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

Total probability = 1: $$\frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1 \implies K = \frac{1}{4}$$.

 Mean ($$\mu$$): $$\sum X \cdot P(X) = \frac{\alpha}{3} - \frac{1}{2}$$.

 $$E(X^2)$$: $$\sum X^2 \cdot P(X) = \frac{\alpha^2}{3} + \frac{5}{2}$$

 Using $$\sigma = \mu + 2 \implies \sigma^2 = \mu^2 + 4\mu + 4$$, and substituting $$\sigma^2 = E(X^2) - \mu^2$$ yields $$\alpha = 6$$.

 $$\mu = \frac{3}{2}$$ and $$\sigma = \frac{7}{2}$$.

 $$\sigma + \mu = \frac{7}{2} + \frac{3}{2} = 5$$