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	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.

Class mark $(x_i)$ = $\frac {\text{Upper \,limit + Lower \,limit}}{2}$

Given that, mean pocket allowance,

Taking 18 as assured mean (a), $d_i$, and $f_id_i$ can be calculated as follows.

**Daily pocket allowance (in Rs) **	Number of children ( $f_i$**) **	**Class mark $\bf{x_i}$ **	$\bf{d_i = x_i -150}$	$\bf{f_id_i}$
11 - 13	7	-6	-6	-42
**13 - 15 **	6	14	-4	-24
15 - 17	9	16	-2	-18
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 **	$\sum f_i$ = 44 + $f$			2 $f$** - 40**

From the table, we obtain

$\sum f_i = 44 +f$
$\sum f_id_i = 2f - 40$

Mean, $\overset{-}{x} = a + (\frac{\sum f_id_i}{\sum f_i})$

     18 = $18 + (\frac{2f - 40 }{44 + f})$

      0  =$\frac{2f - 40 }{44 + f}$

2$f$ - 40 = 0
2$f$ = 40
$f$ = 20

Hence, the missing frequency f is 20.