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(D) Let $X$ be the sum of the numbers on the two dice. The possible values for $X$ are $2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$. The probability distribut

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Let $X$ be the sum of the numbers on the two dice. The possible values for $X$ are $2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$. The probability distribution is as follows: $P(X=2) = \frac{1}{36}, P(X=3) = \frac{2}{36}, P(X=4) = \frac{3}{36}, P(X=5) = \frac{4}{36}, P(X=6) = \frac{5}{36}, P(X=7) = \frac{6}{36}, P(X=8) = \frac{5}{36}, P(X=9) = \frac{4}{36}, P(X=10) = \frac{3}{36}, P(X=11) = \frac{2}{36}, P(X=12) = \frac{1}{36}$. The mean $E(X)$ is given by $\sum X_i P(X_i)$: $E(X) = 2(\frac{1}{36}) + 3(\frac{2}{36}) + 4(\frac{3}{36}) + 5(\frac{4}{36}) + 6(\frac{5}{36}) + 7(\frac{6}{36}) + 8(\frac{5}{36}) + 9(\frac{4}{36}) + 10(\frac{3}{36}) + 11(\frac{2}{36}) + 12(\frac{1}{36})$ $E(X) = \frac{2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12}{36} = \frac{252}{36} = 7$.

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x)$	$a$	$a$	$a$	$b$	$b$	$0.3$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

If two dice are thrown and if $X$ denotes the sum of the numbers that show up on the faces of the dice, then the mean of the random variable $X$ is

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