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(C) The random variable $X$ takes values $x_i \in \{1, 2, 3, 4, 5, 6\}$ with probabilities $P_i = \frac{1}{6}$ for each.Mean $\mu = E(X) = \sum x_i P_

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$\frac{5}{6}$

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(C) The random variable $X$ takes values $x_i \in \{1, 2, 3, 4, 5, 6\}$ with probabilities $P_i = \frac{1}{6}$ for each.Mean $\mu = E(X) = \sum x_i P_i = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = \frac{7}{2}$.$E(X^2) = \sum x_i^2 P_i = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$.Variance $\sigma^2 = E(X^2) - (E(X))^2 = \frac{91}{6} - \left(\frac{7}{2}ight)^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$.Therefore,$\frac{	ext{Variance of } X}{	ext{Mean of } X} = \frac{35/12}{7/2} = \frac{35}{12} 	imes \frac{2}{7} = \frac{5}{6}$.

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

If a random variable $X$ denotes the number that appears on the upper face of a die when it is rolled,then $\frac{\text{Variance of } X}{\text{Mean of } X}$ is equal to

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