Solution:
\( \frac{1}{3}+k+\frac{1}{6}+\frac{1}{4}=1 \Rightarrow k=\frac{1}{4} \).
\( =\frac{\alpha}{3}+k+0-\dfrac{3}{4}=\mu\Rightarrow\mu=\frac{\alpha}{3}-\frac{1}{2}\cdots (i)\).
\(\sigma^2 = \alpha^2 \times \frac{1}{3} + 1^2 \times k + 0^2 \times \frac{1}{6} + (-3)^2 \times \frac{1}{4} - \left( \frac{\alpha}{3} - \frac{1}{2} \right)^2\)

\(\sigma^2 = \frac{\alpha^2}{3} + \frac{1}{4} + \frac{9}{4} - \left( \frac{\alpha^2}{9} + \frac{1}{4} - \frac{\alpha}{3} \right) = \frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4}\)

\(\sigma - \mu = 2 \Rightarrow \sqrt{\frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4}} - \left( \frac{\alpha}{3} - \frac{1}{2} \right) = 2\)

\(\frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4} = \frac{9}{4} + \frac{\alpha^2}{9} + \alpha\)

\(\frac{\alpha^2}{9} - \frac{2\alpha}{3} = 0\)

\(\frac{\alpha}{3} \left[ \frac{\alpha}{3} - 2 \right] = 0\)
\(\alpha=0\) reject or \(\alpha =6\)
(As for \(\alpha = 0\) for \(x=0\), there are 2 different value of \(p(x))\)

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