If the p.m.f. of a r.v. $X$ is then, the standard deviation of $\mathrm{X}$ is (given $p+q=1$ )
If the p.m.f. of a r.v. $X$ is
then, the standard deviation of $\mathrm{X}$ is (given $p+q=1$ )
- $2 \sqrt{q}$
- $\sqrt{2 p q}$
- $2 \sqrt{p}$
- $\sqrt{p q}$
Solution
\begin{array}{|c|c|c|c|}
\hline \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2} \\
\hline 0 & \mathrm{q}^{2} & 0 & 0 \\
\hline 1 & 2 \mathrm{pq} & 2 \mathrm{pq} & 2 \mathrm{pq} \\
\hline 2 & \mathrm{p}^{2} & 2 \mathrm{p}^{2} & 4 \mathrm{p}^{2} \\
\hline Total & & 2 \mathrm{pq}+2 \mathrm{p}^{2} & 2 \mathrm{pq}+4 \mathrm{p}^{2} \\
\hline
\end{array}
$\operatorname{Mean}(\mu)=\mathrm{E}(\mathrm{x})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \quad=2 \mathrm{p}(\mathrm{p}+\mathrm{q})=2 \mathrm{p} \quad \ldots[\because \mathrm{p}+\mathrm{q}=1$, given $]$
Variance $\left(\sigma_{\mathrm{x}}^{2}\right)=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}{ }^{2}-\mu^{2}$
$=2 \mathrm{pq}+4 \mathrm{p}^{2}-4 \mathrm{p}^{2}=2 \mathrm{pq}$
Standard deviation $\left(\sigma_{x}\right)=\sqrt{2 \mathrm{pq}}$
Asked in: MHT CET 2020 (19 Oct Shift 1)
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