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Question

# In the following frequency distribution, ifthe arithmetic mean is 45.6, find out missing frequency. Wages (₹) 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Number of Workers 5 6 7 X 4 3 9

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Solution

## Wages (X) Mid Value (m) No. of worker (f) fm 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 15 25 35 45 55 65 75 5 6 7 X 4 3 9 75 150 245 45 220 195 675 ∑f = 34 + X ∑fm = 1560 + 45X Substituting the values in the formula for mean. $\overline{)X}=\frac{\Sigma fm}{\Sigma f}\phantom{\rule{0ex}{0ex}}45.6=\frac{1560+45X}{34+X}\phantom{\rule{0ex}{0ex}}\mathrm{or},45.6 \left(34 + X\right)= 1560+45 X\phantom{\rule{0ex}{0ex}}\mathrm{or},1550.4+45.6 X= 1560+45 X\phantom{\rule{0ex}{0ex}}\mathrm{or},45.6 X - 45 X = 1560-1550.4\phantom{\rule{0ex}{0ex}}\mathrm{or},0.6X=9.6\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{X}=16$ $X=\frac{9.6}{.6}$ X = 16 Hence, the missing frequency is 16.

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