Formula for Mean with Examples

Here you will learn what is the formula for mean of grouped and ungrouped data and how to find mean with examples.

Let’s begin –

The formula for mean median and mode for grouped and ungrouped frequency distribution is given below.

Formula for Mean : 

(i) For ungrouped distribution : If $x_1$, $x_2$, …… $x_n$ are n values of variate $x_i$ then their mean $\bar{x}$ is defined as

$\bar{x}$ = $x_1 + x_2, …… + x_n \over n$ = ${\sum_{i=1}^{n}x_i}\over n$

$\implies$ $\sum x_i$ = n$\bar{x}$

Example : Neeta and her four friends secured 65, 78, 82, 94 and 71 marks in a test of mathematics. Find the average (arithmetic mean) of their marks.

Solution : Arithmetic mean or average =  $65 + 78 + 82 + 94 + 71\over 5$ = $390\over 5$ = 78

Hence, arithmetic mean = 78

(ii) For ungrouped and grouped frequency distribution :

(a) By Direct Method :

If $x_1$, $x_2$, …… $x_n$ are values of variate with corresponding frequencies $f_1$, $f_2$, …… $f_n$, theb their A.M. is given by

$\bar{x}$ = $f_1x_1 + f_2x_2 + …… + f_nx_n \over f_1 + f_2 + …… + f_n$ = ${\sum_{i=1}^{n}f_ix_i}\over N$, where N = ${\sum_{i=1}^{n}f_i}$

Example : Find the mean of the following freq. dist.

$x_i$	5	8	11	14	17
$f_i$	4	5	6	10	20

Solution : Here N = $\sum f_i$ = 4 + 5 + 6 + 10 + 20 = 45

$\sum f_ix_i$ = 606

$\therefore$ $\bar{x}$ = $\sum f_ix_i\over N$ = $606\over 45$ = 13.47

(b) By short method or assumed mean method :

If the value of $x_i$ are large, then calculation of A.M. by using mean formula is quite tedious and time consuming. In such case we take deviation of variate from an arbitrary point a.

Let      $d_i$ = $x_i$ – a

$\therefore$   $\bar{x}$ = a + $\sum f_id_i\over N$, where a is assumed mean

Example : Find the A.M. of the following freq. dist.

Class Interval	0-50	50-100	100-150	150-200	200-250	250-300
$f_i$	17	35	43	40	21	24

Solution : Let assumed mean a = 175

Class Interval	mid value $(x_i)$	$d_i$ = $x_i – 175$	frequency $f_i$	$f_id_i$
0-50	25	-150	17	-2550
50-100	75	-100	35	-3500
100-150	125	-50	43	-2150
150-200	175	0	40	0
200-250	225	50	21	1050
250-300	275	100	24	2400
			$\sum f_i$ = 180	$\sum f_id_i$ = -4750

Now, a = 175 and N = $\sum f_i$ = 180

$\therefore$   $\bar{x}$ = a + ($\sum f_id_i\over N$) = 175 + $(-4750)\over 180$ = 175 – 26.39 = 148.61

(c) By step deviation method :

Sometime during the application of short method (given above) of finding the A.M. If each deviation $d_i$ are divisible by a common number h(let)

Let   $u_i$ = $d_i\over h$ = $x_i – a\over h$,          where a is assumed mean.

$\therefore$  $\bar{x}$ = a + ($\sum f_iu_i\over N$)h

Example : Find the A.M. of the following freq. dist.

$x_i$	5	15	25	35	45	55
$f_i$	12	18	27	20	17	6

Solution : Let assumed mean a = 35, h = 10

here N = $\sum f_i$ = 100, $u_i$ = $x_i – 35\over 10$

$\sum f_iu_i$ = (12$\times$-3) + (18$\times$-2) + (27$\times$-1) + (20$\times$0) + (17$\times$1) + (6$\times$2)
= -70

$\therefore$   $\bar{x}$ = a + ($\sum f_iu_i\over N$)h = 35 + $(-70)\over 100$$\times$10 = 28