Mathematics - Statistics Question with Solution | TestHub

Old mean$=\frac{x_1+x_2+...x_{14}+5}{15}=8$ (given)

$\Rightarrow x_1+x_2+...x_{14}=115$

Correct mean$=\frac{x_1+x_2+...x_{14}+20}{15}={\overline{x}}_{\mathrm{new}}\Rightarrow {\overline{x}}_{\mathrm{new}}=9$

Given, old variance$=9\Rightarrow \frac{{x_1}^2+{x_2}^2+...{x_{14}}^2+5^2}{15}-8^2=9\Rightarrow {x_1}^2+{x_2}^2+...{x_{14}}^2=1070$

Now ${\mathrm{var}}_{\mathrm{new}}=\frac{{x_1}^2+{x_2}^2+...{x_{14}}^2+{20}^2}{15}-9^2=\frac{1070+400}{15}-81=17$