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  • Please Solve R.D.Sharma class 12 Chapter 31 Mean and Variance of a random Variable Exercise 31.2 Question 1 sub Question 3 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 31 Mean and Variance of a random Variable  Exercise 31.2 Question 1 Sub Question 3 Maths Textbook Solution.

Answers (1)

Answer:

-1,2.9

Hint: Mean E(X)=\sum_{i=1}^{n} x_{i} \cdot p_{i}, E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} \cdot p_{i}

Standard Deviation =\sqrt{var\left ( X \right )}

Given:

\begin{array}{|c|c|c|c|c|} \hline x_{i} & -5 & -4 & 1 & 2 \\ \hline p_{i} & \frac{1}{4} & \frac{1}{8} & \frac{1}{2} & \frac{1}{8} \\ \hline \end{array}

Solution:

\begin{aligned} &=E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ &=-5 \times \frac{1}{4}+(-4) \times \frac{1}{8}+1 \times \frac{1}{2}+2 \times \frac{1}{8} \\ \end{aligned}

\begin{aligned} &=-\frac{5}{4}-\frac{1}{2}+\frac{1}{2}+\frac{1}{4} \\ &=\frac{-5+1}{4}=-1 \\ \end{aligned}

\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} p_{i} \\ &=(-5)^{2} \times \frac{1}{4}+(-4)^{2} \times \frac{1}{8}+1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{8} \\ &=\frac{25}{4}+2+\frac{1}{2}+\frac{1}{2} \\ &=\frac{25+8+2+2}{4} \end{aligned}

\begin{aligned} &=\frac{37}{4} \\ \end{aligned}

\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\\\ &=\frac{37}{4}-1=\frac{33}{4} \\ \end{aligned}

\begin{aligned} &\text { Standard deviation }=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{33}{4}}=2.87=2.9 \end{aligned}

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