Question

The variance of the number obtained on a throw of an unbiased die is:

• A. $\frac{35}{12}$
• B. $\frac{12}{35}$
• C. $\frac{7}{35}$
• D. $\frac{9}{35}$

Answer 1

The correct option is A $\frac{35}{12}$

The sample space of the experiment is $S=\{1,2,3,4,5,6\}$
Let $X$ denote the number obtained on the throw. Then $X$ is a random variable which can take values $1,2,3,4,5$ or $6.$
Also, $P\left(1\right)=P\left(2\right)=P\left(3\right)=P\left(4\right)=P\left(5\right)=P\left(6\right)=\frac{1}{6}$
Therefore, the probability distribution of $X$ is
$\begin{array}{ccccccc}X& \text{1}& \text{2}& \text{3}& \text{4}& \text{5}& \text{6}\\ P\left(X\right)& \frac{1}{6}& \frac{1}{6}& \frac{1}{6}& \frac{1}{6}& \frac{1}{6}& \frac{1}{6}\end{array}$

Now, $E\left(X\right)=\sum _{i=1}^n{x}_{i}p\left(x_i\right)=1\times \frac{1}{6}+2\times \frac{1}{6}+3\times \frac{1}{6}+4\times \frac{1}{6}+5\times \frac{1}{6}+6\times \frac{1}{6}=\frac{21}{6}$
$E\left(X^2\right)=1^2\times \frac{1}{6}+2^2\times \frac{1}{6}+3^2\times \frac{1}{6}+4^2\times \frac{1}{6}+5^2\times \frac{1}{6}+6^2\times \frac{1}{6}=\frac{91}{6}$
$Var\left(X\right)=E\left(X^2\right)-\left(E\right(X){)}^2=\frac{91}{6}-{\left(\frac{21}{6}\right)}^2=\frac{91}{6}-\frac{441}{36}=\frac{35}{12}$