, 3, 9, 27, 81
60, 13, 1.25, –1.6875, –2.421875
1, 2, 4, 8, 16 divergent
10, 8, 6.4, 5.12, 4.096 convergent, in fact heading for 0
1, 3.5, 7.25, 12.875 divergent
3, 2, 0, –4, –12 divergent
6, 0.5, –3.625, –6.71875 convergent, limit is –16
8, –1, 3.5, 1.25, 2.375, 1.8125 convergent, limit 2. This last one
wobbles about above and below the limit before settling. We say a sequence
like this oscillates towards the limit.
OUTCOME 3: NUMERACY/HIGHER
RECURRENCE RELATIONS
?5.3A: Answers
1.
4.
7.
10.
1.12
1.01
1.02
2
2.
5.
8.
1.72
1.075
1.005
3.
6.
9.
1.05
1.965
1.001
Watch that last one. Increasing something by 100% means you are doubling it.
?5.3B: Answers
1.
4.
7.
10.
0.88
2.
0.28
3.
0.95
0.99
5.
0.925
6.
0.035
0.98
8.
0.995
9.
0.999
0 (if you decrease something by 100% you wipe it out!)
?5.4: Answers
1.
U0 = £1,000
U1 = amount at end of year 1 = 1.07 × U0
Un+1 = 1.07 Un
(a)
(b)
2.
U0 = £1.2m = beginning of July
U1 = amount at end of July = 1.05 × U0
Un+1 = 1.05 Un
(a)
(b)
3.
£1.5315 ...million = £1,531,538
Just passes £1.75m at end of March
U0 = 35,000 = amount at 6 a.m.
U1 = amount at 7 a.m. = 1.12 × U0
Un+1 = 1.12 Un
(a)
(b)
(c)
4.
£1,500.73
11 years (£2,104.85)
49,172
12 noon – 69,083; 1p.m. – 77,373 i.e. just over 6 hours
Somewhere between 3 p.m. and 4 p.m.
Un+1 = 0.97 U n ; U0 = 1.4 ppm; U 1 = amount after 1 hour
(a)
12 hours
OUTCOME 3: NUMERACY/HIGHER
135
RECURRENCE RELATIONS
(b)
57 hours to get just below 0.25 ppm
(The nearer the decay factor is to 1, the longer it takes.)
5.
Un+1 = 0.94 U n; U0 = 5 litres; U1 = amount after 1 day
11 days to notice problem, 12th day level dips to 2.37 which is below 2.5
6.
Un+1 = 1.15 Un; U0 = 8 hamsters; U1 = amount after 1 week
At the end of the 7th week of holiday there are 21.78 hamsters, so assuming the
children start buying as soon as they’re back at school there’s room for them.
?5.5: Answers
1.
2.
3.
4.
(1.01)12 = 1.1268 ....; APR = 12.68%
(1.015)12 = 1.1956 ...; APR = 19.56%
(1.0175)12 = 1.2314 ....; APR = 23.14%
(1.0238)12 = 1.3261 ....; APR = 32.61%
Activity 5.6a
1.
Type the first term then press the = or EXE key.
Type in the equation, but replace U n with the ANS key, e.g. instead of typing
0.54 × U n + 3 you would type 0.54 × ANS + 3.
Then press = as often as you need, either keeping count of the number of presses or
writing the answers down.
2.
(a)
(b)
(c)
Limit –10 (exact)
Limit 7.1429 (to 4 d. p. – this is optional, could be to 3 or 2 if you want)
Limit 0.96 (exact)
(d)
Limit –0.5556 to 4 d.p. (in fact, it is the exact fraction
?5.7: Answers
1.
12
2.
5 13
3.
4.
5.
6.
7.
No limit because 1.6 is outside the –1 to +1 range
–100
No limit
1.7143
4.3860
8.
–8 13
136
OUTCOME 3: NUMERACY/HIGHER
5
9
).
RECURRENCE RELATIONS
?5.8: Answers
1.
Monday morning 500
Monday night 150
Tuesday morning 550
Tuesday night 165
Wednesday morning 565
Un+1 = 0.30U n + 400 is convergent because 0.30 is between –1 and 1. Limit 571
tadpoles
2.
1995 – 1,000; 1996 – 950; 1997 – 910; 1998 – 878
Un+1 = 0.80Un + 150 and the limit exists because 0.80 is between –1 and +1.
Limit = 750
3.
5000; 4750; 4462; 4131 (omitting the decimal parts)
The differences between terms are increasing so the sequence is divergent, and this is
confirmed by the relation Un+1 = 1.15Un – 1,000. Eventually the sequence gets to 0
or beyond; too many fish are being taken out.
4.
Un+1 = 0.82Un + 75 with U0 = 500, though, as we know, the limit is independent of
the first term. The limit is 416.67 which means there is approximately 83 hectares
spare capacity.
5.
Un+1 = 0.40Un + 15, U0 = 25
(a)
25 tonnes
(b)
25 tonnes
Exactly the same amount is being picked up as is being dropped.
6.
Un+1 = 0.20Un + 15, U0 = 25
Limit exists; it is 18.75 so the level settles down eventually to between 3.75 (each
morning) and 18.75 (each evening). It never completely falls below 18 tonnes.
7.
Un+1 = 0.05Un + 10 and the limit is 10.53, i.e. the level varies between 0.53 and
10.53.
8.
Un+1 = 0.70Un + 150; U0 is 1st dose at 8 a.m., U 1 is amount in body after 2nd dose,
9 a.m.
9 a.m.
10 a.m.
11 a.m.
12 noon
1 p.m.
2 p.m.
