Let mean and standard deviation of probability distribution $\begin{aligned} \begin{array}{|r|r|c|c|c|}…

The sum of probabilities must equal 1, so K is found from

$\frac{1}{4} + K + \frac{1}{4} + \frac{1}{3} = 1$

The mean is $\mu = \sum xP(X=x) = (-3)\left(\frac{1}{4}\right) + 0\left(\frac{1}{6}\right) + 1\left(\frac{1}{4}\right) + \alpha\left(\frac{1}{3}\right) = -\frac{1}{2} + \frac{\alpha}{3}$.

The second moment is $E(X^2) = 9\left(\frac{1}{4}\right) + 0 + 1\left(\frac{1}{4}\right) + \alpha^2\left(\frac{1}{3}\right) = \frac{5}{2} + \frac{\alpha^2}{3}$.

The variance becomes $\sigma^2 = E(X^2) - \mu^2 = \frac{5}{2} + \frac{\alpha^2}{3} - \left(-\frac{1}{2} + \frac{\alpha}{3}\right)^2 = \frac{9}{4} + \frac{2\alpha^2}{9} + \frac{\alpha}{3}$.

Given $\sigma - \mu = 2$, squaring gives $\sigma^2 = (\mu + 2)^2 = \left(\frac{3}{2} + \frac{\alpha}{3}\right)^2 = \frac{9}{4} + \alpha + \frac{\alpha^2}{9}$.

Equating both expressions for $\sigma^2$ and simplifying leads to $\alpha^2 - 6\alpha = 0$, so $\alpha(\alpha - 6) = 0$.

The solution $\alpha = 0$ is invalid as it creates conflicting probabilities for $x=0$, leaving $\alpha = 6$.

With $\alpha = 6$, $\mu = -\frac{1}{2} + \frac{6}{3} = \frac{3}{2}$, and $\sigma = \mu + 2 = \frac{7}{2}$.

Final answer: $\boxed{\frac{7}{2}}$