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(B) The mean $(\mu)$ of the probability distribution is given by $\mu = \sum x_{i} P_{i}$.$\mu = (0 	imes \frac{25}{36}) + (1 	imes \frac{5}{18}) +

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

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(B) The mean $(\mu)$ of the probability distribution is given by $\mu = \sum x_{i} P_{i}$.$\mu = (0 	imes \frac{25}{36}) + (1 	imes \frac{5}{18}) + (2 	imes \frac{1}{36}) = 0 + \frac{10}{36} + \frac{2}{36} = \frac{12}{36} = \frac{1}{3}$.The variance $(\sigma^2)$ is given by $\sigma^2 = \sum x_{i}^2 P_{i} - \mu^2$.First,calculate $\sum x_{i}^2 P_{i} = (0^2 	imes \frac{25}{36}) + (1^2 	imes \frac{5}{18}) + (2^2 	imes \frac{1}{36}) = 0 + \frac{10}{36} + \frac{4}{36} = \frac{14}{36} = \frac{7}{18}$.Now,$\sigma^2 = \frac{7}{18} - (\frac{1}{3})^2 = \frac{7}{18} - \frac{1}{9} = \frac{7-2}{18} = \frac{5}{18}$.Therefore,the standard deviation $\sigma = \sqrt{\frac{5}{18}} = \sqrt{\frac{5 	imes 2}{18 	imes 2}} = \sqrt{\frac{10}{36}} = \frac{\sqrt{10}}{6}$.Alternatively,$\sigma = \sqrt{\frac{5}{18}} = \frac{1}{3} \sqrt{\frac{5}{2}}$.Thus,option $B$ is correct.

$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

Outcome	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$	$\omega_{4}$	$\omega_{5}$	$\omega_{6}$	$\omega_{7}$
Probability	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

For the probability distribution given by

$x = x_{i}$ $0$ $1$ $2$

$P_{i}$ $\frac{25}{36}$ $\frac{5}{18}$ $\frac{1}{36}$

the standard deviation $(\sigma)$ is

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For the probability distribution given by $x = x_{i}$ $0$ $1$ $2$ $P_{i}$ $\frac{25}{36}$ $\frac{5}{18}$ $\frac{1}{36}$ the standard deviation $(\sigma)$ is