CIMA - C3 Fundamentals Of Business Mathematics

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120 4a: Averages ⏐ Part B Summarising and analysing data

Daily demand

Mid point

Frequency

> 0

≤ 5

> 5

≤ 10

>10

≤ 15

>15

≤ 20

∑f =

∑fx =

190

Arithmetic mean

x =

∑

190

= 9.5 units

Because the assumption that frequencies occur evenly within each class interval is not quite correct in this

example, our approximate mean of 9.5 is not exactly correct, and is in error by 0.25 (9.5 – 9.25).

As the

frequencies become larger, the size of this approximating error should become smaller.

1.5 Example: The arithmetic mean of combined data

Suppose that the mean age of a group of five people is 27 and the mean age of another group of eight people is 32.

How would we find the mean age of the whole group of 13 people?

Arithmetic mean =

itemsof Number

itemsof valuesof Sum

The sum of the ages in the first group is 5

× 27 = 135

The sum of the ages in the second group is 8

× 32 = 256

The sum of all 13 ages is 135 + 256 = 391

The mean age is therefore

391

= 30.08 years.

Question

Mean

The mean weight of 10 units at 5 kgs, 10 units at 7 kgs and 20 units at X kgs is 8 kgs.

The value of X is

Answer

The value of X is

Workings

Mean =

items of Number

items of valuesof Sum

Sum of first 10 units = 5

× 10 = 50 kgs

Sum of second 10 units = 7

× 10 = 70 kgs

Sum of third 20 units = 20

× X = 20X

Part B Summarising and analysing data ⏐ 4a: Averages 121

Sum of all 40 units = 50 + 70 + 20X = 120 + 20X

∴ Arithmetic mean = 8 =

X20120

∴ 8 × 40 = 120 + 20X

320 = 120 + 20X (subtract 120 from both sides)

320 – 120 = 20X

200 = 20X

10 = X (divide both sides by 20)

1.6 The advantages and disadvantages of the arithmetic mean

Advantages of the arithmetic mean

• It is easy to calculate

• It is widely understood

• It is representative of the whole set of data

• It is supported by mathematical theory and is suited to further statistical analysis

Disadvantages of the arithmetic mean

• Its value may not correspond to any actual value. For example, the 'average' family might have 2.3

children, but no family has exactly 2.3 children.

•

An arithmetic mean might be distorted by extremely high or low values. For example, the mean of

3, 4, 4 and 6 is 4.25, but the mean of 3, 4, 4, 6 and 15 is 6.4. The high value, 15, distorts the average

and in some circumstances the mean would be a misleading and inappropriate figure.

Question

Definite variables

For the week ended 15 November, the wages earned by the 69 operators employed in the machine shop of

Mermaid Co were as follows.

Wages

Number of operatives

under $60

$60 and under $70

$70 and under $80

$80 and under $90

$90 and under $100

$100 and under $110

Over $110

Required

Calculate the arithmetic mean wage of the machine operators of Mermaid Co for the week ended 15 November.

122 4a: Averages ⏐ Part B Summarising and analysing data

Answe

The mid point of the range

'under $60' is assumed to be $55 and that of the range over $110 to be $115, since all

other class intervals are $10. This is obviously an approximation which might result in a loss of accuracy, but there

is no better alternative assumption to use. Because wages can vary in steps of 1c, they are virtually a continuous

variable and hence the mid-points of the classes are halfway between their end points.

Mid-point of class

Frequency

165

715

1,200

1,275

950

105

840

115

690

5,835

Arithmetic mean =

∑

835,5

= $84.57

2 The mode

2.1 The modal value

The mode or modal value is an average which means 'the most frequently occurring value'.

2.2 Example: The mode

The daily demand for inventory in a ten day period is as follows.

Demand Number of days

Units

The mode is 7 units, because it is the value which occurs most frequently.

2.3 The mode of a grouped frequency distribution

The mode of a grouped frequency distribution can be calculated from a histogram.

