Let $\mathbf{x}$ 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 .
$\begin{array}{l}
\mathrm{P}(\mathrm{X}=2)=\mathrm{P}(1,1)=\frac{1}{36} \\
\mathrm{P}(\mathrm{X}=3)=\mathrm{P}\{(1,2),(2,1)\}=\frac{2}{36} \\
\mathrm{P}(\mathrm{X}=4)=\frac{3}{36}, \mathrm{P}(\mathrm{X}=5)=\frac{4}{36}: \\
\mathrm{P}(\mathrm{X}=6)=\frac{5}{36} ; \mathrm{P}(\mathrm{X}=7)=\frac{6}{36} ; \\
\mathrm{P}(\mathrm{X}=9)=\frac{4}{36} ; \mathrm{P}(\mathrm{X}=10)=\frac{3}{36} ; \mathrm{P}(\mathrm{X}=11) \\
=\frac{2}{36} ; \\
\mathrm{P}(\mathrm{X}=12)=\frac{1}{36}
\end{array}$
$\therefore \quad$ Probability distribution table is given below
\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c}
\hline \mathbf{X} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline \mathbf{P}(\mathbf{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} \\
\hline
\end{array}
Mean $\bar{X}=\sum X P(X)$
$=\frac{\left[\begin{array}{l}2 \times 1+3 \times 2+4 \times 3+5 \times 4+6 \times 5+7 \times 6+7 \\ 8 \times 5+9 \times 4+10 \times 3+11 \times 2+12 \times 1\end{array}\right]}{36}$
$=\frac{252}{36}=7$
Variance $=\sum X^{2} P(X)-\bar{X}^{2}$
$\begin{aligned}
&\left[\begin{array}{l}
2^{2} \times 1+3^{2} \times 2+4^{2} \times 3+5^{2} \times 4 \\
+6^{2} \times 5+7^{2} \times 6+8^{2} \times 5+9^{2} \times \\
4+10^{2} \times 3+11^{2} \times 2+12^{2} \times 1
\end{array}\right] \\
=& \frac{36}{1974}-7^{2} \\
=& \frac{36}{1974-1764} \\
=& \frac{36}{36}=\frac{35}{6} \\
\therefore & \text { Variance }=\frac{35}{6}
\end{aligned}$
And, $S D=\sqrt{\frac{35}{6}}$