Chapter 5, Problem 5.1LB

Given the following probability distributions:

1. Compute the expected value for each distribution.
2. Compute the standard deviation for each distribution.
3. Compare the results of distributions A and B.

To determine

To find: The expected values for both the distributions.

Answer to Problem 5.1LB

The expected values are 1 and 3 respectively.

Explanation of Solution

Given:

The probability distributions are:

Distribution A		Distribution B
$x_i$	$P\left(X=x_i\right)$	$x_i$	$P\left(X=x_i\right)$
0	0.5	0	0.05
1	0.2	1	0.1
2	0.15	2	0.15
3	0.1	3	0.2
4	0.05	4	0.5

Formula used:

The formula to compute the expected mean is:

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

Here, $P\left(X=x_i\right)$ is the probability associated with the $x_i$ observation.

The formula to compute the standard deviation is:

  $\sigma =\sqrt{{\displaystyle \sum x_i^2\times }P\left(X=x_i\right)-{\left({\displaystyle \sum x_i\times P\left(X=x_i\right)}\right)}^2}$

Calculation:

The expected values for both the distributions can be calculated as:

Distribution A:

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum x_i\times P\left(X=x_i\right)}\\ =0\left(0.5\right)+1\left(0.2\right)+2\left(0.15\right)+3\left(0.1\right)+4\left(0.05\right)\\ =0+0.2+0.30+0.3+0.2\\ =1\end{array}$

Distribution B:

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum x_i\times P\left(X=x_i\right)}\\ =0\left(0.05\right)+1\left(0.1\right)+2\left(0.15\right)+3\left(0.2\right)+4\left(0.5\right)\\ =0+0.1+0.30+0.6+0.2\\ =3\end{array}$

Thus, the required expected values are 1 and 3 respectively.

To determine

To find: The standard deviation for each distribution.

Answer to Problem 5.1LB

The standard deviations are 1.225 and 1.225 respectively.

Explanation of Solution

The standard deviation for each distribution can be computed as:

Distribution A:

  $\begin{array}{c}\sigma =\sqrt{{\displaystyle \sum x_i^2\times }P\left(X=x_i\right)-{\left({\displaystyle \sum x_i\times P\left(X=x_i\right)}\right)}^2}\\ =\sqrt{0^2\left(0.5\right)+1^2\left(0.20\right)+2^2\left(0.15\right)+3^2\left(0.10\right)+4^2\left(0.05\right)-{\left(1\right)}^2}\\ =1.225\end{array}$

Distribution B:

  $\begin{array}{c}\sigma =\sqrt{{\displaystyle \sum x_i^2\times }P\left(X=x_i\right)-{\left({\displaystyle \sum x_i\times P\left(X=x_i\right)}\right)}^2}\\ =\sqrt{0^2\left(0.05\right)+1^2\left(0.1\right)+2^2\left(0.15\right)+3^2\left(0.2\right)+4^2\left(0.5\right)-{\left(3\right)}^2}\\ =1.225\end{array}$

Thus, the required standard deviations are 1.225 and 1.225 respectively.

To determine

To compare: The provided distributions.

Explanation of Solution

From the above two parts, it is clear that the mean of distribution B is greater than the mean of distribution B but there is no difference in the standard deviation of two distributions.