Mean $$(\mu)=\Sigma x_i P\left(x_i\right)$$

Standard deviation $$(\sigma)=\sqrt{\left(\Sigma x_i^2 P\left(x_i\right)\right)-\mu^2}$$

$$\begin{aligned} & \Rightarrow \quad \mu=\frac{1}{3} \alpha+K-\frac{3}{4} \\ & \sigma=\sqrt{\left(\frac{1}{3} \alpha^2+K+0+\frac{9}{4}\right)-\left(\frac{1}{3} \alpha+K-\frac{3}{4}\right)^2} \\ & \because \Sigma P_i=1 \Rightarrow \frac{1}{3}+K+\frac{1}{6}+\frac{1}{4}=1 \\ & \Rightarrow \quad K=\frac{1}{4} \Rightarrow \mu=\frac{1}{3} \alpha-\frac{1}{2} \\ & \because \sigma-\mu=2 \\ & \sigma^2=(\mu+2)^2 \end{aligned}$$

$$\begin{aligned} & \frac{1}{3} \alpha^2+\frac{5}{2}-\mu^2=(\mu+2)^2 \\ & \frac{1}{3} \alpha^2+\frac{5}{2}=\left(\frac{1}{3} \alpha-\frac{1}{2}\right)^2+\left(\frac{1}{3} \alpha+\frac{3}{2}\right)^2 \\ & \Rightarrow \alpha=0,6 \end{aligned}$$

$$\begin{array}{ll} \text { If } \alpha=0, K=\frac{1}{4} & \text { If } \alpha=6, K=\frac{1}{4} \\ \mu=-\frac{1}{2}, \sigma=\frac{3}{2} & \mu=\frac{3}{2}, \sigma=\frac{7}{2} \\ \sigma+\mu=1 & \sigma+\mu=5 \end{array}$$

Both (1) and (5) are correct but according to NTA (5) is correct