Question

The following data gives arithmetic means, standard deviations of three sub-groups . Calculate arithmetic mean and standard deviation of the whole group.

Sub-group	No. of Men	Average Wage (in ₹)	Standard deviation (in ₹)
A	50	61.0	8.0
B	100	70.0	9.0
C	120	80.5	10.0

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ {\overline{\mathit{X}}}_{\mathrm{A}}=61,{\sigma }_{\mathrm{A}}=8,N_{\mathrm{A}}=50 \\ {\overline{\mathit{X}}}_{\mathrm{B}}=70,{\sigma }_{\mathrm{B}}=9,N_{\mathrm{B}}=100 \\ {\overline{\mathit{X}}}_{\mathrm{C}}=80.5,{\sigma }_{\mathrm{C}}=10,N_{\mathrm{C}}=120 \\ \mathbf{Combined}\mathbf{}\mathbf{Mean} \\ {\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B},\mathrm{C}}=\frac{N_{\mathrm{A}}{\overline{\mathit{X}}}_{\mathrm{A}}+N_{\mathrm{B}}{\overline{\mathit{X}}}_{\mathrm{B}}+N_{\mathrm{C}}{\overline{\mathit{X}}}_{\mathrm{C}}}{N_{\mathrm{A}}+N_{\mathrm{B}}+N_{\mathrm{C}}}=\frac{50\times 61+100\times 70+120\times 80.5}{50+100+120} \\ =\frac{3050+7000+9660}{270}=\frac{19710}{270} \\ =73 \end{gathered}$$
Combined Standard Deviation
$$\begin{gathered} d_{\mathrm{A}}^2={\overline{\mathit{X}}}_{\mathrm{A}}-{\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B},\mathrm{C}}={\left(61-73\right)}^2={\left(-12\right)}^2=144 \\ {d}_{\mathrm{B}}^2={\overline{\mathit{X}}}_{\mathrm{B}}-{\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B},\mathrm{C}}={\left(70-73\right)}^2={\left(-3\right)}^2=9 \\ {d}_{\mathrm{C}}^2={\overline{\mathit{X}}}_{\mathrm{C}}-{\overline{\mathit{X}}}_{\mathrm{A},\mathrm{B},\mathrm{C}}=\left(80.5-73\right)={\left(7.5\right)}^2=56.25 \\ {\sigma }_{\mathrm{A},\mathrm{B},\mathrm{C}}=\sqrt{\frac{N_{\mathrm{A}}{\sigma }_{\mathrm{A}}^2+N_{\mathrm{B}}{\sigma }_{\mathrm{B}}^2+N_{\mathrm{C}}{\sigma }_{\mathrm{C}}^2+N_{\mathrm{A}}{d}_{\mathrm{A}}^2+N_{\mathrm{B}}{d}_{\mathrm{B}}^2+N_{\mathrm{C}}{d}_{\mathrm{C}}^2}{N_{\mathrm{A}}+N_{\mathrm{B}}+N_{\mathrm{C}}}} \\ =\sqrt{\frac{50{\left(8\right)}^2+100{\left(9\right)}^2+120{\left(10\right)}^2+50\left(144\right)+100\left(9\right)+120\left(56.25\right)}{50+100+120}} \\ =\sqrt{\frac{3200+8100+12000+7200+900+6750}{270}} \\ =\sqrt{\frac{38150}{270}}=\sqrt{141.296} \\ =11.9 \end{gathered}$$
Thus, arithmetic mean and standard deviation of the whole group are 73 and 11.9 respectively.