There is more than one type of average. Often when someone is working out an averagethey are actually calculating the MEAN.

e.g. 5,8,3,2,8,7,6,1,9,8

Mean = 5 + 8 + 3 + 2 + 8 + 7 + 6 + 1 + 9 + 8 = 59 = 5.9
10 10

(The line means divide by ie 59 divided by 10)

Median

This is the middle number(s) in a list of numbers, once they have been put in ascending or descending order.

1,2,3,5, 6, 7,8,8,8,9

In this case there are two middle numbers 6 and 7. The median is taken to be the MEAN of the two numbers i.e. 6.5.

M ode

The most frequent number in the list. In the above example the mode would be 8 because 8 appears mor times than any other number. In the case of two or more numbers appearing in a list an equal number of times the set of data is said to have NO UNIQUE MODE. It is often easier to spot the mode once the numbers have been put in order.

FINDING THE AVERAGE OF A FREQUENCY DISTRIBUTION

Example

Table to show the number of peas in 20 pea pods

No. of peas in pod
frequency
0
2
1
3
2
8
3
5
4
1
5
1

The mean is calculated by multiplying the number of peas in the pod by the frequency and then dividing by the total frequency.

( 0 x 2) + ( 1 x 3) + ( 2 x 8 ) + ( 3 x 5 ) + ( 4 x 1 ) + ( 5 x 1) = 43 = 2.15
2 + 3 + 8 + 5 + 1 + 1 20

The more usual way of setting out the work is to create an extra column headed

‘f (frequency) x X (number of peas in a pod)’

No. of peas in pod ( x )
frequency
f x X
0
2
0
1
3
3
2
8
16
3
5
15
4
1
4
5
1
5
Total
20
43

 

mean = Sigma f X = 43 = 2.15
Sigma f 20

Sigma means ‘the sum of’ or another way of saying Sigma f would be to say ADD all the frequencies together.

To calculate the MEDIAN you would have to imagine all the pods layed out in order. First would be the 2 pods which contain no peas. Then there would be 3 pods with 1 pea, etc. As there are 20 pods the middle 2 pods would be th tenth and eleveth pods which each contain 2 peas. So the MEDIAN would be 2.

 
0 0 1 1 1 2 2 2 2 2 2 2 2 2 3 3 3 3 4 4

(Number of peas in each pod)

               
                 
       
These are the 10 th and 11 th pods which each contain 2 peas.
         

MODE would be 2 as most pods had two peas in them.

FINDING THE AVERAGE OF A GROUPED FREQUENCY DISTRIBUTION

Mass (Kg) Frequency
40- 3
50- 10
60- 6
70-80 12

This type of table is usually used when you want to draw a cumulative frequency graph ultimately

 

Mass (Kg) Frequency
40 ≤ x < 50 3
50 ≤ x < 60 10
60 ≤ x < 70 6
70 ≤ x < 80 12

This means that the values can equal the lower amount but are below the upper amount. e.g. 40 ≤ x < 50 means that the value of the mass could equal 40 or be any value up too but not including 50.

  

See grouped frequency tables for more information

 

Mass (Kg) Frequency
40-50 3
50-60 10
60-70 6
70-80 12

This could be a poor way of setting out the information unless it was made clear which group the values of 50, 60, 70 would be in.

M ean

To calculate an estimate of the mean it is necessary to find the midpoint of each group or class and assume that all the items within the group have that mass.

e.g.

mass (kg)
midpointX
frequencyf
frequency x midpoint f x X
40 - 50
45
3
135
50 - 60
55
10
550
60 - 70
65
6
390
70 - 80
75
12
900
totals
Sigma f =31
Sigma fx =1975

 

Estimate of mean = S fx = 1975 = 63.709........

 

S f 31

=63.7 kg to 3 significant figures ( See rounding section for more information on significant figures).

Mode

The modal group is 60 - 70

Median

To find the class which contains the median:

Find the total number of values, n, in this case 31

Find the middle value given by n + 1 i n this case 31 + 1 = 16

  2 2

Find which class contains this value. In this case the 16th value is in the class 60 -70 so the median lies within the class 60 - 70.