Deviation of dice rolls

Let $X_1$ be the result of the first die roll, let $X_2$ be the result of the second die roll, and let $X_3$ be the result of the third die roll. Then the variance of three dice rolls will be $\text{Var}[X_1+X_2+X_3]$ . Note that $X_1$ , $X_2$ , and $X_3$ are independent. Now use a property of variance for independent variables:

$\text{Var}[X_1+X_2+X_3]=\text{Var}[X_1]+\text{Var}[X_2]+\text{Var}[X_3]$

The expected value of a die roll is:

$\text{E}[X]=\frac{1}{6}(1)+\frac{1}{6}(2)+\frac{1}{6}(3)+\frac{1}{6}(4)+\frac{1}{6}(5)+\frac{1}{6}(6)=\frac{7}{2}$ .

The variance of a die roll is:

$\text{Var}[X]=\frac{1}{6}\left(1-\frac{7}{2}\right)^2+\frac{1}{6}\left(2-\frac{7}{2}\right)^2+\frac{1}{6}\left(3-\frac{7}{2}\right)^2+\frac{1}{6}\left(4-\frac{7}{2}\right)^2+\frac{1}{6}\left(5-\frac{7}{2}\right)^2+\frac{1}{6}\left(6-\frac{7}{2}\right)^2=\frac{35}{12}$

Thus, the variance of the dice roll sum is:

$\text{Var}[X_1+X_2+X_3]=3\times \frac{35}{12}=\frac{35}{4}$

The standard deviation is the square root of the variance, and so:

$\sigma(X_1+X_2+X_3)=\frac{\sqrt{35}}{2}\approx \boxed{2.958}$