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

Two fair dice are rolled then
Sample space

$S=\left\{\begin{array}{cccccc}(1,1),& (1,2),& (1,3),& (1,4),& (1,5)& (1,6),\\ (2,1),& (2,2),& (2,3),& (2,4),& (2,5)& (2,6),\\ (3,1),& (3,2),& (3,3),& (3,4),& (3,5),& (3,6),\\ (4,1),& (4,2),& (4,3),& (4,4),& (4,5),& (4,6),\\ (5,1),& (5,2),& (5,3),& (5,4),& (5,5),& (5,6),\\ (6,1),& (6,2),& (6,3),& (6,4),& (6,5)& (6,6),\end{array}\right\}$

Total number of possible outcomes$=36$

$X:$ Denote the sum of the numbers obtained when two fair dice are rolled.
Hence, $X$ can take any value of

$2,3,4,5,6,7,8,9,10,11\text{or}12.$

For$X=2,$ the possible outcome is $(1,1).$

$\Rightarrow P\left(X\right)=\frac{1}{36}$

For $X=3,$ the possible outcomes are $(1,2)$
and $(2,1)$

$\Rightarrow P\left(X\right)=\frac{2}{36}=\frac{1}{18}$

For$X=4,$ the possible outcomes are

$(1,3),(2,2)\text{and}(3,1)$

$\Rightarrow P\left(X\right)=\frac{3}{36}=\frac{1}{12}$

For $X=5,$ the possible outcomes are

$(1,4),(4,1),(2,3)\text{and}(3,2)$

$\Rightarrow P\left(X\right)=\frac{4}{36}=\frac{1}{9}$

For $X=6,$ the possible outcomes are

$(1,5),(5,1),(2,4),(4,2)\text{and}(3,3).$

$\Rightarrow P\left(X\right)=\frac{5}{36}$

For $X=7,$ the possible outcomes are

$(1,6),(6,1),(2,5),(5,2),(3,4)\text{and}(4,3).$

$\Rightarrow P\left(X\right)=\frac{6}{36}=\frac{1}{6}$

For $X=8,$ The possible outcomes are
$(2,6),(6,2),(3,5),(5,3)\text{and}(4,4).$

$\Rightarrow P\left(X\right)=\frac{5}{36}$
For $X=9,$ the possible outcomes are

$(5,4),(4,5),(3,6)\text{and}(6,3)$

$\Rightarrow P\left(X\right)=\frac{4}{36}=\frac{1}{9}$

For$X=10,$ the possible outcomes are

$(5,5),(4,6)and(6,4).$

$\Rightarrow P\left(X\right)=\frac{3}{36}=\frac{1}{12}$

For $X=11,$ the possible outcomes are $(6,5)$
and $(5,6)$

$\Rightarrow P\left(X\right)=\frac{1}{18}$

For $X=12,$ the possible outcome is $(6,6).$

$\Rightarrow P\left(X\right)=\frac{1}{36}$

Hence,the required probability distribution is,

$\begin{array}{ccccccc}X& 2& 3& 4& 5& 6& 7\\ P\left(X\right)& \frac{1}{36}& \frac{1}{18}& \frac{1}{12}& \frac{1}{9}& \frac{5}{36}& \frac{1}{6}\end{array}$

$\begin{array}{cccccc}X& 8& 9& 10& 11& 12\\ P\left(X\right)& \frac{5}{36}& \frac{1}{9}& \frac{1}{12}& \frac{1}{18}& \frac{1}{36}\end{array}$

The Mean of $X$ is given by

$\mu =E\left(X\right)={\sum }_{i=1}^n{x}_i{p}_i$

Putting the value of $x_i{p}_i$

$\Rightarrow E\left(X\right)=2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{9}$

$+6\times \frac{5}{36}+7\times \frac{1}{6}+8\times \frac{5}{36}+9\times \frac{1}{9}$

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

$\Rightarrow E\left(X\right)=\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1$

$+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}$

$\Rightarrow E\left(X\right)=\frac{1+3+6+10+15+21+20+18+15+11+6}{18}$

$\Rightarrow E\left(X\right)=\frac{126}{18}$

$\therefore E\left(X\right)=7$

We know that variance of $X$ is

𝑉𝑎𝑟$\left(X\right)=E\left(X^2\right)-\left(E\right(X){)}^2$ ... $\left(1\right)$
Here,$E\left(X\right)=7$
Finding $E\left(X^2\right)$
$E\left(X^2\right)={\sum }_{i=1}^n{x}_i^2.p\left(x_i\right)$

$\Rightarrow E\left(X^2\right)=4\times \frac{1}{36}+9\times \frac{1}{18}+16\times \frac{1}{12}+25\times \frac{1}{9}$
$+36\times \frac{5}{36}+49\times \frac{1}{6}+64\times \frac{5}{36}+81\times \frac{1}{9}$
$+100\times \frac{1}{12}+121\times \frac{1}{18}+144\times \frac{1}{36}$

$\Rightarrow E\left(X^2\right)=\frac{1}{9}+\frac{1}{2}+\frac{4}{3}+\frac{25}{9}+5$
$+\frac{49}{6}+\frac{80}{9}+9+\frac{25}{3}+\frac{121}{18}+4$

$\Rightarrow E\left(X^2\right)=\frac{2+9+24+50+90+147+160+162+150+121+72}{18}$

$\Rightarrow E\left(X^2\right)=\frac{987}{18}$

$\Rightarrow E\left(X^2\right)=54.833$
Now, putting the value of $E\left(x^2\right)\&E\left(x\right)$ in $\left(1\right)$

$\Rightarrow Var\left(X\right)=54.833-(7{)}^2$

$\Rightarrow Var\left(X\right)=54.833-49$

$\therefore Var\left(X\right)=5.833$

$\because$ Standard deviation $=\sqrt{Var\left(X\right)}$

Put value of $Var\left(X\right)$

Standard deviation$=\sqrt{5.833}$
Standard deviation$=2.415$