Chapter 5, Problem 1PS

Given the following probability distribution:

a. Compute the expected value for each distribution.

b. Compute the standard deviation for each distribution.

c. Compare the results of distribution A and B.

To determine

Find the expected value for distribution A and distribution B.

Answer to Problem 1PS

The expected value of distribution A is 1 and distribution B is 3.

Explanation of Solution

Calculation:

The table of probability distributions for distribution A and B is given.

The formula for calculating expected value is,

$E\left(X\right)={\displaystyle \sum _{i=1}^N{x}_{i}P\left(X=x_i\right)}$

Here, $x_i$ is the ith value of $X$ , $P\left(X=x_i\right)$ is the probability of occurrence of the ith value of $X$ , and $N$ is the number of values in the discrete variable $X$ .

The expected value of distribution A is calculated by substituting the values in the above formula. It is calculated as,

$\begin{array}{c}E\left(X\right)=\left(0\times 0.50\right)+\left(1\times 0.20\right)+\left(2\times 0.15\right)+\left(3\times 0.10\right)+\left(4\times 0.05\right)\\ =0+0.20+0.30+0.30+0.20\\ =1\end{array}$

$$

The expected value of distribution B is,

$\begin{array}{c}E\left(X\right)=\left(0\times 0.05\right)+\left(1\times 0.10\right)+\left(2\times 0.15\right)+\left(3\times 0.20\right)+\left(4\times 0.50\right)\\ =0+0.10+0.30+0.60+2\\ =3\end{array}$

Thus, the expected value of distribution A and distribution B are 1 and 3, respectively.

To determine

Find the standard deviation of distribution A and B.

Answer to Problem 1PS

The standard deviation for both distributions A and B is 1.22474.

Explanation of Solution

Calculation:

The standard deviation is calculated using the formula,

$\sigma =\sqrt{{\displaystyle \sum _{i=1}^N{\left(x_i-E\left(X\right)\right)}^{2}P\left(X=x_i\right)}}$

Here, $x_i$ is the ith value of $X$ , $P\left(X=x_i\right)$ is the probability of occurrence of the ith value of $X$ , and $N$ is the number of values in the discrete variable $X$ .

For distribution A, $E\left(X\right)=1$ . So, the standard deviation for distribution A is calculated as,

$\begin{array}{c}\sigma =\sqrt{\begin{array}{l}\left[{\left(0-1\right)}^2\times 0.50\right]+\left[{\left(1-1\right)}^2\times 0.20\right]+\left[{\left(2-1\right)}^2\times 0.15\right]\\ +\left[{\left(3-1\right)}^2\times 0.10\right]+\left[{\left(4-1\right)}^2\times 0.05\right]\end{array}}\\ =\sqrt{1.5}\\ =1.22474\end{array}$

For distribution B, $E\left(X\right)=3$ . So, the standard deviation for distribution B is calculated as,

$\begin{array}{c}\sigma =\sqrt{\left[\begin{array}{l}\left[{\left(0-3\right)}^2\times 0.05\right]+\left[{\left(1-3\right)}^2\times 0.10\right]+\left[{\left(2-3\right)}^2\times 0.15\right]\\ +\left[{\left(3-3\right)}^2\times 0.20\right]+\left[{\left(4-3\right)}^2\times 0.50\right]\end{array}\right]}\\ =\sqrt{1.5}\\ =1.22474\end{array}$

Thus, the standard deviation of distribution A and B is same, that is, 1.2247.

To determine

Compare the results of the two distributions.

Answer to Problem 1PS

The mean value for distribution B is higher than distribution A but the standard deviation is same for both the distributions.

Explanation of Solution

From part a, the expected value for distribution A and B are obtained as 1 and 3, respectively. From part b, the standard deviation for both the distributions is obtained as 1.2247.

Thus, it can be said that the mean value or the expected value of distribution B is greater than that of distribution A. However, both of them have the same spread as the value of standard deviation is same.