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Name: Explain solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise multiple choice questions question 5 maths textbook solution
Uploaded: Sat, 2021-09-11 23:48
Description: Explain solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise multiple choice questions question 5 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise multiple choice questions question 5 maths textbook solution

Answers (1)

Answer:

$\frac{3}{4}$

Hint:

To solve this we add all variables

Given:

$\begin{array}{ccccc} X & 0 & 1 & 2 & 3 \\ P(X) & k & 3 k & 3 k & k \end{array}$

Solution:

$\begin{aligned} As we know, \sum_{i=0}^{n} P\left(X_{i}\right)=1 \end{aligned}$

$\begin{aligned} &P(X=0)+P(X=1)+P(X=2)+P(X=3)=1 \\ &k+3 k+3 k+k=1 \\ &8 k=1 \\ &k=\frac{1}{8} \end{aligned}$

As we know variance, Var= $\sum X^{2}P\left ( X \right )-\left ( \sum XP\left ( X \right ) \right )^{2}$

$X$	$P\left ( X \right )$	$XP\left ( X \right )$	$X^{2}P\left ( X \right )$
$0$	$\frac{1}{8}$	$0$	$0$
$1$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
$2$	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
$3$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$