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(A) When two fair dice are rolled,$6 	imes 6 = 36$ outcomes are possible.The probability distribution of the sum $X$ is:$X$$2$$3$$4$$5$$6$$7$$8$$9$$1

Vedclass · Accepted Answer

Variance $= 5.833$,Standard Deviation $= 2.415$

Vedclass · Answer

(A) When two fair dice are rolled,$6 	imes 6 = 36$ outcomes are possible.The probability distribution of the sum $X$ is:$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}$$\frac{1}{36}$$E(X) = \sum X_i P(X_i) = 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}) = 7$.$E(X^2) = \sum X_i^2 P(X_i) = 4(\frac{1}{36}) + 9(\frac{2}{36}) + 16(\frac{3}{36}) + 25(\frac{4}{36}) + 36(\frac{5}{36}) + 49(\frac{6}{36}) + 64(\frac{5}{36}) + 81(\frac{4}{36}) + 100(\frac{3}{36}) + 121(\frac{2}{36}) + 144(\frac{1}{36}) = \frac{1974}{36} = \frac{329}{6} \approx 54.833$.$	ext{Var}(X) = E(X^2) - [E(X)]^2 = 54.833 - 49 = 5.833$.$	ext{Standard Deviation} = \sqrt{	ext{Var}(X)} = \sqrt{5.833} \approx 2.415$.

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(a)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$k$	$3k$	$5k$	$7k$	$8k$	$k$

$X$	$30$	$10$	$-10$
$P(X)$	$\frac{1}{5}$	$A$	$B$

Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of $X$.

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