Question

Find the varaiable of the number obtained on a throw of an unbiased die.

Answer 1

Given :

A unbiased die is thrown, then sample space is

$S=\{1,2,3,4,5,6\}$

Let $X$ be the random variable denoted a number on the thrown. Then

$X=1\text{or}2\text{or}3\text{or}4\text{or}5\text{or}6$

Probability distribution of $X$ are

$P(X=1)=\frac{1}{6}$, $P(X=2)=\frac{1}{6}$

$P(X=3)=\frac{1}{6}$, $P(X=4)=\frac{1}{6}$

$P(X=5)=\frac{1}{6}$, $P(X=6)=\frac{1}{6}$

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

We know that ,

Mean or expectation of $P\left(x_i\right)$ is

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

$E\left(X\right)=1\times \frac{1}{6}+2\times \frac{1}{6}+3\times \frac{1}{6}+4\times \frac{1}{6}+5\times \frac{1}{6}+6\times \frac{1}{6}$

$E\left(X\right)=\frac{21}{6}$
Now,

$E\left(X^2\right)=1^2\times \frac{1}{6}+2^2\times \frac{1}{6}+3^2\times \frac{1}{6}+4^2\times \frac{1}{6}+5^2\times \frac{1}{6}+6^2\times \frac{1}{6}$

$E\left(X^2\right)=\frac{91}{6}$

Variance of $X$ is

$Var\left(X\right)=E\left(X^2\right)-\left(E\right(X){)}^2$

$Var\left(X\right)=\frac{91}{6}-{\left(\frac{21}{6}\right)}^2=\frac{91}{6}-\frac{441}{36}$

$\therefore Var\left(X\right)=\frac{35}{12}$