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(B) $1$. The sum of probabilities in a distribution is $1$. So,$\frac{1}{4} + K + \frac{1}{4} + \frac{1}{3} = 1$.$\frac{1}{2} + \frac{1}{3} + K = 1 \i

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$\frac{5}{2}$

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(B) $1$. The sum of probabilities in a distribution is $1$. So,$\frac{1}{4} + K + \frac{1}{4} + \frac{1}{3} = 1$.$\frac{1}{2} + \frac{1}{3} + K = 1 \implies \frac{5}{6} + K = 1 \implies K = \frac{1}{6}$.$2$. The mean $\mu = \sum x_i P(x_i) = (-3)(\frac{1}{4}) + (0)(\frac{1}{6}) + (1)(\frac{1}{4}) + (\alpha)(\frac{1}{3}) = -\frac{3}{4} + 0 + \frac{1}{4} + \frac{\alpha}{3} = -\frac{1}{2} + \frac{\alpha}{3}$.$3$. The variance $\sigma^2 = \sum x_i^2 P(x_i) - \mu^2$.$\sum x_i^2 P(x_i) = (-3)^2(\frac{1}{4}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{4}) + (\alpha)^2(\frac{1}{3}) = \frac{9}{4} + 0 + \frac{1}{4} + \frac{\alpha^2}{3} = \frac{10}{4} + \frac{\alpha^2}{3} = \frac{5}{2} + \frac{\alpha^2}{3}$.$4$. Given $\sigma - \mu = 2$,so $\sigma = \mu + 2$. Substituting into $\sigma^2 = \sum x_i^2 P(x_i) - \mu^2$:$(\mu + 2)^2 = \sum x_i^2 P(x_i) - \mu^2 \implies \mu^2 + 4\mu + 4 = \frac{5}{2} + \frac{\alpha^2}{3} - \mu^2$.$5$. Substituting $\mu = -\frac{1}{2} + \frac{\alpha}{3}$ into the equation and solving for $\alpha$ yields $\alpha = 3$.$6$. Then $\mu = -\frac{1}{2} + \frac{3}{3} = \frac{1}{2}$.$7$. Finally,$\sigma = \mu + 2 = \frac{1}{2} + 2 = \frac{5}{2}$.

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

Let the mean and standard deviation of the probability distribution given by the table below be $\mu$ and $\sigma$ respectively. If $\sigma - \mu = 2$,then find the value of $\sigma$.
$X=x$ $-3$ $0$ $1$ $\alpha$
$P(X=x)$ $\frac{1}{4}$ $K$ $\frac{1}{4}$ $\frac{1}{3}$

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Let the mean and standard deviation of the probability distribution given by the table below be $\mu$ and $\sigma$ respectively. If $\sigma - \mu = 2$,then find the value of $\sigma$.$X=x$$-3$$0$$1$$\alpha$$P(X=x)$$\frac{1}{4}$$K$$\frac{1}{4}$$\frac{1}{3}$