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Question

# Find the mean age from the following frequency distribution: Age (in years) 25-29 30-34 35-39 40-44 45-49 50-54 55-59 No. of person 4 14 22 16 6 5 3

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Solution

## Converting the series into exclusive form, we get: Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-42\right)}{5}$ $\left({f}_{i}×{u}_{i}\right)$ 24.5-29.5 4 27 −3 −12 29.5-34.5 14 32 −2 −28 34.5-39.5 22 37 −1 −22 39.5-44.5 16 42 = A 0 0 44.5-49.5 6 47 1 6 49.5-54.5 5 52 2 10 54.5-59.5 3 57 3 9 $\sum {f}_{i}=70$ $\sum \left({f}_{i}×{u}_{i}\right)=-37$ $\text{Now,}A=42,h=5,\sum {f}_{i}=70\mathrm{and}\sum \left({f}_{{}_{i}}×{u}_{i}\right)=-37\phantom{\rule{0ex}{0ex}}\therefore \text{Mean,}\overline{x}=A+\left\{h×\frac{\sum \left({f}_{{}_{i}}×{u}_{i}\right)}{\sum {f}_{i}}\right\}\phantom{\rule{0ex}{0ex}}\text{=42+}\left\{5×\frac{\left(-37\right)}{70}\right\}\phantom{\rule{0ex}{0ex}}\text{=42-2}\text{.64}\phantom{\rule{0ex}{0ex}}\text{=39}\text{.36}\phantom{\rule{0ex}{0ex}}\therefore \overline{x}=39.36\phantom{\rule{0ex}{0ex}}\therefore \text{Mean age=39}\text{.36 years}$

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