Q60 / 90 · JEE Main 2024 (05 Apr Shift 2) · Mathematics
Let the mean and the standard deviation of the probability distribution
$\begin{array}{|c|c|c|c|c|}
\hline \mathrm{X} & \mathrm{\alpha} & \mathrm{1} & 0 & -3 \\
\hline \mathrm{P}(\mathrm{X}) & \frac{1}{3} & \mathrm{~K} & \frac{1}{6} & \frac{1}{4} \\
\hline
\end{array}$
be $\mu$ and $\sigma$, respectively. If $\sigma-\mu=2$, then $\sigma+\mu$ is equal to________
Correct answer: 5
Solution
$\begin{aligned} & \frac{1}{3}+\mathrm{k}+\frac{1}{6}+\frac{1}{4}=1 \quad \Rightarrow \mathrm{k}=\frac{1}{4} \\ & \mu=\frac{\alpha}{3}+\frac{1}{4}-\frac{3}{4} \\ & \mu=\frac{\alpha}{3}-\frac{1}{2} \\ & \sigma=\sqrt{\left(\alpha^2 \frac{1}{3}+\frac{1}{4}+9 \frac{1}{4}\right)-\left(\frac{\alpha}{3}-\frac{1}{2}\right)^2} \\ & \sigma=\sqrt{\frac{2 \alpha^2}{9}+\frac{\alpha}{3}+\frac{9}{4}} \\ & \sigma=\mu+2 \\ & \sigma^2=(\mu+2)^2 \Rightarrow \frac{2 \alpha^2}{9}+\frac{\alpha}{3}+\frac{9}{4}=\frac{\alpha^2}{9}+\frac{9}{4}+\alpha \\ & \frac{\alpha^2}{9}-\frac{2 \alpha}{3}=0 \\ & \alpha=0,(\text { reject }) \text { or } \alpha=6 \\ & (\because \mathrm{x}=0 \text { is already given) } \\ & \Rightarrow \sigma+\mu=2 \mu+2 \\ & =5\end{aligned}$
Hence, the correct answer is 5.