Let X denotes the sum of the numbers obtained when two fair dice are rolled.
Question:

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

Solution:

When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

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

$P(X=3)=P(1,2)+P(2,1)=\frac{2}{36}=\frac{1}{18}$

$P(X=4)=P(1,3)+P(2,2)+P(3,1)=\frac{3}{36}=\frac{1}{12}$

$P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=\frac{4}{36}=\frac{1}{9}$

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

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

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

$P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=\frac{4}{36}=\frac{1}{9}$

$P(X=10)=P(4,6)+P(5,5)+P(6,4)=\frac{3}{36}=\frac{1}{12}$

$P(X=11)=P(5,6)+P(6,5)=\frac{2}{36}=\frac{1}{18}$

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

Therefore, the required probability distribution is as follows.

Then, $\mathrm{E}(\mathrm{X})=\sum \mathrm{X}_{i} \cdot \mathrm{P}\left(\mathrm{X}_{i}\right)$

$=2 \times \frac{1}{36}+3 \times \frac{1}{18}+4 \times \frac{1}{12}+5 \times \frac{1}{9}+6 \times \frac{5}{36}+7 \times \frac{1}{6}$

$+8 \times \frac{5}{36}+9 \times \frac{1}{9}+10 \times \frac{1}{12}+11 \times \frac{1}{18}+12 \times \frac{1}{36}$

$=\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}$

$=7$

$\mathrm{E}\left(\mathrm{X}^{2}\right)=\sum \mathrm{X}_{i}^{2} \cdot \mathrm{P}\left(\mathrm{X}_{i}\right)$

$=4 \times \frac{1}{36}+9 \times \frac{1}{18}+16 \times \frac{1}{12}+25 \times \frac{1}{9}+36 \times \frac{5}{36}+49 \times \frac{1}{6}$

$+64 \times \frac{5}{36}+81 \times \frac{1}{9}+100 \times \frac{1}{12}+121 \times \frac{1}{18}+144 \times \frac{1}{36}$

$=\frac{1}{9}+\frac{1}{2}+\frac{4}{3}+\frac{25}{9}+5+\frac{49}{6}+\frac{80}{9}+9+\frac{25}{3}+\frac{121}{18}+4$

$=\frac{987}{18}=\frac{329}{6}=54.833$

Then, $\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}$

$=54.833-(7)^{2}$

$=54.833-49$

$=5.833$

$\therefore$ Standard deviation $=\sqrt{\operatorname{Var}(X)}$

$=\sqrt{5.833}$

$=2.415$