Question

For a group of 100 males, mean and standard deviation of their daily wages are ₹ 36 and ₹ 9 respectively. For a group of 50 females, it is ₹ 45 and ₹ 6. Find the standard deviation for the whole group.

Answer 1

Calculating combined standard deviation,

$$\begin{gathered} { \\ }_{1,2}^{\sigma }=\sqrt{\frac{N_1{\sigma }_1^2+N_2{\sigma }_2^2+N_1{d}_1^2+N_2{d}_2^2}{N_1+N_2}} \\ where, \\ {N}_1=100N_2=50 \\ {\overline{X}}_1=36{\overline{X}}_2=45 \\ {\sigma }_1=9{\sigma }_2=6 \end{gathered}$$

To calculate combined standard deviation, we need to find the combined mean of the observations.

Thus, Combined mean is

$$\begin{gathered} {\overline{X}}_{12}=\frac{N_1{\overline{X}}_1+N_2{\overline{X}}_2}{N_1+N_2} \\ =\frac{100\left(36\right)+50\left(45\right)}{150} \\ =\frac{3600+2250}{150}=Rs39 \\ {d}_1^2={\left({\overline{X}}_1-{\overline{X}}_{12}\right)}^2={\left(36-39\right)}^2={\left(-3\right)}^2=9 \\ {d}_2^2={\left({\overline{X}}_2-{\overline{X}}_{12}\right)}^2={\left(45-39\right)}^2={\left(6\right)}^2=36 \\ Thus,Combineds\tan darddeviationwillbe:{ \\ }_{1,2}^{\mathrm{\sigma }}=\sqrt{\frac{N_1{\sigma }_1^2+N_2{\sigma }_2^2+N_{1}d{}_1^2+N_{2}d{}_2^2}{N_1+N_2}} \\ or,{\sigma }_{1,2}=\sqrt{\frac{100{\left(9\right)}^2+50{\left(6\right)}^2+100\left(9\right)+50\left(36\right)}{150}} \\ or,{\sigma }_{1,2}=\sqrt{\frac{8100+1800+900+1800}{150}} \\ or,{\sigma }_{1,2}=\sqrt{\frac{12600}{150}} \\ or,{\sigma }_{1,2}=\sqrt{84} \\ \Rightarrow {\sigma }_{1,2}=9.16 \end{gathered}$$

Hence, the combined standard deviation is Rs 9.16