Q78 / 180 · KCET 2018 · Mathematics
For the probability distribution given by
$\begin{array}{|c|l|l|l|}
\hline \mathrm{x}=\mathrm{x}_{\mathrm{i}} & 0 & 1 & 2 \\
\hline \mathrm{P}_{\mathrm{i}} & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\
\hline
\end{array}$
the standard deviation \( (\sigma) \) is
$\begin{array}{|c|l|l|l|} \hline \mathrm{x}=\mathrm{x}_{\mathrm{i}} & 0 & 1 & 2 \\ \hline \mathrm{P}_{\mathrm{i}} & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array}$
the standard deviation \( (\sigma) \) is
- A\( \sqrt{\frac{1}{3}} \)
- Correct answerB\( \frac{1}{3} \sqrt{\frac{5}{2}} \)
- C\( \sqrt{\frac{5}{36}} \)
- DNone of the above
Solution
Probability distribution given by
\begin{array}{|l|l|l|l|}
\hline \boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}} & \mathbf{0} & \mathbf{1} & \mathbf{2} \\
\hline \mathrm{P} & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\
\hline
\end{array}
We know that, \( M^{2}+\sigma^{2}=\sum x_{i}^{2} P\left(x=x_{i}\right) \)
Here
\( M=\Sigma x_{i} \cdot P\left(x-x_{i}\right)=\frac{5}{18}+\frac{2}{36}=\frac{6}{18}=\frac{1}{3} \)
So, \( \frac{1}{9}+\sigma^{2}=\frac{5}{18}+\frac{4}{36}=\frac{7}{18} \)
\( \Rightarrow \sigma^{2}=\frac{7}{18}-\frac{2}{18}=\frac{5}{18} \)
Therefore \( \sigma=\frac{1}{3} \sqrt{\frac{5}{2}} \)
\begin{array}{|l|l|l|l|}
\hline \boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}} & \mathbf{0} & \mathbf{1} & \mathbf{2} \\
\hline \mathrm{P} & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\
\hline
\end{array}
We know that, \( M^{2}+\sigma^{2}=\sum x_{i}^{2} P\left(x=x_{i}\right) \)
Here
\( M=\Sigma x_{i} \cdot P\left(x-x_{i}\right)=\frac{5}{18}+\frac{2}{36}=\frac{6}{18}=\frac{1}{3} \)
So, \( \frac{1}{9}+\sigma^{2}=\frac{5}{18}+\frac{4}{36}=\frac{7}{18} \)
\( \Rightarrow \sigma^{2}=\frac{7}{18}-\frac{2}{18}=\frac{5}{18} \)
Therefore \( \sigma=\frac{1}{3} \sqrt{\frac{5}{2}} \)
Hence, the correct answer is option: 'B'.