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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 6 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise 31.2 Question 6 Maths Textbook Solution.

Answers (1)

Answer:

3,\frac{3}{4},\frac{\sqrt{3}}{2}

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:

Three coins are tossed

Solution:

S=\left \{ HHH,HHT,HTH,THH,TTH,THT,HTT,TTT \right \}

Let ‘X’ be the random variable for tail

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

\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8} \\ &=\frac{3}{8}+\frac{6}{8}+\frac{3}{8} \\ &=\frac{12}{8}=\frac{3}{2} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \end{aligned}

\begin{aligned} &=0^{2} \times \frac{1}{8}+1^{2} \times \frac{3}{8}+2^{2} \times \frac{3}{8}+3^{2} \times \frac{1}{8} \\ &=0+\frac{3}{8}+\frac{12}{8}+\frac{9}{8} \\ &=\frac{24}{8}=3 \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=3-\left(\frac{3}{2}\right)^{2}=3-\frac{9}{4} \\ &=\frac{3}{4} \\ \end{aligned}

\begin{aligned} &\text { Standard deviation }=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \end{aligned}

 

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