Question:medium

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 pocket11 - 1313 - 1515 - 1717 - 1919 - 2121 - 2323 - 25
Number of workers76913f54

Updated On: Jan 13, 2026
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Solution and Explanation

The class mark (\(x_i\)) for each interval is calculated using the formula:

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

The mean pocket allowance is given.

With an assumed mean of 18 (a), \(d_i\) and \(f_id_i\) are computed 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

12

-138

-966

13 - 15

6

14

-136

-816

15 - 17

9

16

-134

-1206

17 - 19

13

18

-132

-1716

19 - 21

\(f\)

20

-130

-130\(f\)

21 - 23

5

22

-128

-640

23 - 25

4

24

-126

-504

Total

\(\sum f_i\) = 44 + \(f\)

 

 

-130\(f\) - 5846

From the table, we have:

\(\sum f_i = 44 +f\)
\(\sum f_id_i =  -130f - 5846\)

The mean is calculated as:

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

         18 = 18 + \((\frac{-130f - 5846}{44 + f})\)

          0 = \(\frac{-130f - 5846}{44 + f}\)

 -130\(f\) - 5846 = 0
         -130\(f\) = 5846
          \(f\) = -45

Therefore, the missing frequency f is -45.

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