Let the mean and the standard deviation of the probability distribution X α\\ 1 0 -3 --- --- --- --- --- P(X) frac13 K frac16 frac14 be μ and σ , respectively. If σ-μ=2 , then σ+μ is equal to

Mathematics · Probability

JEE Main 2024 — 5 April, Shift 2 — Question 21

Let the mean and the standard deviation of the probability distribution

X	$\alpha\\$	1	0	-3
P(X)	$\frac13$	K	$\frac16$	$\frac14$

be $\mu$ and $\sigma$ , respectively. If $\sigma-\mu=2$ , then $\sigma+\mu$ is equal to $\qquad$

Numerical answer — enter this value.

$\frac{1}{3}+\mathrm{k}+\frac{1}{6}+\frac{1}{4}=1 \quad \Rightarrow \mathrm{k}=\frac{1}{4}$ ,

$\mu=\frac{\alpha}{3}+\frac{1}{4}-\frac{3}{4}$

$\mu=\frac{\alpha}{3}-\frac{1}{2}$

$\sigma=\sqrt{\left(\alpha^{2} \frac{1}{3}+\frac{1}{4}+9 \frac{1}{4}\right)-\left(\frac{\alpha}{3}-\frac{1}{2}\right)^{2}}$

$\sigma=\sqrt{\frac{2 \alpha^{2}}{9}+\frac{\alpha}{3}+\frac{9}{4}}$

$\sigma=\mu+2$

$\sigma^{2}=(\mu+2)^{2} \Rightarrow \frac{2 \alpha^{2}}{9}+\frac{\alpha}{3}+\frac{9}{4}=\frac{\alpha^{2}}{9}+\frac{9}{4}+\alpha$

$\frac{\alpha^{2}}{9}-\frac{2 \alpha}{3}=0$

$\alpha=0$ , (reject) or $\alpha=6$

$(\because \mathrm{x}=0$ is already given $)$

$\Rightarrow \sigma+\mu=2 \mu+2$

$=5$

Answer key and solution verified before publishing.

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