T F

RWAR

T F

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Part B Summarising and analysing data ⏐ 4a: Averages 123

2.4 Example: Finding the mode from a histogram

Consider the following grouped frequency distribution

Class interval Frequency

0 and less than 10 0

10 and less than 20 50

20 and less than 30 150

30 and less than 40 100

(a) The modal class (the one with the highest frequency) is '20 and less than 30'. But how can we find a single

value to represent the mode?

(b) What we need to do is draw a histogram of the frequency distribution.

The modal class is always the class with the tallest bar. This may not be the class with the highest

frequency if the classes do not all have the same width.

Step 1 Join with a straight line the top left hand corner of the bar for the modal class and the top left

hand corner of the next bar to the right.

Step 2 Join with a straight line the top right hand corner of the bar for the modal class and the top

right hand corner of the next bar to the left.

(d) Where these two lines intersect, we find the

estimated modal value. In this example it is approximately

26.7.

(e) We are assuming that the frequencies occur evenly within each class interval but this may not always be

correct. It is unlikely that the 150 values in the modal class occur evenly. Hence

the mode in a grouped

frequency distribution is only an estimate.

124 4a: Averages ⏐ Part B Summarising and analysing data

You cannot draw a histogram in a computer based assessment so questions may ask you to interpret a diagram.

2.5 The advantages and disadvantages of the mode

Advantages of the mode

• It is easy to find

• It is not influenced by a few extreme values

• It can be used for data which are not even numerical (unlike the mean and median)

• It can be the value of an actual item in the distribution

Disadvantages of the mode

• It may be unrepresentative; it takes no account of a high proportion of the data, only representing

the most common value

• It does not take every value into account

• There can be two or more modes within a set of data

• If the modal class is only very slightly bigger than another class, just a few more items in this other

class could mean a substantially different result, suggesting some instability in the measure

3 The median

3.1 The middle item of a distribution

The median of a set of ungrouped data is found by arranging the items in ascending or descending order of value,

and selecting the item in the middle of the range.

A list of items in order of value is called an array.

The median is the value of the middle member of an array. The middle item of an odd number of items is

calculated as the

)1n( +

item.

3.2 Example: The median

(a) The median of the following nine values:

8 6 9 12 15 6 3 20 11

is found by taking the middle item (the fifth one) in the array:

3 6 6 8 9 11 12 15 20

The median is 9.

(b) Consider the following array.

1 2 2 2 3 5 6 7 8 11

The median is 4 because, with an even number of items, we have to take the arithmetic mean of the two

middle ones (in this example, (3 + 5)/2 = 4).

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Assessment

focus point

Part B Summarising and analysing data ⏐ 4a: Averages 125

Question

Median (1)

The following times taken to produce a batch of 100 units of Product X have been noted.

21 mins, 17 mins, 24 mins, 11 mins, 37 mins, 27 mins,

20 mins, 15 mins, 17 mins, 23 mins, 29 mins, 30 mins

24 mins, 18 mins, 17 mins, 21 mins, 24 mins, 20 mins

What is the median time?

Answe

The times can be arranged as follows.

11, 15, 17, 17, 17, 18, 20, 20, 21, 21, 23, 24, 24,

24, 27, 29 30, 37

There are eighteen items which is an even number, therefore the median is the arithmetic mean of the two middle

items (ie ninth and tenth items) = 21 mins.

Question

Median (2)

The following scores are observed for the times taken to complete a task, in minutes.

12, 34, 14, 15, 21, 24, 9, 17, 11, 8

What is the median score?

A 14.00

B 14.10

C 14.50

D 14.60

Answe

The first thing to do is to arrange the scores in order of magnitude.

8, 9, 11, 12, 14, 15, 17, 21, 24, 34

There are ten items, and so median is the arithmetic mean of the 5

and 6

items.

1514 +

= 14.50

The correct answer is therefore C.

You could have eliminated options B and D straight away. Since there are ten items, and they are all whole

numbers, the average of the 5

and 6

items is either going to be a whole number (14.00) or 'something and a half'

(14.50).

126 4a: Averages ⏐ Part B Summarising and analysing data

3.3 Finding the median of an ungrouped frequency distribution

The median of an ungrouped frequency distribution is found in a similar way. Consider the following distribution.

