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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise 31.2 Question 8 Maths Textbook Solution.

Provide Solution For R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise 31.2 Question 8 Maths Textbook Solution.

Answers (1)

Answer:

2.53,1.96

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:

Let X be the random variable which denotes the minimum of two numbers which appears

Solution:

\begin{aligned} &S=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\\ \end{aligned}

          \begin{aligned} &\begin{array}{|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(X) & \frac{11}{36} & \frac{9}{36} & \frac{7}{36} & \frac{5}{36} & \frac{3}{36} & \frac{1}{36} \\ \hline \end{array} \end{aligned}

\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=1 \times \frac{11}{36}+2 \times \frac{9}{36}+3 \times \frac{7}{36}+4 \times \frac{5}{36}+5 \times \frac{3}{36}+6 \times \frac{1}{36} \\ &=\frac{91}{36}=2.53 \\ \end{aligned}

\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=1^{2} \times \frac{11}{36}+2^{2} \times \frac{9}{36}+3^{2} \times \frac{7}{36}+4^{2} \times \frac{5}{36}+5^{2} \times \frac{3}{36}+6^{2} \times \frac{1}{36} \\ &=\frac{301}{36} \\ &=8.36 \end{aligned}

\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=8.36-2.53^{2} \\ &=1.96 \end{aligned}

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