Chapter 8.BMO, Problem 1BMO

To determine

The probability distribution of the random variable $X$.

Answer to Problem 1BMO

Solution:

$x$	$-3$	$-2$	$0$	$1$	$2$	$3$
$P\left(X=x\right)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Explanation of Solution

Given:

The value taken on by a random variable $X$ and the frequency of their occurrence are shown in the following table.

$x$	$-3$	$-2$	$0$	$1$	$2$	$3$
Frequency of occurrence	$4$	$8$	$20$	$24$	$16$	$8$

Table $\left(1\right)$

Approach:

Let, $x_1,x_2,\mathrm{...}{x}_n$, are the values of random variable $X$, then the probability for $P\left(x_i\right)$ is,

$P\left(X=x_i\right)=\frac{x_i}{{\displaystyle \sum x_i}}$

Calculation:

Consider table $\left(1\right)$, the sum of all the frequencies of occurrence is,

$\begin{array}{rll}{\displaystyle \sum x_i}& =& 4+8+20+24+16+8\\ & =& 80\end{array}$

Therefore, the sum of all frequencies of occurrence is $80$.

Consider table $\left(1\right)$, the probability of occurrence of $-3$ is,

$\begin{array}{rll}P\left(X=-3\right)& =& \frac{4}{80}\\ & =& 0.05\end{array}$

The probability of occurrence of $-2$ is,

$\begin{array}{rll}P\left(X=-2\right)& =& \frac{8}{80}\\ & =& 0.1\end{array}$

The probability of occurrence of $0$ is,

$\begin{array}{rll}P\left(X=0\right)& =& \frac{20}{80}\\ & =& 0.25\end{array}$

The probability of occurrence of $1$ is,

$\begin{array}{rll}P\left(X=1\right)& =& \frac{24}{80}\\ & =& 0.3\end{array}$

The probability of occurrence of $2$ is,

$\begin{array}{rll}P\left(X=2\right)& =& \frac{16}{80}\\ & =& 0.2\end{array}$

The probability of occurrence of $3$ is,

$\begin{array}{rll}P\left(X=3\right)& =& \frac{8}{80}\\ & =& 0.1\end{array}$

Therefore, the probability distribution for random variable $X$ is,

$x$	$-3$	$-2$	$0$	$1$	$2$	$3$
$P\left(X=x\right)$	$0.05$	$0.1$	$0.25$	$0.3$	$0.2$	$0.1$

Table $\left(2\right)$

Conclusion:

Hence, the probability distribution for the random variable $X$ is shown in table $\left(2\right)$.