Correct answer: 5
\(\frac{1}{3}+k+\frac{1}{6}+\frac{1}{4}=1 \)
\( \Rightarrow k=\frac{1}{4}\)
\(\mu=\frac{\alpha}{3}+\frac{1}{4}-\frac{3}{4}\)
\(\mu=\frac{\alpha}{3}-\frac{1}{2}\)
\(\sigma=\sqrt{\left(\alpha^{2} \frac{1}{3}+\frac{1}{4}+9 \frac{1}{4}\right)-\left(\frac{\alpha}{3}-\frac{1}{2}\right)^{2}}\)
\(\sigma=\sqrt{\frac{2 \alpha^{2}}{9}+\frac{\alpha}{3}+\frac{9}{4}}\)
\(\sigma=\mu+2\)
\(\sigma^{2}=(\mu+2)^{2} \Rightarrow \frac{2 \alpha^{2}}{9}+\frac{\alpha}{3}+\frac{9}{4}=\frac{\alpha^{2}}{9}+\frac{9}{4}+\alpha\)
\(\frac{\alpha^{2}}{9}-\frac{2 \alpha}{3}=0\)
\(\alpha=0\), (reject) or \(\alpha=6\)
(\(\because \mathrm{x}=0\) is already given)
\(\Rightarrow \sigma+\mu=2 \mu+2\)
\(=5\)
