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

: Sample space consists of 36 elements,

$X$ denote the random variable which represents the sum of the numbers on the two dice

$\therefore$ $X$ can assume the values $2,3,4, \ldots \ldots ,$ or 12

Now, $P(X = 2) = P((1,1)) = \cfrac{1}{{36}}$

$P(X = 3) = P((1,2),(2,1)) = \cfrac{2}{{36}}$

$P(X = 4) = P((1,3),(2,2),(3,1)) = \cfrac{3}{{36}}$

$P(X = 5) = P((1,4),(2,3),(3,2),(4,1)) = \cfrac{4}{{36}}$

$P(X = 6) = P((1,5),(2,4),(3,3),(4,2),(5,1)) = \cfrac{5}{{36}}$

$P(X = 7) = P((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)) = \cfrac{6}{{36}}$

$P(X = 8) = P((2,6),(3,5),(4,4),(5,3),(6,2)) = \cfrac{5}{{36}}$

$P(X = 9) = P((3,6),(4,5),(5,4),(6,3)) = \cfrac{4}{{36}}$

$P(X = 10) = P((4,6),(5,5),(6,4)) = \cfrac{3}{{36}}$

$P(X = 11) = P((5,6),(6,5)) = \cfrac{2}{{36}}$

$P(X = 12) = P((6,6)) = \cfrac{1}{{36}}$

Hence, the probability distribution is

$\therefore$ Mean $= \mu = E(X) = \sum X P(X)$
$= 2 \times \cfrac{1}{{36}} + 3 \times \cfrac{2}{{36}} + 4 \times \cfrac{3}{{36}} + 5 \times \cfrac{4}{{36}} + 6$

$\times \cfrac{5}{{36}} + 7 \times \cfrac{6}{{36}} + 8 \times \cfrac{5}{{36}} + 9 \times \cfrac{4}{{36}} + 10 \times \cfrac{3}{{36}} + 11 \times \cfrac{2}{{36}} + 12 \times \cfrac{1}{{36}}$

$= \cfrac{1}{{36}}(2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)$

$= \cfrac{1}{{36}}(252) = 7$

Hence, the required mean = 7.

Variance $= \sigma _x^2 = E\left( {{X^2}} \right) - {[E(X)]^2}$

$= {(2)^2} \times \left( {\cfrac{1}{{36}}} \right) + {(3)^2} \times \left( {\cfrac{2}{{36}}} \right) + {(4)^2} \times \left( {\cfrac{3}{{36}}} \right) + {(5)^2} \times \left( {\cfrac{4}{{36}}} \right)$

$+ {(6)^2} \times \left( {\cfrac{5}{{36}}} \right) + {(7)^2} \times \left( {\cfrac{6}{{36}}} \right) + {(8)^2} \times \left( {\cfrac{5}{{36}}} \right) + {(9)^2} \times \left( {\cfrac{4}{{36}}} \right)$

$+ {(10)^2} \times \left( {\cfrac{3}{{36}}} \right) + {(11)^2} \times \left( {\cfrac{2}{{36}}} \right) + {(12)^2} \times \left( {\cfrac{1}{{36}}} \right) - {(7)^2}$

$= \cfrac{1}{{36}}[1974] - 49 = \cfrac{{329}}{6} - 49 = \cfrac{{329 - 294}}{6} = \cfrac{{35}}{6} = 5.833$

Standard deviation $= \sqrt {\sigma _x^2} = {\sigma _x} = 2.415$