Value

Frequency

Cumulative frequency

The median would be the (35 + 1)/2 = 18

item. The 18

item has a value of 16, as we can see from the cumulative

frequencies in the right hand column of the above table.

3.4 Finding the median of a grouped frequency distribution

The median of a grouped frequency distribution can be established from an ogive.

Finding the median of a grouped frequency distribution from an ogive is best explained by means of an example.

3.5 Example: The median from an ogive

Construct an ogive of the following frequency distribution and hence establish the median.

Class

Frequency

Cumulative frequency

≥ 340, < 370

≥ 370, < 400

≥ 400, < 430

≥ 430, < 460

≥ 460, < 490

Solution

T F

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Part B Summarising and analysing data ⏐ 4a: Averages 127

The median is at the 1/2 × 40 = 20

item. Reading off from the horizontal axis on the ogive, the value of the median

is approximately $380.

Note that,

because we are assuming that the values are spread evenly within each class, the median

calculated is only approximate.

Again, you cannot be asked to draw an ogive in a computer based assessment so you just need to know how to

obtain the median from an ogive.

3.6 The advantages and disadvantages of the median

Advantages of the median

• It is easy to understand

• It is unaffected by extremely high or low values

• It can be the value of an actual item in the distribution

Disadvantages of the median

• It fails to reflect the full range of values

• It is unsuitable for further statistical analysis

• Arranging data into order of size can be tedious

If you are asked to find the median of a set of ungrouped data, remember to arrange the items in order of value

first and then to count the number of items in the array. If you have an even number of items, the median may not

be the value of one of the items in the data set. The median of an even number of items is found by calculating the

arithmetic mean of the two middle items.

• The arithmetic mean, mode and median of a grouped frequency distribution can only be estimated

approximately

• Each type of average has a number of advantages and disadvantages that you need to be aware of

Assessment

focus point

Important!

Assessment

focus point

128 4a: Averages ⏐ Part B Summarising and analysing data

Chapter Roundup

• The arithmetic mean is the best known type of average and is widely understood. It is used for further statistical

analysis.

• The arithmetic mean of ungrouped data = sum of items ÷ number of items.

• The arithmetic mean of grouped data, x =

∑

where n is the number of values recorded, or the

number of items measured.

• The mode or modal value is an average which means 'the most frequently occurring value'.

• The mode of a grouped frequency distribution can be calculated from a histogram.

• The median is the value of the middle member of an array. The middle item of an odd number of items is

calculated as the

)1(n

item.

• The median of a grouped frequency distribution can be established from an ogive.

Quick Quiz

1 Insert the formulae in the box below into the correct position.

(a) The arithmetic mean of ungrouped data =

(b) The arithmetic mean of grouped data = or

•

∑

•

∑

•

∑

2 What is the name given to the average which means 'the most frequently occurring value'?

Arithmetic mean

Median

Mode

3 The mean weight of a group of components has been calculated as 133.5. The individual weights of the

components were 143, 96, x, 153.5, 92.5, y, 47. When y = 4x;

What is the value of x?

Part B Summarising and analysing data ⏐ 4a: Averages 129

4 Calculate the mid-points for both discrete and continuous variables in the table below.

Class interval Mid-point Mid-point

(Discrete data) (Continuous data)

25 < 30

30 < 35

35 < 40

40 < 45

45 < 50

50 < 55

55 < 60

60 < 65

5 (a) The mode of a grouped frequency distribution can be found from a(n) histogram/ogive.

(b) The median of a grouped frequency distribution can be found from a(n) histogram/ogive.

6 A group of children have the following ages in years: 10, 8, 6, 9, 13, 12, 7, 11.

What is the median age?

Answers to Quick Quiz

1 (a)

∑

(b)

∑

f∑

∑

2 Mode

Mean =

Total

So Total = 7 × 133.5

= 934.5

934.5 = 143 + 96 + x + 153.5 + 92.5 + y + 47

934.5 = 532 + x + y

y = 4x

So 934.5 = 532 + x + 4x

5x = 934.5 – 532

5x = 402.5

x = 80.5