Question

Of the two sets of income distribution of five and seven persons respectively, as given below calculate standard deviation:

(i) Income (₹)	4,000	4,200	4,400	4,600	4,800
(ii) Income (₹)	3,000	4,000	4,200	4,400	4,600	4,800	5,800

Answer 1

For Income Group I

X1	$x_1\mathit{=}{X}_1\mathit{-}\overline{X_1}$	${\mathit{x}}_{\mathbf{I}}^{\mathbf{2}}$
4000 4200 4400 4600 4800	−400 −200 0 200 400	160000 40000 0 40000 160000
∑X1 = 22,000		∑${\mathit{x}}_{\mathbf{I}}^{\mathbf{2}}$ = 4,00,000

$$\begin{gathered} {\overline{X}}_1=\frac{{\sum }_{}{X}_1}{n_1}=\frac{22000}{5}=4400 \\ \mathrm{Standard}\mathrm{Deviation}\left({\sigma }_1\right)=\sqrt{\frac{{\sum }_{}{x}_1^2}{n_1}}=\sqrt{\frac{4,00,000}{5}}=282.84 \\ \mathrm{Thus},\mathrm{Standard}\mathrm{Deviation}\mathrm{for}\mathrm{the}\mathrm{first}\mathrm{group}\mathrm{of}\mathrm{income}\mathrm{is}\mathrm{Rs}282.84 \end{gathered}$$

For Income Group II

X2	$x_2\mathit{=}{X}_2\mathit{-}\overline{X_2}$	${\mathit{x}}_{\mathbf{2}}^{\mathbf{2}}$
3000 4000 4200 4400 4600 4800 5800	−1400 −400 −200 0 200 400 1400	1960000 160000 40000 0 40000 160000 1960000
30,800		∑${\mathit{x}}_{\mathbf{2}}^{\mathbf{2}}$ = 4320000

$$\begin{gathered} {\overline{X}}_2=\frac{{\sum }_{}{X}_2}{n_2}=\frac{30,800}{7}=4,400 \\ \mathrm{Standard}\mathrm{Deviation}\left({\sigma }_2\right)=\sqrt{\frac{{\sum }_{}{x}_2^2}{n_2}}=\sqrt{\frac{4,32,000}{7}}=785.58 \\ \mathrm{Thus},\mathrm{Standard}\mathrm{Deviation}\mathrm{for}\mathrm{the}\mathrm{second}\mathrm{group}\mathrm{of}\mathrm{income}\mathrm{is}\mathrm{Rs}785.58 \end{gathered}$$