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(C) First, we calculate $E(X)$ and $E(X^2)$ using the given distribution:$E(X) = \Sigma x P(X=x) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}

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$\frac{113}{12}$

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(C) First, we calculate $E(X)$ and $E(X^2)$ using the given distribution:$E(X) = \Sigma x P(X=x) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}{6}) + (2)(\frac{1}{3}) = -\frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3} = \frac{-2+0+1+4}{6} = \frac{3}{6} = \frac{1}{2}$$E(X^2) = \Sigma x^2 P(X=x) = (-1)^2(\frac{1}{3}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{6}) + (2)^2(\frac{1}{3}) = \frac{1}{3} + 0 + \frac{1}{6} + \frac{4}{3} = \frac{2+0+1+8}{6} = \frac{11}{6}$Now, we find $\operatorname{Var}(X) = E(X^2) - (E(X))^2 = \frac{11}{6} - (\frac{1}{2})^2 = \frac{11}{6} - \frac{1}{4} = \frac{22-3}{12} = \frac{19}{12}$Finally, we calculate $6 E(X^2) - \operatorname{Var}(X)$:$6 E(X^2) - \operatorname{Var}(X) = 6(\frac{11}{6}) - \frac{19}{12} = 11 - \frac{19}{12} = \frac{132-19}{12} = \frac{113}{12}$

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

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$0.10$	$0.12$	$0.20$	$0.08$	$0.07$	$0.05$

$X=x$	$1$	$2$	$3$
$P(X=x)$	$1/5$	$2/5$	$2/5$

The probability distribution of a discrete random variable $X$ is given below. If $E(X^2) = \Sigma x^2 P(X=x)$, then $6 E(X^2) - \operatorname{Var}(X) =$
$X=x$ $-1$ $0$ $1$ $2$
$P(X=x)$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$

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The probability distribution of a discrete random variable $X$ is given below. If $E(X^2) = \Sigma x^2 P(X=x)$, then $6 E(X^2) - \operatorname{Var}(X) =$$X=x$$-1$$0$$1$$2$$P(X=x)$$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$