If X is a random variable with the following | vedclass

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(A) The mean $E(X)$ is calculated as $\sum x_i P(x_i) = (-3 	imes \frac{1}{6}) + (6 	imes \frac{1}{2}) + (9 	imes \frac{1}{3}) = -0.5 + 3 + 3 = 5.5

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$\frac{65}{4}$

Vedclass · Answer

(A) The mean $E(X)$ is calculated as $\sum x_i P(x_i) = (-3 	imes \frac{1}{6}) + (6 	imes \frac{1}{2}) + (9 	imes \frac{1}{3}) = -0.5 + 3 + 3 = 5.5 = \frac{11}{2}$.The expectation of the square $E(X^2)$ is $\sum x_i^2 P(x_i) = ((-3)^2 	imes \frac{1}{6}) + (6^2 	imes \frac{1}{2}) + (9^2 	imes \frac{1}{3}) = (9 	imes \frac{1}{6}) + (36 	imes \frac{1}{2}) + (81 	imes \frac{1}{3}) = 1.5 + 18 + 27 = 46.5 = \frac{93}{2}$.The variance $Var(X)$ is given by $E(X^2) - [E(X)]^2 = \frac{93}{2} - (\frac{11}{2})^2 = \frac{93}{2} - \frac{121}{4} = \frac{186 - 121}{4} = \frac{65}{4}$.

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X = x)$	$k$	$\frac{k}{3}$	$\frac{k}{4}$	$\frac{k}{2}$	$\frac{k}{2}$

If $X$ is a random variable with the following probability distribution:
$X=x$ $-3$ $6$ $9$
$P(X=x)$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{1}{3}$

Then the variance of $X$ is:

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If $X$ is a random variable with the following probability distribution:$X=x$$-3$$6$$9$$P(X=x)$$\frac{1}{6}$$\frac{1}{2}$$\frac{1}{3}$Then the variance of $X$ is: