Chapter 9, Problem 1RE

(a)

To determine

To graph: The histogram of the given probability distribution.

Answer to Problem 1RE

Solution:

The histogram is obtained as:

Explanation of Solution

Given:

The data points for $P\left(x\right)$ and x are given as:

x	0	1	2	3
P(x)	0.22	0.54	0.16	0.08

Calculation:

Plot the given data to sketch the histogram.

The terms $P\left(x\right)$ and x are plotted on the y and x, respectively.

The histogram is obtained as:

Conclusion:

The histogram is obtained as:

(b)

To determine

To calculate: The expected value of the probability distribution.

Answer to Problem 1RE

Solution:

The expected value of the probability distribution is 1.10.

Explanation of Solution

Given:

The data points for $P\left(x\right)$ and x are given as:

x	0	1	2	3
P(x)	0.22	0.54	0.16	0.08

Formula used:

$E\left(x\right)={\displaystyle \sum x_i{p}_i}$

Here, $E$ is the expected value of the random variable $x$ and $p$ is the cross-ponding probability.

Calculation:

The random variable x is multiplied with its probability to obtain the product.

In the same way, for all values of x, the products obtained are summed together to obtain the expected value. It is defined as a balance point of the probability distribution.

Thus, the expected value is given as:

$\begin{array}{ccc}\hfill E\left(x\right)& =& x_1{p}_1+x_2{p}_2+x_3{p}_3+\cdots +x_n{p}_n.\hfill \\ \hfill \text{Since}n& =& 4,\text{therefore,}\text{the}\text{expected}\text{value}\text{is:}\hfill \\ \hfill E\left(x\right)& =& x_1{p}_1+x_2{p}_2+x_3{p}_3+x_4{p}_4\hfill \\ \hfill & =& \left(0\times 0.22\right)+\left(1\times 0.54\right)+\left(2\times 0.16\right)+\left(3\times 0.08\right)\hfill \\ \hfill & =& 0+0.54+0.32+0.24\hfill \\ \hfill & =& \mathrm{1.10.}\hfill \end{array}$