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  • Explain Solution R.D. Sharma Class 12 Chapter 31Mean and Variance of a Random Variable Exercise 31.2 Question 22 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 31Mean and Variance of a Random Variable  Exercise 31.2 Question 22 maths Textbook Solution.

Answers (1)

Answer:

(i) c=\frac{1}{5}  (ii) Mean=3.2 (iii) Variance=2.9

Hint:

P\left ( X \right )=1

Given:

\begin{array}{|c|c|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(X) & 4 c^{2} & 3 c^{2} & 2 c^{2} & c^{2} & c & 2 c \\ \hline \end{array}

Solution:

(i) We know that

\begin{aligned} &4 c^{2}+3 c^{2}+2 c^{2}+c^{2}+c+2 c=1 \\ &10 c^{2}+3 c-1=0 \\ &10 c^{2}+5 c-2 c-1=0 \\ &5 c(2 c+1)-1(2 c+1)=0 \\ &(2 c+1)(5 c-1)=0 \\ &c=\frac{1}{5} \because c \geq 0 \end{aligned}

(ii) Mean E\left ( X \right )=\sum_{i-1}^{n}X_{i}P_{i}

\begin{aligned} &=3 c^{2}+4 c^{2}+3 c^{2}+4 c+10 c \\ &=10 c^{2}+14 c \\ &=10 \times \frac{1}{25}+14 \times \frac{1}{5} \\ &=\frac{10}{25}+\frac{14}{5} \\ &=\frac{16}{5}=3.2 \end{aligned}

\begin{aligned} &\text { (iii) } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=14-(3.32)^{2} \\ &=2.9 \end{aligned}

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