Question

Calculate standard deviation for the given data:

Age (in yrs)	20−25	25−30	30−35	35−40	40−45	45−50	50−55
No. of workers	17	11	8	5	4	3	2

Answer 1

Age (X)	No. of Workers (f)	Mid-Values (m)	fm	m2	$\mathit{f}\mathbf{\times }{\mathit{m}}^{\mathbf{2}}\mathbf{=}\mathit{f}{\mathit{m}}^{\mathbf{2}}$
20−25	17	22.5	382.5	506.25	8606.25
25−30	11	27.5	302.5	756.25	8318.75
30−35	8	32.5	260	1056.25	8450
35−40	5	37.5	187.5	1406.25	7031.25
40−45	4	42.5	170	1806.25	7225
45−50	3	47.5	142.5	2256.25	6768.75
50−55	2	52.5	105	2756.25	5512.5
	Σf=50		Σfm=1550		Σfm2=51912.5

$$\begin{gathered} Mean,\overline{X}=\frac{\Sigma fm}{\Sigma f}=\frac{1550}{50}=31 \\ \mathrm{Standard}\mathrm{deviation},\left(\sigma \right)=\sqrt{\frac{\Sigma f{m}^2}{\Sigma f}-{\left(\overline{X}\right)}^2} \\ or,\left(\sigma \right)=\sqrt{\frac{51912.5}{50}-{\left(31\right)}^2} \\ or,\left(\sigma \right)=\sqrt{1038.25-961} \\ or,\left(\sigma \right)=\sqrt{77.25} \\ \Rightarrow \left(\sigma \right)=8.79\mathrm{years} \end{gathered}$$

Hence, standard deviation of the above series is 8.79 years.