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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 1 Maths Textbook Solution.

Answers (1)

Answer:3.1,0.7

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}(x)

Given:

\begin{array}{|c|c|c|c|} \hline x_{i} & 2 & 3 & 4 \\ \hline p_{i} & 0.2 & 0.5 & 0.3 \\ \hline \end{array}

Solution:

Mean:=E(X)=\sum_{i=1}^{n} x_{i} p_{i}

          \begin{aligned} &=2 \times 0.2+3 \times 0.5+4 \times 0.3 \\ &=0.4+1.5+1.2 \\ &=3.1 \end{aligned}

           \begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} \cdot p_{i} \\ &=2^{2} \times 0.2+3^{2} \times 0.5+4^{2} \times 0.3 \\ &=0.8+4.5+4.8 \\ &=10.1 \\ \end{aligned}

           \begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=10.1-3.1^{2} \\ &=10.1-9.61 \\ &=0.49 \end{aligned}

Standard deviation =\sqrt{var\left ( X \right )}=\sqrt{0.49}=0.7

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