Question

For the probability distribution given by

$\begin{array}{cccc}X=x_i& 0& 1& 2\\ P_i& \frac{25}{36}& \frac{5}{18}& \frac{1}{36}\end{array}$

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

The correct option is B $\frac{1}{3}\sqrt{\frac{5}{2}}$

$\text{Mean},E\left(X\right)=\sum P_i{x}_i=\frac{5}{18}+2\times \frac{1}{36}=\frac{1}{3}$
$\text{Variance},{\sigma }^2=E\left(X^2\right)-\left(E\right(X){)}^2$
$=1\times \frac{5}{18}+4\times \frac{1}{36}-{\left(\frac{1}{3}\right)}^2=\frac{5}{18}$
$\therefore \text{Standard deviation},\sigma =\sqrt{\frac{5}{18}}=\frac{1}{3}\sqrt{\frac{5}{2}}$