# 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