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The probability distribution of a random variable $X$ is given by \begin{array}{|c|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 \\
\hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{5} & \frac{2}{5} & \frac{2}{5} \\
\hline
\end{array}
then the variance of $\mathrm{X}$ is
Options:
\hline \mathrm{X}=x & 0 & 1 & 2 \\
\hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{5} & \frac{2}{5} & \frac{2}{5} \\
\hline
\end{array}
then the variance of $\mathrm{X}$ is
Solution:
1182 Upvotes
Verified Answer
The correct answer is:
$\frac{14}{25}$
\begin{array}{|c|c|c|c|}
\hline \mathrm{x}_{1} & \mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right) & \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}
$$
\begin{aligned}
\therefore \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} &=\frac{6}{5} \quad \text { and } \quad \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2}=\frac{10}{5}=2 \\
\text { Variance } &=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2}-\left(\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^{2} \\
&=2-\left(\frac{6}{5}\right)^{2} \quad=2-\frac{36}{25}=\frac{14}{25}
\end{aligned}
$$
\hline \mathrm{x}_{1} & \mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right) & \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}
$$
\begin{aligned}
\therefore \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} &=\frac{6}{5} \quad \text { and } \quad \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2}=\frac{10}{5}=2 \\
\text { Variance } &=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2}-\left(\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^{2} \\
&=2-\left(\frac{6}{5}\right)^{2} \quad=2-\frac{36}{25}=\frac{14}{25}
\end{aligned}
$$
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