255
328
379
415.965
441.175
458.175
The 1p.m. dose is the
first one which doesn’t
drop to below 300.
OUTCOME 3: NUMERACY/HIGHER
137
RECURRENCE RELATIONS
9.
Hourly sequence is Un+1 = (0.95)6 × Un which means that the initial 100 mg drops to
73.51 mg, a decay factor of 0.7351.
The 6-hourly sequence is therefore Un+1 = 0.7351U n + 100
The 9th dose takes you to 353.84 so it must have been 253.84 immediately
beforehand.
The limit is 377.5 so the dose varies between 277.5 and 377.5 and, since 360 mg is
fatal, the spy is in serious trouble.
10.
(a)
(b)
11.
Un+1 = 0.78U n + 600
The limit is 2,727.27 but this is after 600 ml has been added, so the range of values is
2,127 up to 2,727. So it drops below 2,500 and the policy is unsafe.
12.
Un+1 = 0.80Un + 35 for which the limit is 175 cm, but this is after the 35 cm growth,
so the policy will keep the plant below 180 cm.
138
30,000
Un+1 = 1.15Un – 40,0000; U0 = 200,000
7 culls take the number down to 89,332.
OUTCOME 3: NUMERACY/HIGHER
TUTOR ASSIGNMENTS
Mean and Standard Deviation
T1
The table shows how many minutes, to the nearest
tenth of a minute, it took for a task to be completed
by a number of clerical staff.
Calculate the mean and standard deviation of the
data.
Time (min)
10.0-10.4
10.5-10.9
11.0-11.4
11.5-11.9
12.0-12.4
12.5-12.9
13.0-13.4
TOTAL
No. of staff
5
12
23
10
5
2
1
58
OUTCOME 3: NUMERACY/HIGHER
139
TUTOR ASSIGNMENTS
Median and Quartiles
T2
Calculate the median, the other two quartiles, and the semi-interquartile range, for the
following:
1.
31 22 45 24 30 33 40 19 27 22 34 26 24
2.
The table shows how many faulty light bulbs there are in boxes of six.
No. faulty
No. of boxes
3.
0
47
2
20
3
8
4
5
5
3
6
1
This table shows the value of sales made by 80 sales persons working for a company
during a period of time:
Value of sales (£000)
No. of salespersons
140
1
53
25-30
1
OUTCOME 3: NUMERACY/HIGHER
30-35
14
35-40
23
40-45
21
45-50
15
50-55
6
TUTOR ASSIGNMENTS
Transposition of Formulae
T3
Transpose these formulae:
ab
3c
1.
W=
2.
P = ab –
3.
A=
to
(a)
a
(b)
c
c
to
(a)
a
(b)
c
5 – pq
q
to
(a)
p
(b)
q
4.
R = 3π(t – 2m) to
(a)
t
(b)
m
5.
Z =4
(a)
x
(b)
y
1
2
x –y
y
to
OUTCOME 3: NUMERACY/HIGHER
141
TUTOR ASSIGNMENTS
Correlation and Regression
T4
The table shows ten mythical local authorities with the population of each and the amount
spent on a certain aspect of health care.
Authority
A
B
C
D
E
F
G
H
I
J
Population
145,000
236,000
95,000
142,000
178,000
263,000
74,000
137,000
89,000
312,000
Amount spent
(£m)
7.2
9.5
3.6
4.2
9.3
10.1
3.9
5.6
3.3
11.4
(a)
Draw the scatter diagram on squared paper.
(b)
Make up a table which will allow you to estimate the amount spent for a given
population (so get your x and y the right way round).
(c)
Calculate the equation of the line of regression and also the value of the correlation
coefficient.
(d)
Draw the regression line on your scatter diagram.
(e)
Use the line of regression to make a point estimate about the amount which ‘should’
be spent by local authorities with the following populations:
(i)
(ii)
(f)
142
153,000
406,000
Make comments about the reliability of each of these estimates, making sure you use
one of your previous calculations to back up your argument.
OUTCOME 3: NUMERACY/HIGHER
TUTOR ASSIGNMENTS
Recurrence Relations
T5
1.
2.
3.
A painting appreciates in value by 15% each year. It is initially worth £1 million.
(a)
Write down a formula which will allow you to home in on its value after any
number of years. Then find its value after (i) 3 years (ii) 5 years.
(b)
Write down a recurrence relation which ties the value each year to the value
the previous year. How many years is it before the painting trebles in value?
At the beginning of November a person borrows £4,000 and agrees to pay back
£500 at the beginning of each month after this (i.e. first repayment at the beginning of
December). However, during every month the amount owed increases by 2% of the
value at the start of the month.
(a)
Writing down the amount owing every month as you go, calculate during
which month the loan will be eventually paid off. In other words, number
crunch your way through the problem.
(b)
Write down the problem in the form of a recurrence relation which allows
you the use of the ANS key.
There are an estimated 20 million brownling fish in the sea. During the breeding
season there is an estimated 8% increase in the number of fish. After this comes the
fishing season and fishermen are allowed to catch 1,750,000 fish.
(a)
(b)
(c)
4.
Write down the number of fish in the sea each year in the form of a recurrence
relation.
How many fish will be in the sea after four years?
Does the relation have a limit? If so, find it and say what it means; if not, why
not? And what are the implications for the fish?
In November a turkey farmer has 6,000 turkeys. Each following Christmas he sells
off 20% of them and buys in 2,000 chicks which he then brings on for the next
season.
(a)
(b)
Write down a recurrence relation for the number of turkeys.
Find the farmer’s long-term position if he continues this policy.
OUTCOME 3: NUMERACY/HIGHER
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