Question

Let $X$ denotes the sum of the numbers obtained when two fair dice are rolled. Then the standard deviation of $X$ is:

• A. $1.414$
• B. $2.415$
• C. $1.715$
• D. $1.2$

Answer 1

The correct option is B $2.415$

Given two fair dice are rolled
When two fair dice are rolled then number of observations will be $6\times 6=36.$
$X$ denotes the sum of the numbers obtained when two fair dice are rolled. Hence, $X$ can take any value of $2,3,4,5,6,7,8,9,10,11$ or $12.$
For $X=2,$ the possible observation is $(1,1).$
$P\left(X\right)=\frac{1}{36}$
For $X=3,$ the possible observations are (1, 2) and $(2,1)$
$P\left(X\right)=\frac{2}{36}=\frac{1}{18}$
For $X=4,$ the possible observations are $(1,3),(2,2)$ and $(3,1).$
$P\left(X\right)=\frac{3}{36}=\frac{1}{12}$
For $X=5,$ the possible observations are $(1,4),(2,3),(3,2)$ and $(4,1).$
$P\left(X\right)=\frac{4}{36}=\frac{1}{9}$
For $X=6,$ the possible observations are $(1,5),(2,4),(3,3),(4,2)$ and $(5,1).$
$P\left(X\right)=\frac{5}{36}$
For $X=7,$ the possible observations are $(1,6),(2,5),(3,4),(4,3)(5,2)$ and $(6,1).$
$P\left(X\right)=\frac{6}{36}=\frac{1}{6}$
For $X=8,$ the possible observations are $(2,6),(3,5),(4,4)(5,3)$ and $(6,2).$
$P\left(X\right)=\frac{5}{36}$
For $X=9,$ the possible observations are $(3,6),(4,5)(5,4)$ and $(6,3).$
$P\left(X\right)=\frac{4}{36}=\frac{1}{9}$
For $X=10,$ the possible observations are $(4,6),(5,5)$ and $(6,4).$
$P\left(X\right)=\frac{3}{36}=\frac{1}{12}$
For X=11, the possible observations are $(5,6)$ and $(6,5).$
$P\left(X\right)=\frac{2}{36}=\frac{1}{18}$
For $X=12,$ the possible observations are $(6,6).$
$P\left(X\right)=\frac{1}{36}$
Hence, the required probability distribution is,
$\begin{array}{cccccccccccc}\text{X}& \text{2}& \text{3}& \text{4}& \text{5}& \text{6}& \text{7}& \text{8}& \text{9}& \text{10}& \text{11}& \text{12}\\ \text{P(X)}& \text{1/36}& \text{1/18}& \text{1/12}& \text{1/9}& \text{5/36}& \text{1/6}& \text{5/36}& \text{1/9}& \text{1/12}& \text{1/18}& \text{1/36}\end{array}$
Therefore $E\left(X\right)$ is:
$E\left(X\right)=\sum _{i=1}^n{x}_{i}p\left(x_i\right)$
Now by substituting the values we get
$=2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{9}+6\times \frac{5}{36}+7\times \frac{1}{6}+8\times \frac{5}{36}+9\times \frac{1}{9}+10\times \frac{1}{12}+11\times \frac{1}{18}+12\times \frac{1}{36}$
$=\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}$
$=\frac{1+3+6+10+15+21+20+18+15+11+6}{18}=\frac{126}{18}$
$E\left(X\right)=7$
And $E\left(x^2\right)$ is
$E\left(X^2\right)=\sum _{i=1}^n{x}_i^{2}p\left(x_i\right)$
$=4\times \frac{1}{36}+9\times \frac{1}{18}+16\times \frac{1}{12}+25\times \frac{1}{9}+36\times \frac{5}{36}+49\times \frac{1}{6}+64\times \frac{5}{36}+81\times \frac{1}{9}+100\times \frac{1}{12}+121\times \frac{1}{18}+144\times \frac{1}{36}$
$=\frac{1}{9}+\frac{1}{2}+\frac{4}{3}+\frac{25}{9}+5+\frac{49}{6}+\frac{80}{9}+9+\frac{25}{3}+\frac{121}{18}+4$
$\frac{2+9+24+50+90+147+160+162+150+121+72}{18}=\frac{987}{18}$
$\Rightarrow E\left(X^2\right)=54.833$
Then Variance, $Var\left(X\right)=E\left(X^2\right)-\left(E\right(X){)}^2=54.833-(7^2)54.833-49=5.833$
And standard deviaation $=\sqrt{Var\left(X\right)}=\sqrt{5.833}=2.415$