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Question

(A) When two dice are rolled, the sum $X$ can take values from $2$ to $12$. The probability distribution is as follows:$X: 2, 3, 4, 5, 6, 7, 8, 9, 10,

Vedclass · Accepted Answer

$\frac{15}{2}$

Vedclass · Answer

(A) When two dice are rolled, the sum $X$ can take values from $2$ to $12$. The probability distribution is as follows:$X: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$$P(X): \frac{1}{36}, \frac{2}{36}, \frac{3}{36}, \frac{4}{36}, \frac{5}{36}, \frac{6}{36}, \frac{5}{36}, \frac{4}{36}, \frac{3}{36}, \frac{2}{36}, \frac{1}{36}$The mean $\mu = E(X) = \sum X_i P(X_i) = \frac{2(1)+3(2)+4(3)+5(4)+6(5)+7(6)+8(5)+9(4)+10(3)+11(2)+12(1)}{36} = \frac{252}{36} = 7$.Now, calculate the probabilities:$P(X $P(X > 9) = P(X=10) + P(X=11) + P(X=12) = \frac{3+2+1}{36} = \frac{6}{36}$.$P(X = 7) = \frac{6}{36}$.Substituting these values into the expression $\mu + P(X  9) + P(X = 7)$:$= 7 + \frac{6}{36} + \frac{6}{36} + \frac{6}{36} = 7 + \frac{18}{36} = 7 + \frac{1}{2} = \frac{15}{2}$.

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$1/6$	$k$	$1/4$	$k$	$1/6$

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

Two dice are rolled. If a random variable $X$ denotes the sum of the numbers on them and $\mu$ denotes the mean of $X$, then $\mu+P(X < 5)+P(X>9)+P(X=7)=$

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