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

Answers (1)

Answer:

2,1,1

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 coin tossed four times

Solution:

X denotes the occurrence of head

Sample space \left \{ \text { HHHH,HHHT,HHTH,HHTT,HTHH,HTHT,HTTH,HTTT,THHH,THHT,THTH,THTT, TTHH, TTHT, TTTH, TTTT \} } \right \}

\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16} \\ \hline \end{array}\\ \end{aligned}

\begin{aligned} &\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0 \times \frac{1}{16}+1 \times \frac{1}{4}+2 \times \frac{3}{8}+3 \times \frac{1}{4}+4 \times \frac{1}{16} \\ &=\frac{1}{4}+\frac{3}{4}+\frac{3}{4}+\frac{1}{4} \\ &=2 \end{aligned} \end{aligned}

\begin{aligned} &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=0 \times \frac{1}{16}+1^{2} \times \frac{1}{4}+2^{2} \times \frac{3}{8}+3^{2} \times \frac{1}{4}+4^{2} \times \frac{1}{16} \\ &=0+\frac{1}{4}+\frac{3}{2}+\frac{9}{4}+1 \\ &=5 \\ \end{aligned}

\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=5-2^{2} \\ &=5-4 \\ &=1 \end{aligned}

Standard deviation =\sqrt{\operatorname{Var}(X)}=\sqrt{1}=1

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