The probability distribution of a random variable $X$ is given by $\begin{array}{|c|c|c|c|} \hline X=x & 0 & 1 & 2 \ \hline P(X=x) & \frac{1}{5} & \frac{2}{5} & \frac{2}{5} \ \hline \end{array}$ then the variance of $X$ is

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Aicharya · Accepted Answer

$\begin{array}{|c|c|c|c|} \hline \mathrm{x}_{1} & \mathrm{p}\left(\mathrm{x}_{\mathrm{i}}ight) & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2} \ \hline 0 & \frac{1}{5} & 0 & 0 \ \hline 1 & \frac{2}{5} & \frac{2}{5} & \frac{2}{5} \ \hline 2 & \frac{2}{5} & \frac{4}{5} & \frac{8}{5} \ \hline \end{array} $ $\begin{aligned} 	herefore \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} &=\frac{6}{5} \quad 	ext { and } \quad \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2}=\frac{10}{5}=2 \ 	ext { Variance } &=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2}-\left(\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}ight)^{2} \ &=2-\left(\frac{6}{5}ight)^{2} \quad=2-\frac{36}{25}=\frac{14}{25} \end{aligned}$

The probability distribution of a random variable $X$ is given by $\begin{array}{|c|c|c|c|} \hline X=x & 0 &…

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