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If a discrete random variable X has probability distribution as follows
$\begin{array}{|c|c|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 & 3 \\
\hline \mathrm{P}[\mathrm{X}=x] & \mathrm{k} & 3 \mathrm{k} & 3 \mathrm{k} & \mathrm{k} \\
\hline
\end{array}$
Then $\operatorname{var}(\mathrm{X})=$
Options:
$\begin{array}{|c|c|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 & 3 \\
\hline \mathrm{P}[\mathrm{X}=x] & \mathrm{k} & 3 \mathrm{k} & 3 \mathrm{k} & \mathrm{k} \\
\hline
\end{array}$
Then $\operatorname{var}(\mathrm{X})=$
Solution:
2561 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{4}$
Here $\mathrm{k}+3 \mathrm{k}+3 \mathrm{k}+\mathrm{k}=8 \mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{8}$
$\begin{aligned} \sum \mathrm{P}_{\left(\mathrm{x}_{1}\right)} \mathrm{x}_{\mathrm{i}} &=0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}=\frac{12}{8}=\frac{3}{2} \text { and } \\ \sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}^{2} &=\left(0 \times \frac{1}{8}\right)+\left(1 \times \frac{3}{8}\right)+\left(4 \times \frac{3}{8}\right)+\left(9 \times \frac{1}{8}\right)=\frac{24}{8}=3 \\ \text { Variance } &=\mathrm{V}(\mathrm{X}) \\ &=\sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}^{2}-\left[\sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}\right]^{2} \\ &=3-\left(\frac{3}{2}\right)^{2}=3-\frac{9}{4}=\frac{3}{4} \end{aligned}$
$\begin{aligned} \sum \mathrm{P}_{\left(\mathrm{x}_{1}\right)} \mathrm{x}_{\mathrm{i}} &=0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}=\frac{12}{8}=\frac{3}{2} \text { and } \\ \sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}^{2} &=\left(0 \times \frac{1}{8}\right)+\left(1 \times \frac{3}{8}\right)+\left(4 \times \frac{3}{8}\right)+\left(9 \times \frac{1}{8}\right)=\frac{24}{8}=3 \\ \text { Variance } &=\mathrm{V}(\mathrm{X}) \\ &=\sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}^{2}-\left[\sum \mathrm{P}_{\left(\mathrm{x}_{i}\right)} \mathrm{x}_{\mathrm{i}}\right]^{2} \\ &=3-\left(\frac{3}{2}\right)^{2}=3-\frac{9}{4}=\frac{3}{4} \end{aligned}$
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