The probability distribution of a random variable \(\mathrm{X}\) is

Then \(\operatorname{Var}(\mathrm{X})=\)

\(\therefore \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} =\frac{1}{\mathrm{n}}+\frac{2}{\mathrm{n}}+\frac{3}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}}{\mathrm{n}} \)
\( =\frac{1+2+3+\ldots \mathrm{n}}{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{2(\mathrm{n})}=\frac{\mathrm{n}+1}{2}\)

\(\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\frac{1}{\mathrm{n}}+\frac{4}{\mathrm{n}}+\frac{9}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}^2}{\mathrm{n}} \)
\( \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\frac{1}{\mathrm{n}}+\frac{4}{\mathrm{n}}+\frac{9}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}^2}{\mathrm{n}} \)
\( =\frac{1+4+9+\ldots \mathrm{n}^2}{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}}=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6} \)
\( \therefore \operatorname{Var}(\mathrm{x})=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2 \)
\( =\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\left[\frac{(\mathrm{n}+1)}{2}\right]^2=\frac{2 \mathrm{n}^2+3 \mathrm{n}+1}{6}-\frac{\mathrm{n}^2+2 \mathrm{n}+1}{4} \)
\( =\frac{4 \mathrm{n}^2+6 \mathrm{n}+2-3 \mathrm{n}^2-6 \mathrm{n}-3}{12}=\frac{\mathrm{n}^2-1}{12}\)