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

Let 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, 11or 12.(as 1 can't be the sum of two numbers on fair dice)

$P(X=2)=P\left[1,1\right]=\frac{1}{36},P(X=3)=P\left[(1,2),(2,1)\right]=\frac{2}{36}$

$P(X=4)=P\left[(1,3),(2,2),(3,1)\right]=\frac{3}{36}$

$P(X=5)=P\left[(1,4),(2,3),(3,2),(4,1)\right]=\frac{4}{36}$

$P(X=6)=P\left[(1,5),(2,4),(3,3),(4,2),(5,1)\right]=\frac{5}{36}$

$P(X=7)=P\left[(1,6),(2,5),(3,4),(4,3),(5,2)(6,1)\right]=\frac{6}{36}$

$P(X=8)=P\left[(2,6),(3,5),(4,4),(5,3),(6,2)\right]=\frac{5}{36}$

$P(X=9)=P\left[(3,6),(4,5),(5,4),(6,3)\right]=\frac{4}{36}$

$P(X=10)=P\left[(4,6),(5,5),(6,4)\right]=\frac{3}{36}$

$P(X=11)=P\left[(5,6),(6,5)\right]=\frac{2}{36},P(X=12)=P\left[6,6\right]=\frac{1}{36}$

$\begin{array}{cccccccccccc}X& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ P\left(X\right)& \frac{1}{36}& \frac{2}{36}& \frac{3}{36}& \frac{4}{36}& \frac{5}{36}& \frac{6}{36}& \frac{5}{36}& \frac{4}{36}& \frac{3}{36}& \frac{2}{36}& \frac{1}{36}\end{array}$

Mean X $=\sum XP\left(X\right)=\frac{\left[\begin{array}{c}2\times 1+3\times 2+4\times 3+5\times 4+6\times 5+7\times 6\\ +8\times 5+9\times 4+10\times 3+11\times 2+12\times 1\end{array}\right]}{36}=\frac{252}{36}=7$

Variance X $\sum X^{2}P\left(X\right)-(Mean{)}^2$

$\frac{\left[\begin{array}{c}{2}^2\times 1+3^2\times 2+4^2\times 3+5^2\times 4+6^2\times 5+7^2\times 6\\ +8^2\times 5+9^2\times 4+{10}^2\times 3+{11}^2\times 2+{12}^2\times 1\end{array}\right]}{36}-7^2=\frac{1974}{36}-49=\frac{1947-1764}{36}=\frac{210}{36}=\frac{35}{6}$
Hence, $SD=\sqrt{Variance}=\sqrt{\frac{35}{6}}=2.415$