If m and {sigma ^2} are the mean and variance | vedclass

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(C) The mean $m$ is calculated as $E(X) = \sum x_i P(x_i)$.$m = (0 	imes \frac{1}{3}) + (1 	imes \frac{1}{2}) + (2 	imes 0) + (3 	imes \frac{1}{6}

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$m = {\sigma ^2} = 1$

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(C) The mean $m$ is calculated as $E(X) = \sum x_i P(x_i)$.$m = (0 	imes \frac{1}{3}) + (1 	imes \frac{1}{2}) + (2 	imes 0) + (3 	imes \frac{1}{6}) + (4 	imes 0) = 0 + \frac{1}{2} + 0 + \frac{1}{2} + 0 = 1$.The variance ${\sigma ^2}$ is calculated as $E(X^2) - [E(X)]^2$.First,calculate $E(X^2) = \sum x_i^2 P(x_i)$.$E(X^2) = (0^2 	imes \frac{1}{3}) + (1^2 	imes \frac{1}{2}) + (2^2 	imes 0) + (3^2 	imes \frac{1}{6}) + (4^2 	imes 0) = 0 + \frac{1}{2} + 0 + \frac{9}{6} + 0 = \frac{1}{2} + \frac{3}{2} = 2$.Now,${\sigma ^2} = E(X^2) - m^2 = 2 - (1)^2 = 2 - 1 = 1$.Thus,$m = 1$ and ${\sigma ^2} = 1$.

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

If $m$ and ${\sigma ^2}$ are the mean and variance of a random variable $X$,whose distribution is given by:
$X=x$ $0$ $1$ $2$ $3$ $4$
$P(X=x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$ $0$

,then:

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If $m$ and ${\sigma ^2}$ are the mean and variance of a random variable $X$,whose distribution is given by:$X=x$$0$$1$$2$$3$$4$$P(X=x)$$\frac{1}{3}$$\frac{1}{2}$$0$$\frac{1}{6}$$0$,then: