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# 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 Allowence(in Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-35 Number of children 7 6 9 13 f 5 4

Given, $$\overline{x} \ = \ 18$$

 Class interval Number of children (frequency ($$f_i$$) ) Mid-point $$(x_i)$$ $$f_i u_i$$ 11-13 7 12 84 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 f 20 20f 21-23 5 22 110 23-25 4 24 96 Total $$\sum f_i \ = \ 44 \ + \ f$$ $$\sum f_i x_i \ = \ 752 \ + \ 20f$$

Thus, $$\ \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i}$$

$$\Rightarrow \ 18 \ = \ \frac{752 \ + \ 20f}{44 \ + \ f}$$

$$\Rightarrow \ 18(44 \ + \ f) \ = \ (752 \ + \ 20f)$$

$$\Rightarrow \ 792 \ + \ 18f \ = \ 752 \ + \ 20f$$

$$\Rightarrow \ 792 \ - \ 752 \ = \ 20f \ - \ 18f \$$
$$\Rightarrow \ 40 \ = \ 2f$$

$$\Rightarrow \ f \ = \ 20$$

So, the missing frequency, f = 20.