Question

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

Answer 1

When two fair dice are rolled, $6\times 6=36$ 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, $E\left(X\right)=\sum X_i\cdot P\left(X_i\right)$
$=2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{9}+6\times \frac{5}{36}+7\times \frac{1}{6}+8\times \frac{5}{36}+9\times \frac{1}{9}+10\times \frac{1}{12}+11\times \frac{1}{18}+12\times \frac{1}{36}$
$=\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}$
$=7$
$E\left(X^2\right)=\sum X_i^2\cdot P\left(X_i\right)$
$=4\times \frac{1}{36}+9\times \frac{1}{18}+16\times \frac{1}{12}+25\times \frac{1}{9}+36\times \frac{5}{36}+49\times \frac{1}{6}+64\times \frac{5}{36}+81\times \frac{1}{9}+100\times \frac{1}{12}+121\times \frac{1}{18}+144\times \frac{1}{36}$
$=\frac{1}{9}+\frac{1}{2}+\frac{4}{3}+\frac{25}{9}+5+\frac{49}{6}+\frac{80}{9}+9+\frac{25}{3}+\frac{121}{18}+4$
$=\frac{987}{18}=\frac{329}{6}=54.833$
Then, $Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2$
$=54.833-{\left(7\right)}^2$
$=54.833-49$
$=5.833$
$\therefore$ Standard deviation $=\sqrt{Var\left(X\right)}$
$=\sqrt{5.833}$
$=2.415$