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(A) For a probability distribution,the sum of probabilities is $1$:$\frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1$$K + \frac{4+2+3}{12} = 1 \Rightar

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(A) For a probability distribution,the sum of probabilities is $1$:$\frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1$$K + \frac{4+2+3}{12} = 1 \Rightarrow K + \frac{9}{12} = 1 \Rightarrow K = 1 - \frac{3}{4} = \frac{1}{4}$.The mean $\mu = \sum X P(X)$:$\mu = \alpha(\frac{1}{3}) + 1(\frac{1}{4}) + 0(\frac{1}{6}) + (-3)(\frac{1}{4}) = \frac{\alpha}{3} + \frac{1}{4} - \frac{3}{4} = \frac{\alpha}{3} - \frac{1}{2}$.The variance $\sigma^2 = \sum X^2 P(X) - \mu^2$:$\sum X^2 P(X) = \alpha^2(\frac{1}{3}) + 1^2(\frac{1}{4}) + 0^2(\frac{1}{6}) + (-3)^2(\frac{1}{4}) = \frac{\alpha^2}{3} + \frac{1}{4} + \frac{9}{4} = \frac{\alpha^2}{3} + \frac{10}{4} = \frac{\alpha^2}{3} + \frac{5}{2}$.$\sigma^2 = (\frac{\alpha^2}{3} + \frac{5}{2}) - (\frac{\alpha}{3} - \frac{1}{2})^2 = \frac{\alpha^2}{3} + \frac{5}{2} - (\frac{\alpha^2}{9} - \frac{\alpha}{3} + \frac{1}{4}) = \frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4}$.Given $\sigma - \mu = 2$,so $\sigma = \mu + 2$. Squaring both sides:$\sigma^2 = (\mu + 2)^2 = \mu^2 + 4\mu + 4$.Substituting $\mu = \frac{\alpha}{3} - \frac{1}{2}$:$\frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4} = (\frac{\alpha}{3} - \frac{1}{2})^2 + 4(\frac{\alpha}{3} - \frac{1}{2}) + 4$$\frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4} = \frac{\alpha^2}{9} - \frac{\alpha}{3} + \frac{1}{4} + \frac{4\alpha}{3} - 2 + 4$$\frac{\alpha^2}{9} - \frac{2\alpha}{3} = 0 \Rightarrow \alpha(\frac{\alpha}{9} - \frac{2}{3}) = 0$.Since $\alpha 
eq 0$ (as $X=0$ is already a distinct value),$\alpha = 6$.Then $\mu = \frac{6}{3} - \frac{1}{2} = 2 - 0.5 = 1.5$.Since $\sigma = \mu + 2 = 1.5 + 2 = 3.5$.Therefore,$\sigma + \mu = 3.5 + 1.5 = 5$.

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

$X = x$	$-1$	$0$	$1$	$2$
$F(X = x)$	$0.3$	$0.7$	$0.8$	$1$

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$0.2$	$0.3$	$0.15$	$0.25$	$0.1$

Let the mean and the standard deviation of the probability distribution be $\mu$ and $\sigma$,respectively. If $\sigma - \mu = 2$,then $\sigma + \mu$ is equal to:

$X$ $\alpha$ $1$ $0$ $-3$

$P(X)$ $\frac{1}{3}$ $K$ $\frac{1}{6}$ $\frac{1}{4}$

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Let the mean and the standard deviation of the probability distribution be $\mu$ and $\sigma$,respectively. If $\sigma - \mu = 2$,then $\sigma + \mu$ is equal to: $X$ $\alpha$ $1$ $0$ $-3$ $P(X)$ $\frac{1}{3}$ $K$ $\frac{1}{6}$ $\frac{1}{4}$