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

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

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

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

Then:

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