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Measures of Average and of Dispersion

CIMA Free Mock Exam

An average tells us the typical value of a set of data; a measure of dispersion tells us how spread out the data are around that average.

This chapter shows how data are set out in a frequency distribution – ungrouped (discrete) or grouped (continuous) – and presented in a bar chart, pie chart, histogram or ogive. It then calculates the three averages (arithmetic mean, median and mode) and the four measures of dispersion (range, variance, standard deviation and coefficient of variation) for both ungrouped and grouped data.

1 Introduction

It is often of interest to know the average of a set of data. For example we may have asked a sample of people what their wages are, and want to know what the average wage is. In this chapter we will look at different ways we can calculate an average. Additionally, even if we have calculated the average wage, it might be of interest to know whether all of the sample had a wage close to the average or whether some earned a lot more and some a lot less than the average. This is known as the dispersion and we will look at different ways of measuring this.

2 Frequency distributions

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A frequency distribution is a table showing the number of observations of each variable. They may be discrete variables which can only consist of certain values, or continuous variables where we group the variables.

The syllabus calls these ungrouped data (each possible value listed separately, as with discrete variables) and grouped data (values collected into classes, as with continuous variables). Grouping loses some detail, but it makes a long list of different values manageable.

Discrete variables (ungrouped data)

A company has recorded the number of complaints received per week over the last year, and has produced the following table:

Number of complaints

Frequency (weeks)

0

12

1

16

2

20

3

4

Total

52

The number of complaints can only be 0, 1, 2 or 3, so every possible value is listed.

Continuous variables (grouped data)

A company has recorded the total amount paid to employees each week over the last year, and has produced the following table:

Total paid ($)

Frequency (weeks)

0 – under 500

1

500 – under 1,000

4

1,000 – under 1,500

8

1,500 – under 2,000

19

2,000 – under 2,500

14

2,500 – under 3,000

6

Total

52

The total paid could be any amount between $0 and $3,000, so listing every value would be pointless. Instead the amounts are grouped into classes of $500.

3 Ways of presenting data

In many cases, management do not need to see the actual numbers (and indeed the actual numbers may confuse them). Often a chart or graph can present the information more clearly.

You need to be aware of four charts:

Chart

How it is drawn

Bar chart

One bar for each value or class; the length of the bar shows the frequency. The bars are separate.

Pie chart

A circle cut into slices, one for each value or class. Each slice is the same proportion of the circle as the class is of the total: angle = frequency ÷ total frequency × 360°.

Histogram

Like a bar chart, but for grouped data: the classes run along the horizontal axis and the bars touch. It is the AREA of each bar (width × height) that represents the frequency – so if one class is twice as wide as the others, its bar is drawn at half the height its frequency would suggest.

Ogive

A graph of cumulative frequency (the running total of the frequencies). Each cumulative frequency is plotted against the upper end of its class, and the points are joined with straight lines.

A company has recorded the total amount paid to employees each week over the last year, and has produced the following table:

Total paid ($)

Frequency (weeks)

0 – under 500

1

500 – under 1,000

4

1,000 – under 1,500

8

1,500 – under 2,000

19

2,000 – under 2,500

14

2,500 – under 3,000

6

Total

52

Present the above table in the form of

  1. a bar chart

  2. a pie chart

  3. a histogram

  4. an ogive

Show answerHide answer

(a) Bar chart – one bar per class, its length showing the number of weeks.

Weekly pay – bar chart (frequency in weeks)05101520$0–500$500–1,000$1,000–1,500$1,500–2,000$2,000–2,500$2,500–3,000

(b) Pie chart – each class as a share of the 52 weeks:

Total paid ($)

Frequency

% of total

Angle

0 – under 500

1

1.9%

6.9°

500 – under 1,000

4

7.7%

27.7°

1,000 – under 1,500

8

15.4%

55.4°

1,500 – under 2,000

19

36.5%

131.5°

2,000 – under 2,500

14

26.9%

96.9°

2,500 – under 3,000

6

11.5%

41.5°

Total

52

100%

360°

For example, $2,000 – under $2,500: 14 ÷ 52 × 360° = 96.9°. (The rounded figures do not add exactly to 100% and 360°.)

Weekly pay – pie chart (share of 52 weeks)7.7%15.4%36.5%26.9%11.5%$0–500$500–1,000$1,000–1,500$1,500–2,000$2,000–2,500$2,500–3,000

(c) Histogram – the classes are all $500 wide, so the height of each bar is simply its frequency, and the bars touch.

Weekly pay ($) – histogram (frequency in weeks)05101520050010001500200025003000

(d) Ogive – first find the cumulative frequencies, then plot each against the upper end of its class:

Total paid ($)

Frequency

Cumulative frequency

under 500

1

1

under 1,000

4

5

under 1,500

8

13

under 2,000

19

32

under 2,500

14

46

under 3,000

6

52

Weekly pay ($) – ogive (cumulative frequency in weeks)0204060500100015002000250030000

4 Measures of average

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You need to be aware of the following different measures of determining the average of a set of observations:

Arithmetic mean

This is calculated by adding up all of the observations and dividing by the number of observations

With a frequency distribution, multiply each value (x) by its frequency (f), add up the results and divide by the total frequency: mean = Σfx ÷ Σf, where Σ means ‘the total of’.

Median

This is the centrally occurring observation when all of the observations are arranged in order of magnitude

With n observations, the median is the value of the (n + 1) ÷ 2th observation – there are as many observations below it as above it. Use the cumulative frequencies to find which value it is. If n is even the position ends in a half (for example the 7½th): take the value halfway between the two observations either side.

Mode

This is the most frequently occurring observation

There can be more than one mode if two values share the highest frequency.

The mean is what most people mean by ‘the average’, but it can be distorted by a few extreme values. If 99 employees earn about $100 a week and one earns $10,000, the mean is far above what nearly everyone is paid; the median (the middle earner) or the mode (the most common wage) is more meaningful. That is why the median is often used for wages.

A company has recorded the number of complaints received per week over the past thirteen weeks, and has produced the following table:

Number of complaints

Frequency (weeks)

0

1

1

6

2

4

3

2

Total

13

Calculate:

  1. the arithmetic mean

  2. the median

  3. the mode

Show answerHide answer

Number of complaints (x)

Frequency (f)

fx

Cumulative frequency

0

1

0

1

1

6

6

7

2

4

8

11

3

2

6

13

Total

13

20

(a) Arithmetic mean = Σfx ÷ Σf = 20 ÷ 13 = 1.54 complaints a week

(b) Median = value of the (13 + 1) ÷ 2 = 7th observation. The cumulative frequencies show that observations 2 to 7 are all 1, so the median = 1 complaint

(c) Mode = most frequently occurring observation (frequency 6) = 1 complaint

Grouped data

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With grouped data we do not know the exact value of each observation, only its class. So:

  • Mean – each class is represented by its mid-point (for $500 – under $1,000, use $750). This assumes the observations are spread evenly through the class, so the mean is an approximation. Then use Σfx ÷ Σf as before.

  • Median – with grouped data we do not know the individual values, so the median is the value at the halfway point of the total frequency, n ÷ 2 (the (n + 1) ÷ 2 rule is for individual observations, as in Example 2). The cumulative frequencies show which class that point falls in (the median class). For a single figure, assume the observations are spread evenly through that class: median = bottom of the median class + (n ÷ 2 − cumulative frequency below the class) ÷ frequency of the median class × class width. An ogive drawn accurately and read at a cumulative frequency of n ÷ 2 gives the same figure.

  • Mode – for grouped data you will be asked only for the modal class: the class with the highest frequency.

A company has recorded the total amount paid to employees each week over the last year, and has produced the following table:

Total paid ($)

Frequency (weeks)

0 – under 500

1

500 – under 1,000

4

1,000 – under 1,500

8

1,500 – under 2,000

19

2,000 – under 2,500

14

2,500 – under 3,000

6

Total

52

Calculate:

  1. the arithmetic mean

  2. the median

  3. the mode

Show answerHide answer

Total paid ($)

Mid-point (x)

Frequency (f)

fx

Cumulative frequency

0 – under 500

250

1

250

1

500 – under 1,000

750

4

3,000

5

1,000 – under 1,500

1,250

8

10,000

13

1,500 – under 2,000

1,750

19

33,250

32

2,000 – under 2,500

2,250

14

31,500

46

2,500 – under 3,000

2,750

6

16,500

52

Total

52

94,500

(a) Arithmetic mean = 94,500 ÷ 52 = $1,817 (to the nearest $)

(b) Median: with grouped data, find the value at the halfway point of the 52 observations, 52 ÷ 2 = 26. 13 observations are below $1,500 and 32 below $2,000, so the median class is $1,500 to $2,000. To put a figure on it, assume the 19 observations in that class are spread evenly: the halfway point is 26 − 13 = 13 observations into the class, so median = $1,500 + (13 ÷ 19 × $500) = $1,500 + $342 = $1,842 (to the nearest $)

(c) Modal class (highest frequency, 19) = $1,500 to $2,000

In the lecture that works Example 3, the tutor finds the median at the (52 + 1) ÷ 2 = 26.5th observation, gets $1,855 and says an accurate ogive gives the same. For grouped data use the halfway point, 52 ÷ 2 = 26, as in Answer 3: the median is $1,842, and that is also the figure an accurately drawn ogive gives at a cumulative frequency of 26. The (n + 1) ÷ 2 rule is for individual observations, as in Example 2.

5 Measures of dispersion

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Dispersion is looking at the spread of the observations.

You need to be aware of the following measures of dispersion:

Range

This is simply the difference between the highest and the lowest of the observations

For grouped data, take the top of the highest class less the bottom of the lowest class. The range is easy to find but is distorted by a single extreme value: if 49 weeks have 0 to 3 complaints and one week has 50, the range is 50, which gives a false picture of the spread.

Variance

Here we measure the differences between the observations and the arithmetic mean, square the differences, and then take the average of these squared differences.

Variance = Σf(x − mean)² ÷ Σf. The differences are squared because otherwise the positive and negative differences would cancel out and their average would always be zero. For grouped data, x is the mid-point of each class.

Standard deviation

This is the square root of the variance

Squaring changes the units (‘squared complaints’ or ‘squared dollars’); taking the square root brings the measure back to the units of the data. The standard deviation is the measure of spread you will use most – it is central to the normal distribution in the next chapter.

Coefficient of variation

The standard deviation divided by the arithmetic mean

On its own, a standard deviation of 0.84 complaints or $580 does not tell us whether the spread is large or small – that depends on the size of the data. Dividing by the mean gives a relative measure, so the spread of data sets with very different means can be compared.

For the information in Example 2, calculate:

  1. the range

  2. the variance

  3. the standard deviation

  4. the coefficient of variation

Show answerHide answer

From Example 2, the mean = 1.54.

x

f

fx

x − mean

(x − mean)²

f(x − mean)²

0

1

0

−1.54

2.37

2.37

1

6

6

−0.54

0.29

1.74

2

4

8

+0.46

0.21

0.84

3

2

6

+1.46

2.13

4.26

Total

13

20

9.21

(a) Range = 3 − 0 = 3 complaints

(b) Variance = 9.21 ÷ 13 = 0.71

(c) Standard deviation = √0.71 = 0.84 complaints

(d) Coefficient of variation = 0.84 ÷ 1.54 = 0.55

In the lecture that works Example 4 (earlier in this section), 4 × 0.21 is worked as 0.17 instead of 0.84, so the lecture’s variance (0.66), standard deviation (0.81) and coefficient of variation (0.53) are slightly too low. The correct figures are those in Answer 4 above.

YouTube video

For the information in Example 3, calculate:

  1. the range

  2. the variance

  3. the standard deviation

  4. the coefficient of variation

Show answerHide answer

From Example 3, the mean = $1,817; x is the mid-point of each class.

Total paid ($)

x

f

fx

x − mean

(x − mean)²

f(x − mean)²

0 – under 500

250

1

250

−1,567

2,455,489

2,455,489

500 – under 1,000

750

4

3,000

−1,067

1,138,489

4,553,956

1,000 – under 1,500

1,250

8

10,000

−567

321,489

2,571,912

1,500 – under 2,000

1,750

19

33,250

−67

4,489

85,291

2,000 – under 2,500

2,250

14

31,500

+433

187,489

2,624,846

2,500 – under 3,000

2,750

6

16,500

+933

870,489

5,222,934

Total

52

94,500

17,514,428

(a) Range = $3,000 − $0 = $3,000

(b) Variance = 17,514,428 ÷ 52 = 336,816 (in squared dollars)

(c) Standard deviation = √336,816 = $580

(d) Coefficient of variation = 580 ÷ 1,817 = 0.32

The weekly pay is much more spread out in dollars than the complaints are in numbers, but relative to its mean the spread is smaller: 0.32 against 0.55.

6 Test your knowledge

Two quick checks before you move on: work through the flashcards to fix this chapter’s terms and definitions, then sit the objective questions for exam-style practice. Both mark themselves and explain the answers as you go.

Practice questions

Measures of Average and of Dispersion

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