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(D) Let $X_1, X_2, X_3$ be the random variables representing the numbers appearing on the three dice respectively.Each $X_i$ can take values from the

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$10.5$

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(D) Let $X_1, X_2, X_3$ be the random variables representing the numbers appearing on the three dice respectively.Each $X_i$ can take values from the set $\{1, 2, 3, 4, 5, 6\}$ with equal probability $P(X_i = k) = \frac{1}{6}$ for $k \in \{1, 2, 3, 4, 5, 6\}$.The mean (expected value) of a single die is $E[X_i] = \sum_{k=1}^{6} k \cdot P(X_i = k) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$.Let $S$ be the sum of the numbers appearing on the three dice,so $S = X_1 + X_2 + X_3$.By the linearity of expectation,the mean of the sum is $E[S] = E[X_1 + X_2 + X_3] = E[X_1] + E[X_2] + E[X_3]$.Substituting the mean of a single die,we get $E[S] = 3.5 + 3.5 + 3.5 = 10.5$.Therefore,the mean of the sum of the numbers is $10.5$.

$X$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

If three dice are thrown,then the mean of the sum of the numbers appearing on them is

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