Question

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f.

Daily pocket allowance (in Rs	11 - 13	13 - 15	15 - 17	17 - 19	19 - 21	21 - 23	23 - 25
Number of workers	7	6	9	13	f	5	4

Answer 1

To find the class mark (x_i) for each interval, the following relation is used.

Given that, mean pocket allowance, `barx `= Rs 18

Taking 18 as assured mean (a), d_i and f_id_i are calculated as follows.

Daily pocket allowance (in Rs)	Number of children fi	Class mark xi	di = xi − 18	fidi
11­ −13	7	12	-6	-42
13 − 15	6	14	-4	-24
15 − 17	9	16	-2	-8
17 −19	13	18	0	0
19 − 21	f	20	2	2 f
21 − 23	5	22	4	20
23 − 25	4	24	6	24
Total	`sumf_i=44+f `			2f − 40

From the table, we obtain

`sumf_i = 44+f`

`sumf_iu_i = 2f - 40`

`barx = a+(sumf_id_i)/(sumf_i)`

`= 18 = 18 + ((2f - 40)/(44+f))`

`0 = ((2f - 40)/(44+f))`

2f - 40 = 0

2f = 40

f = 20

Hence, the missing frequency, f, is 20.