Answer:
Option B
Explanation:
Ley x denote the sum of the numbers obtained when two fair dice are rolled
So, X may have values 2,3,4,5,6,7,8,9,10,11 and 12
$P(X=2)=P(1,1)=\frac{1}{36}$
$P(X=3)=P{(1,2),(2,1)}=\frac{2}{36}$
$P(X=4)=\frac{3}{36}$, $P(X=5)=\frac{4}{36}$
$P(X=6)=\frac{5}{36}$, $P(X=7)=\frac{6}{36}$
$P(X=8)=\frac{5}{36}$,
$P(X=9)=\frac{4}{36}$,$P(X=10)=\frac{3}{36}$,$P(X=11)=\frac{2}{36}$,
$P(X=12)=\frac{1}{36}$,
$\therefore$ Probability distribution table is given below
$ Mean \overline{X}= \sum XP (X)$
=$\frac{\left[2\times1+3\times2+4\times3+5\times4+6\times5+7\times6+8\times5+9\times4+10\times3+11\times2+12\times1\right]}{36}$
$=\frac{252}{36}=7$
Variance = $\sum X^{2} P(X)-\overline{X}^{2}$
$\frac{\left[2^{2}\times1+3^{2}\times2+4^{2}\times3+5^{2}\times4+6^{2}\times5+7^{2}\times6+
8^{2}\times5+9^{2}\times4+10^{2}\times3+11^{2}\times2+12^{2}\times1\right]}{36}$-$7^{2}$
=$\frac{1974}{36}-49$
=$\frac{1974-1764}{36}$
$\frac{210}{36}=\frac{35}{6}$
$\therefore$ Variance =$\frac{35}{6}$
and SD=$ \sqrt{\frac{35}{6}}$
