The variance of the random variable X having | vedclass

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(D) The variance of a random variable $X$ is given by the formula:$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$First,we calculate the expected value $E(

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$\frac{5}{3}$

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(D) The variance of a random variable $X$ is given by the formula:$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$First,we calculate the expected value $E(X) = \sum P_i x_i$:$E(X) = (-2) 	imes \frac{1}{6} + (-1) 	imes \frac{1}{6} + 0 	imes \frac{1}{3} + 1 	imes \frac{1}{6} + 2 	imes \frac{1}{6}$$E(X) = -\frac{2}{6} - \frac{1}{6} + 0 + \frac{1}{6} + \frac{2}{6} = 0$Next,we calculate $E(X^2) = \sum P_i x_i^2$:$E(X^2) = (-2)^2 	imes \frac{1}{6} + (-1)^2 	imes \frac{1}{6} + 0^2 	imes \frac{1}{3} + 1^2 	imes \frac{1}{6} + 2^2 	imes \frac{1}{6}$$E(X^2) = 4 	imes \frac{1}{6} + 1 	imes \frac{1}{6} + 0 + 1 	imes \frac{1}{6} + 4 	imes \frac{1}{6}$$E(X^2) = \frac{4+1+0+1+4}{6} = \frac{10}{6} = \frac{5}{3}$Finally,the variance is:$\operatorname{Var}(X) = \frac{5}{3} - (0)^2 = \frac{5}{3}$

$X = k$	$-2$	$-1$	$0$	$1$	$2$
$P(X = k)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

$X$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

The variance of the random variable $X$ having the following distribution is:
$X = k$ $-2$ $-1$ $0$ $1$ $2$
$P(X = k)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$

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The variance of the random variable $X$ having the following distribution is:$X = k$$-2$$-1$$0$$1$$2$$P(X = k)$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$