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Explain solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise multiple choice questions question 10 maths textbook solution

Answers (1)

Answer:

$E\left ( X \right )=4$

Hint:

To solve this we use $\sum_{k=1}^{3}X.P\left ( X \right )$

Given:

$X$	$30$	$10$	$-10$
$P\left ( X \right )$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{1}{2}$

Solution:

$E\left ( X \right )=\sum_{k=1}^{3}X.P\left ( X \right )$

$=30\times \frac{1}{5}+10\times \frac{3}{10}+\left ( -10 \right )\frac{1}{2}$

$=6+3-5$

$=9-5$

$=4$

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