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.