Let the mean and the standard deviation of the probability distribution array |c|c|c|c|c| X & α & 1 & 0 & -3 P(X) & (1)/(3) & ~K & (1)/(6) & (1)/(4) array be μ and σ , respectively. If σ-μ=2 , then σ+μ is equal

Correct answer: 0 aligned & (1)/(3)+k+(1)/(6)+(1)/(4)=1 k=(1)/(4) & μ=(α)/(3)+(1)/(4)-(3)/(4) & μ=(α)/(3)-(1)/(2) & σ= (α^2 (1)/(3)+(1)/(4)+9 (1)/(4))-((α)/(3)-(1)/(2))^2 & σ= (2 α^2)/(9)+(α)/(3)+(9)/(4) & σ=μ+2 & σ^2=(μ+2)^2 (2 α^2)/(9)+(α)/(3)+(9)/(4)=(α^2)/(9)+(9)/(4)+α & (α^2)/(9)-(2 α)/(3)=0 & α=0,( reject ) or α=6 & ( x=0 is already given) & σ+μ=2 μ+2 & =5 aligned

JEE Main 20245 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Statistics

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Statistics

Let the mean and the standard deviation of the probability...

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}

\mu and

\sigma

, respectively. If

\sigma-\mu=2

, then

\sigma+\mu

is equal to________

Correct answer

Step-by-step 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{\text{ or } } \alpha=6 \\ & (\because \mathrm{x}=0 \text{ is already given) } \\ & \Rightarrow \sigma+\mu=2 \mu+2 \\ & =5\end{aligned}

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