Mathematics - Statistics Question with Solution | TestHub

$\begin{array}{rl}& \text{ Mean }(\overline{\mathrm{x}})=10\\ & \Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}=10\\ & \Sigma {\mathrm{x}}_{\mathrm{i}}=10\times 20=200\end{array}$
If 8 is replaced by 12, then $\Sigma x_i=200-8+12=204$ $\therefore$ Correct mean $(\overline{\mathrm{x}})=\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}$ $=\frac{204}{20}=10.2$ $\because$ Standard deviation $=2$ $\therefore$ Variance $=(\text{ S.D. }{)}^2=2^2=4$ $\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}-{\left(\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}\right)}^2=4$ $\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}-(10{)}^2=4$ $\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}=104$ $\Rightarrow \Sigma x_i^2=2080$
Now, replaced ' 8 ' observations by ' 12 ' Then, $\Sigma {\mathrm{x}}_{\mathrm{i}}^2=2080-8^2+12^2=2160$ $\therefore$ Variance of removing observations $\begin{array}{rl}& \Rightarrow \frac{\Sigma x_i^2}{20}-{\left(\frac{\Sigma x_i}{20}\right)}^2\\ & \Rightarrow \frac{2160}{20}-(10.2{)}^2\\ & \Rightarrow 108-104.04\\ & \Rightarrow 3.96\end{array}$
Correct standard deviation $=\sqrt{3.96}$