The mean and variance of n values of a variable x are 0 and ${\sigma }^2$, respectively. If the variable y = x²  then mean of y is 

Solution:

Given that, mean = 0 and variance $={\mathrm{\sigma }}^2\left(\text{ for }\mathrm{x}\right)$ Therefore, variance $={\mathrm{\sigma }}^2=\frac{{\mathrm{\Sigma x}}^2}{\mathrm{n}}-{\left(\frac{\mathrm{\Sigma x}}{\mathrm{n}}\right)}^2$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{\sigma }}^2=\frac{{\mathrm{\Sigma x}}^2}{\mathrm{n}}-0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow {\mathrm{\sigma }}^2=\frac{{\mathrm{\Sigma x}}^2}{\mathrm{n}}\\ \text{ Now, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{y}={\mathrm{x}}^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\Sigma y}={\mathrm{\Sigma x}}^2\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\overline{\mathrm{y}}=\frac{{\mathrm{\Sigma x}}^2}{\mathrm{n}}={\mathrm{\sigma }}^2\end{array}$