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(D) When two dice are rolled, the total number of outcomes is $6 	imes 6 = 36$. Let $X$ be the absolute difference of the numbers on the two dice. Th

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$\frac{35}{18}$

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(D) When two dice are rolled, the total number of outcomes is $6 	imes 6 = 36$. Let $X$ be the absolute difference of the numbers on the two dice. The possible values of $X$ are $0, 1, 2, 3, 4, 5$. The probability distribution is as follows:| $X$ | $P(X)$ | $P_i X_i$ ||---|---|---|| $0$ | $6/36$ | $0$ || $1$ | $10/36$ | $10/36$ || $2$ | $8/36$ | $16/36$ || $3$ | $6/36$ | $18/36$ || $4$ | $4/36$ | $16/36$ || $5$ | $2/36$ | $10/36$ |The mean $\mu$ is given by $\sum P_i X_i$:$\mu = 0 + \frac{10}{36} + \frac{16}{36} + \frac{18}{36} + \frac{16}{36} + \frac{10}{36}$$\mu = \frac{10 + 16 + 18 + 16 + 10}{36} = \frac{70}{36} = \frac{35}{18}$.

Two dice are rolled. If a random variable $X$ is defined as the absolute difference of the two numbers that appear on them, then the mean of $X$ is

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