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  • Explain Solution R.D. Sharma Class 12 Chapter 31Mean and Variance of a Random Variable Exercise 31.2 Question 19 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 31Mean and Variance of a Random Variable  Exercise 31.2 Question 19 maths Textbook Solution.

Answers (1)

Answer:

\frac{14}{9}

Hint:

Mean E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}

Given:

Positive integer 2, 3,4,5,6 and 7

Solution:

\begin{aligned} &P(X=3)=\frac{1}{15} \\ &P(X=4)=\frac{2}{15} \\ &P(X=5)=\frac{3}{15} \\ &P(X=6)=\frac{4}{15} \\ &P(X=7)=\frac{5}{15} \end{aligned}

\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline X & 3 & 4 & 5 & 6 & 7 \\ \hline P(X) & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{4}{15} & \frac{5}{15} \\ \hline \end{array}\\ \end{aligned}

\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i}\\ \end{aligned}

\begin{aligned} &\begin{aligned} &=\frac{85}{15}=\frac{17}{3} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=3^{2} \times \frac{1}{15}+4^{2} \times \frac{2}{15}+5^{2} \times \frac{3}{15}+6^{2} \times \frac{4}{15}+7^{2} \times \frac{5}{15} \\ &=\frac{101}{3} \end{aligned}\\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \end{aligned}

=\frac{101}{3}-\left ( \frac{17}{3} \right )^{2}

=\frac{14}{9}

 

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