If m and ${\mathrm{\sigma }}^2$ are the mean and variance of the random variable X, whose distribution is given by
 

X = x	0	1	2	3
P(X = x)	$\frac{1}{3}$	$\frac{1}{2}$	0	$\frac{1}{6}$

 

Given 

Mean(m)$=\sum _{\mathrm{i}=0}^3{\mathrm{x}}_{\mathrm{i}}{\mathrm{p}}_{\mathrm{i}}$

                $$\begin{gathered} =0\left(\frac{1}{3}\right)+1\left(\frac{1}{2}\right)+2\left(0\right)+3\left(\frac{1}{6}\right) \\ =\frac{1}{2}+\frac{1}{2} \\ =1 \end{gathered}$$

Variance $\left({\sigma }^2\right)=\sum _{\mathrm{i}=0}^3{\left({\mathrm{x}}_{\mathrm{i}}-\mathrm{m}\right)}^2 {\mathrm{p}}_{\mathrm{i}}$ where m is mean

$$\begin{gathered} =\frac{1}{3}{(0-1)}^2+\frac{1}{2}{(1-1)}^2+0{(2-1)}^2+\frac{1}{6}{(3-1)}^2 \\ =\frac{1}{3}+0+0+\frac{2}{3} \\ =1 \end{gathered}$$

$\therefore \mathrm{m}=1, {\mathrm{\sigma }}^2=1$