Question

For the probability distribution given by $\begin{array}{cc}X=x_i& 0\\ P.& \frac{25}{36}\end{array}|$ $\begin{array}{c}1\\ 5\\ 18\end{array}$ $\left|\begin{array}{c}2\\ 1\\ 36\end{array}\right|$ the standard deviation $\left(\sigma \right)$ is?

• A. $\sqrt{\frac{1}{3}}$
• B. $\frac{1}{3}\sqrt{\frac{5}{2}}$
• C. $\sqrt{\frac{5}{36}}$
• D. None of the above

Answer 1

Answer: (B)

$\frac{1}{3}\sqrt{\frac{5}{2}}$

$M^2+{\sigma }^2=\sum x_i^{2}P(x=x_i)$

$M=\sum x_{i}P(x=x_i)=\frac{5}{18}+\frac{2}{36}=\frac{6}{18}=\frac{1}{3}$

$\therefore \frac{1}{9}+{\sigma }^2=\frac{5}{18}+\frac{4}{36}=\frac{7}{18}$

$\therefore {\sigma }^2=\frac{7}{18}-\frac{2}{18}=\frac{5}{18}$

$\therefore S.D=\frac{1}{3}\sqrt{\frac{5}{2}}$.