The probability distribution of a random variable $\mathrm{X}$ is Then $\operatorname{Var}(\mathrm{X})=$

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Aicharya · Accepted Answer

$\begin{aligned} 	herefore \quad \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & =\frac{1}{\mathrm{n}}+\frac{2}{\mathrm{n}}+\frac{3}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}}{\mathrm{n}} \ & =\frac{1+2+3+\ldots \mathrm{n}}{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{2(\mathrm{n})}=\frac{\mathrm{n}+1}{2}\end{aligned}$ $\begin{aligned} & \quad \sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\frac{1}{\mathrm{n}}+\frac{4}{\mathrm{n}}+\frac{9}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}^2}{\mathrm{n}} \ & \quad \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\frac{1}{\mathrm{n}}+\frac{4}{\mathrm{n}}+\frac{9}{\mathrm{n}}+\ldots .+\frac{\mathrm{n}^2}{\mathrm{n}} \ & \quad=\frac{1+4+9+\ldots \mathrm{n}^2}{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}}=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6} \ & 	herefore \quad \operatorname{Var}(\mathrm{x})=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}ight)^2 \ & \quad=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\left[\frac{(\mathrm{n}+1)}{2}ight]^2=\frac{2 \mathrm{n}^2+3 \mathrm{n}+1}{6}-\frac{\mathrm{n}^2+2 \mathrm{n}+1}{4} \ & =\frac{4 \mathrm{n}^2+6 \mathrm{n}+2-3 \mathrm{n}^2-6 \mathrm{n}-3}{12}=\frac{\mathrm{n}^2-1}{12}\end{aligned}$

The probability distribution of a random variable $\mathrm{X}$ is Then $\operatorname{Var}(\mathrm{X})=$

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