Mathematics - Statistics Question with Solution | TestHub

Incorrect mean is

$\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=20$

$\Rightarrow \sum _{i=1}^{10}{x}_i=200 ....\left(\mathrm{i}\right)$

So, corrected mean is

$\frac{\sum _{i=1}^{10}{x}_i-50+40}{10}=\frac{200-10}{10}=19$

Incorrect variance is

$\frac{\sum _{i=1}^{10}{x}_i^2}{10}-400=64$

$\Rightarrow \sum _{i=1}^{10}{x}_i^2=4640$

Now, corrected variance is

$=\frac{\sum _{i=1}^9{x}_i^2+{40}^2-{50}^2}{10}-{\left(19\right)}^2$

$$\begin{gathered} =\frac{4640-900}{10}-{\left(19\right)}^2 \\ =374-361 \\ =13 \end{gathered}$$