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  • Need Solution for R D Sharma Maths Class 12 Chapter 31 Mean and Variance of a Random Variable Exercise 31 Point 2 Question 10 Maths Textbook Solution

Need Solution for R.D.Sharma Maths Class 12 Chapter 31 Mean and Variance of a Random Variable Exercise 31.2  Question 10  Maths Textbook Solution.

Answers (1)

Answer:

7,11.67

Hint:

E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}

Given:

A fair dice is tossed

Solution:

\begin{aligned} &\begin{aligned} &S=\{1,2,3,4,5,6\} \\ &X=\{2,4,6,8,10,12\} \end{aligned}\\ &\begin{array}{|c|c|c|c|c|c|c|} \hline X & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline P(X) & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ \hline \end{array} \end{aligned}

\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=\frac{1}{6}[2+4+6+8+10+12] \\ &=\frac{42}{6}=7 \\ \end{aligned}

\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=\frac{1}{6}\left[2^{2}+4^{2}+6^{2}+8^{2}+10^{2}+12^{2}\right] \\ &=\frac{1}{6}[4+16+36+64+100+144] \\ &=60.67 \\ \end{aligned}

\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\\\ &=60.67-49 \\\\ &=11.67 \end{aligned}

 

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