Definition:
A value which is used in this way to
represent the distribution is called an ‘average’. Since the average lies in the centre of a distribution, they
are called ‘measures of central tendency’.
They are also known as ‘measures of location’.
Types of Averages:
(a) Arithmetic Mean,
(b) Geometric Mean,
(c) Harmonic Mean,
(d) Median, and
(e) Mode.
(a)
Arithmetic Mean (AM):
1. AM is defined as the value obtained by dividing the sum of the values by their number.
2.
It is expressed as follows for sample data:
3.
For population data:
4.
The above formulae are for ungrouped data and they cannot be applied to
grouped data. For grouped data, the formula for AM is as follows:
or
Where x (all
the values) falling in a class are assumed to be equal to the class mark or mid
point of that class. So on the sum
of the values in k^{th} class would be f_{k}x_{k},
and the sum of values in all the k classes would be equal to
.
The total
number of values is the sum of class frequencies, i.e.,
.
Example:
Class Boundaries 
Frequency 
9.519.5 
5 
19.529.5 
8 
29.539.5 
13 
39.549.5 
19 
49.559.5 
23 
59.569.5 
15 
69.579.5 
7 
79.589.5 
5 
89.599.5 
3 
99.5109.5 
2 
Total 
100 
Calculate Mean.
Solution:
Class Boundaries 
f 
x
(MidPoint) 
fx 
9.519.5 
5 
14.5 
72.5 
19.529.5 
8 
24.5 
196 
29.539.5 
13 
34.5 
448.5 
39.549.5 
19 
44.5 
845.5 
49.559.5 
23 
54.5 
1253.5 
59.569.5 
15 
64.5 
967.5 
69.579.5 
7 
74.5 
521.5 
79.589.5 
5 
84.5 
422.5 
89.599.5 
3 
94.5 
283.5 
99.5109.5 
2 
104.5 
209 
Total 
100 

5220 
Alternate Formulae for Computing
Mean:
1. The computation of AM using the grouped data formula is easily provided that the values x and f are not large.
2. If the values are large, considerable time can be saved by taking deviations from an assumed or guessed mean.
3. If A is an assumed or guessed mean and D denotes the deviations of x from A (i.e., D = x – A) then x = A + D.
4.
The AM can be expressed as follows:
 for ungrouped data
 for grouped data
Example:
Class Boundaries 
Frequency 
9.519.5 
5 
19.529.5 
8 
29.539.5 
13 
39.549.5 
19 
49.559.5 
23 
59.569.5 
15 
69.579.5 
7 
79.589.5 
5 
89.599.5 
3 
99.5109.5 
2 
Total 
100 
Solution:
Class
Boundaries 
f 
X 
D
= x – A 
fD 
9.519.5 
5 
14.5 
40 
200 
19.529.5 
8 
24.5 
30 
240 
29.539.5 
13 
34.5 
20 
260 
39.549.5 
19 
44.5 
10 
190 
49.559.5 
23 
54.5 
0 
0 
59.569.5 
15 
64.5 
10 
150 
69.579.5 
7 
74.5 
20 
140 
79.589.5 
5 
84.5 
30 
150 
89.599.5 
3 
94.5 
40 
120 
99.5109.5 
2 
104.5 
50 
100 
Total 
100 


230 
Although any class mark can be taken as
assumed mean, we take the class mark 54.5 as A, because it corresponds to the
largest frequency. See the above
table. We have A = 54.5, n = 100 and
= 230.
Second Alternative Method for
Computing Mean:
,
 for ungrouped
data
 for grouped data
Example:
Class
Boundaries 
Frequency 
9.519.5 
5 
19.529.5 
8 
29.539.5 
13 
39.549.5 
19 
49.559.5 
23 
59.569.5 
15 
69.579.5 
7 
79.589.5 
5 
89.599.5 
3 
99.5109.5 
2 
Total 
100 
Solution:
Class
Boundaries 
f 
X 
D
= x – A 

fu 
9.519.5 
5 
14.5 
40 
4 
20 
19.529.5 
8 
24.5 
30 
3 
24 
29.539.5 
13 
34.5 
20 
2 
26 
39.549.5 
19 
44.5 
10 
1 
19 
49.559.5 
23 
54.5 
0 
0 
0 
59.569.5 
15 
64.5 
10 
1 
15 
69.579.5 
7 
74.5 
20 
2 
14 
79.589.5 
5 
84.5 
30 
3 
15 
89.599.5 
3 
94.5 
40 
4 
12 
99.5109.5 
2 
104.5 
50 
5 
10 
Total 
100 



