If the p.m.f. of a r.v. \(X\) is

then, the standard deviation of \(\mathrm{X}\) is (given \(p+q=1\) )

(B)
\(\begin{array}{|c|c|c|c|}
\hline \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2} \\
\hline 0 & \mathrm{q}^{2} & 0 & 0 \\
\hline 1 & 2 \mathrm{pq} & 2 \mathrm{pq} & 2 \mathrm{pq} \\
\hline 2 & \mathrm{p}^{2} & 2 \mathrm{p}^{2} & 4 \mathrm{p}^{2} \\
\hline \text{Total} & & 2 \mathrm{pq}+2 \mathrm{p}^{2} & 2 \mathrm{pq}+4 \mathrm{p}^{2} \\
\hline
\end{array}\)
\(\operatorname{Mean}(\mu)=\mathrm{E}(\mathrm{x})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \quad=2 \mathrm{p}(\mathrm{p}+\mathrm{q})=2 \mathrm{p} \quad\)\(\ldots[\because \mathrm{p}+\mathrm{q}=1\), 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 \(\left(\sigma_{x}\right)=\sqrt{2 \mathrm{pq}}\)