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Question

# Calculate the Mean and Standard Deviation from the following distribution. Age (years) 15−19 20−24 25−29 30−34 35−39 40−44 No. of Persons 4 20 38 24 10 4

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Solution

## In order to calculate the mean and standard deviation, we first need to convert the inclusive series into exclusive series as given below: Age (X) Frequency (f) Mid-Values (m) fm m2 fm2 14.5−19.5 4 17 68 289 1156 19.5−24.5 20 22 440 484 9680 24.5−29.5 38 27 1026 729 27702 29.5−34.5 24 32 768 1024 24576 34.5−39.5 10 37 370 1369 13690 39.5−44.5 4 42 168 1764 7056 Σf=100 Σfm=2840 Σfm2=83860 $Mean,\overline{X}=\frac{\Sigma fm}{\Sigma f}=\frac{2840}{100}=28.4years\phantom{\rule{0ex}{0ex}}\mathrm{Standard}\mathrm{deviation},\left(\sigma \right)=\sqrt{\frac{\Sigma f{m}^{2}}{\Sigma f}-{\left(\overline{X}\right)}^{2}}\phantom{\rule{0ex}{0ex}}or,\left(\sigma \right)=\sqrt{\frac{83860}{100}-{\left(28.4\right)}^{2}}\phantom{\rule{0ex}{0ex}}or,\left(\sigma \right)=\sqrt{838.6-806.56}\phantom{\rule{0ex}{0ex}}or,\left(\sigma \right)=\sqrt{32.04}\phantom{\rule{0ex}{0ex}}⇒\left(\sigma \right)=5.66years$ Hence, mean age of the persons is 28.4 years and standard deviation is 5.66 years.

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