23 
A = 54.5; n = 100; h = 10 and
= 23
Weighted Arithmetic Mean (WAM):
WAM is used to find average of certain values which are not of equal importance.
The numerical
values are called ‘weights’, and denoted as w_{1}, w_{2}, ….. w_{k}.
WAM is expressed as follows:
Example:
Sectors 
Expenditure (All
figures in Rs. Billion) 
Weight 
General Public Service 
503 
46 
Development 
272 
25 
Defence 
223 
20 
Public Order Safety 
19 
3 
Education 
17 
3 
Health 
4 
2 
Housing 
1 
1 
Environment Protection 
1 
1 
Calculate Weighted Average Mean.
Solution:
Sectors 
X 
W 
WX 
General Public Service 
503 
46 
23138 
Development 
272 
24 
6528 
Defence 
223 
20 
4460 
Public Order Safety 
19 
3 
57 
Education 
17 
3 
51 
Health 
4 
2 
8 
Housing 
1 
1 
1 
Environment Protection 
1 
1 
1 
Total 

100 
34244 
Properties of Arithmetic Mean:
or
is a minimum, if a =
.
If x = a (a constant), then = a
If y = x ± a, then = ± a
If y = bx, then = b
If y =
,
then
=
1. GM is defined only for nonzero positive values. It is the nth root of the product of n values in the data.
2.
It can be expressed as follows:
Where
= x_{1} × x_{2} ×
x_{3} × ………. × x_{n}.
Alternate Method for Computing
Geometric Mean:
Example:
Find the GM of the following data:
5,6,7,8,2,3,1,10,13,11
Solution:
Where
= x_{1} × x_{2} ×
x_{3} × ………. × x_{n}.
Geometric Mean for Grouped Data:
For grouped data, the GM is computed as
below:
Taking logarithm of both sides:
Weighted Geometric Mean:
Example:
Class Boundaries 
f 
9.519.5 
5 
19.529.5 
8 
29.539.5 
13 
39.549.5 
19 
49.559.5 
23 
59.569.5 
15 
69.579.5 
7 
79.589.5 
5 
89.599.5 
3 
99.5109.5 
2 
Total 
100 
Solution:
Class
Boundaries 
f 
x 
log x 
f
log x 
9.519.5 
5 
14.5 
1.1614 
8.07 
19.529.5 
8 
24.5 
1.3892 
11.1136 
29.539.5 
13 
34.5 
1.5378 
19.9914 
39.549.5 
19 
44.5 
1.6484 
31.3196 
49.559.5 
23 
54.5 
1.7364 
39.9372 
59.569.5 
15 
64.5 
1.8096 
27.144 
69.579.5 
7 
74.5 
1.8722 
13.1054 
79.589.5 
5 
84.5 
1.9269 
9.6345 
89.599.5 
3 
94.5 
1.9754 
5.9262 
99.5109.5 
2 
104.5 
2.0191 
4.0382 
Total 
100 


170.2801 
HM is defined only for nonzero positive
values. It is the reciprocal of
mean of reciprocals of values. More
briefly, HM, of a set of n values x_{1}, x_{2}, ….. , x_{n},
is the reciprocal of the AM of the reciprocals of the values.
Thus:
 for ungrouped data
Harmonic Mean for Grouped Data:
The reciprocal of the class marks (in
case of grouped data) will be
,
,…….,
. Since the reciprocals occur with
frequencies f_{1}, f_{2}, ….. , f_{k},
the total value of the reciprocals in the first class is
, in second class
, ….. , and in the k^{th} class is
. The sum of reciprocals in all the
k classes would be:
Weighted Harmonic Mean:
Example:
Find HM of the values 1, 2 and 3.
Solution:
Example:
Class Boundaries 
Frequency 
9.519.5 
5 
19.529.5 
8 
29.539.5 
13 
39.549.5 
19 
49.559.5 
23 
59.569.5 
15 
69.579.5 
7 
79.589.5 
5 
89.599.5 
3 
99.5109.5 
2 
Total 
100 
Solution:
Class
Boundaries 
f 
x 


9.519.5 
5 
14.5 
0.06897 
0.34485 
19.529.5 
8 
24.5 
0.04082 
0.32656 
29.539.5 
13 
34.5 
0.02899 
0.37687 
39.549.5 
19 
44.5 
0.02247 
0.42693 
49.559.5 
23 
54.5 
0.01835 
0.42205 
59.569.5 
15 
64.5 
0.01550 
0.2325 
69.579.5 
7 
74.5 
0.01342 
0.09394 
79.589.5 
5 
84.5 
0.01183 
0.05915 
89.599.5 
3 
94.5 
0.01058 
0.03174 
99.5109.5 
2 
104.5 
0.00957 
0.01914 
Total 
100 


2.33373 
Relation between AM, GM and HM: