QJEEM24Mad61f: JEE Main 2024 Mathematics - Class 11 JEEM24Mad61f Solution: Let the mean and the standard deviation of the probability…

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