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.

$\begin{matrix} Daily pocket allowance (in Rs) & 11 - 13 & 13 - 15 & 15 - 17 & 17 - 19 & 19 - 21 & 21 - 23 & 23 - 25 Number of workers & 7 & 6 & 9 & 13 & f & 5 & 4 \end{matrix}$

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Solution

We may find class mark $x_{i}$ for each interval by using the relation:

$x_{i} = \frac{Upper class limit + lower class limit}{2}$

Given that mean pocket allowance $x_{i}$ = Rs.18

Now taking 18 as assured mean $a$ we may calculate $d_{i}$ and $f_{i} d_{i}$ as following.

$\begin{matrix} Daily pocket allowance (in Rs) & Number of workers f_{i} & Class mark x_{i} & d_{i} = x_{i} - 18 & f_{i} d_{i} 11 - 13 & 7 & 12 & - 6 & - 42 13 - 15 & 6 & 14 & - 4 & - 24 15 - 17 & 9 & 16 & - 2 & - 18 17 - 19 & 13 & 18 & 0 & 0 19 - 21 & f & 20 & 2 & 2 f 21 - 23 & 5 & 22 & 4 & 20 23 - 25 & 4 & 24 & 6 & 24 T o t a l & \sum f_{i} = 44 + f & 2 f - 40 \end{matrix}$

From the table we may obtain:

$\sum f_{i} = 44 + f \sum f_{i} d_{i} = 2 f - 40 ¯ x = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}} 18 = 18 + (\frac{2 f - 40}{44 + f}) 2 f - 40 = 0 2 f = 40 f = 20$

Hence, the missing frequency 'f' is 20.

Suggest Corrections

Class:	11−13	13−15	15−17	17−19	19−21	21−23	23−25
Frequency:	7	6	9	13	−	5	4

Class interval:	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Frequency:	$7$	$6$	$9$	$13$	---	$5$	$4$