For the following probability distribution,th | vedclass

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(B) The mean of the random variable $X$ is given by $E(X) = \sum x_i P(x_i)$.$E(X) = (2 	imes 0.2) + (3 	imes 0.5) + (4 	imes 0.3) = 0.4 + 1.5 + 1.

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$0.7$

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(B) The mean of the random variable $X$ is given by $E(X) = \sum x_i P(x_i)$.$E(X) = (2 	imes 0.2) + (3 	imes 0.5) + (4 	imes 0.3) = 0.4 + 1.5 + 1.2 = 3.1$.The expected value of $X^2$ is $E(X^2) = \sum x_i^2 P(x_i)$.$E(X^2) = (2^2 	imes 0.2) + (3^2 	imes 0.5) + (4^2 	imes 0.3) = (4 	imes 0.2) + (9 	imes 0.5) + (16 	imes 0.3) = 0.8 + 4.5 + 4.8 = 10.1$.The variance is $Var(X) = E(X^2) - [E(X)]^2$.$Var(X) = 10.1 - (3.1)^2 = 10.1 - 9.61 = 0.49$.The standard deviation is $\sigma = \sqrt{Var(X)} = \sqrt{0.49} = 0.7$.

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.2$
$\omega_{3}$	$0.3$
$\omega_{4}$	$0.4$
$\omega_{5}$	$0.5$
$\omega_{6}$	$0.6$
$\omega_{7}$	$0.7$

For the following probability distribution,the standard deviation of the random variable $X$ is:

$X$ $2$ $3$ $4$

$P(X=x)$ $0.2$ $0.5$ $0.3$

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For the following probability distribution,the standard deviation of the random variable $X$ is: $X$ $2$ $3$ $4$ $P(X=x)$ $0.2$ $0.5$ $0.3$