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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 7 maths textbook solution

Answers (1)

Answer:

$-1.8$

Hint:

To solve this equation we use $\sum X.P\left ( X \right )$ formula

Given:

$\begin{array}{cccccc} X & -4 & -3 & -2 & -1 & 0 \\ P(X) & 0.1 & 0.2 & 0.3 & 0.2 & 0.2 \end{array}$

Solution:

$E\left ( X \right )=\sum X.P\left ( X \right )$

$X$	$P\left ( X \right )$	$X.P\left ( X \right )$
$-4$	$0.1$	$-0.4$
$-3$	$0.2$	$-0.6$
$-2$	$0.3$	$-0.6$
$-1$	$0.2$	$-0.2$
$0$	$0.2$	$0$

$E\left ( X \right )=\sum X.P\left ( X \right )$

$=-0.4-0.6-0.6-0.2$

$=-1.8$

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