Mathematics - Statistics Question with Solution | TestHub

$\begin{array}{rl}& \frac{1}{3}+\mathrm{k}+\frac{1}{6}+\frac{1}{4}=1\Rightarrow \mathrm{k}=\frac{1}{4}\\ & \mu =\frac{\alpha }{3}+\frac{1}{4}-\frac{3}{4}\\ & \mu =\frac{\alpha }{3}-\frac{1}{2}\\ & \sigma =\sqrt{\left({\alpha }^2\frac{1}{3}+\frac{1}{4}+9\frac{1}{4}\right)-{\left(\frac{\alpha }{3}-\frac{1}{2}\right)}^2}\\ & \sigma =\sqrt{\frac{2{\alpha }^2}{9}+\frac{\alpha }{3}+\frac{9}{4}}\\ & \sigma =\mu +2\\ & {\sigma }^2=(\mu +2{)}^2\Rightarrow \frac{2{\alpha }^2}{9}+\frac{\alpha }{3}+\frac{9}{4}=\frac{{\alpha }^2}{9}+\frac{9}{4}+\alpha \\ & \frac{{\alpha }^2}{9}-\frac{2\alpha }{3}=0\\ & \alpha =0,(\text{ reject })\text{ or }\alpha =6\\ & (\because \mathrm{x}=0\text{ is already given) }\\ & \Rightarrow \sigma +\mu =2\mu +2\\ & =5\end{array}$