Chapter 9, Problem 50E

To determine

To find: The probabilities for the provided five intervals.

Answer to Problem 50E

Solution: The probabilities for the provided intervals are given in the following table:

Interval	Probability
$$ $\left[X\le -0.6\right]$	0.2743
$$ $\left[-0.6<X\le -0.1\right]$	0.1859
$\left[0.1<X\le -0.1\right]$	0.0796
$\left[0.1<X\le 0.6\right]$	0.1859
$\left[X>0.6\right]$	0.2743

Explanation of Solution

Calculation: From Table A, in the left side z-score table, look for $-0.6$ in the row and 0.00 in the column and the intercept term is found, i.e., 0.2743. Thus, the probability of the interval $\left[X\le -0.6\right]$ is 0.2743. The expected number of frequency can be calculated as

$\begin{array}{c}{E}_i=n\times p\\ =500\times 0.2743\\ =137.15\end{array}$

From Table A, the probability less than $-0.1$, look for $-0.1$ in the row and 0.00 in the column and the intercept term is found to be 0.4602. Similarly, the probability for less than $-0.6$ is 0.2743. The required probability further can be calculated as given below:

$\begin{array}{c}\left[-0.6<X\le -0.1\right]=\left[X\le -0.1\right]-\left[X\le -0.6\right]\\ =0.4602-0.2743\end{array}$

$=0.1859$

The expected number of frequency can be calculated as

$\begin{array}{c}{E}_{{}_{{}_i}}=n\times p\\ =500\times 0.1859\\ =92.95\end{array}$

To obtain the probability less than 0.1, use Table A, look for 0.1 in the row and 0.00 in the column and the intercept term is found to be 0.5398. Similarly, the probability for less than $-0.1$ is 0.4602. The required probability further can be calculated as given below:

$\begin{array}{c}P\left[0.1<X\le -0.1\right]=P\left[X\le 0.1\right]-P\left[X\le -0.1\right]\\ =0.5398-0.4602\\ =0.0796\end{array}$

The expected number of frequency is calculated as

$\begin{array}{c}{E}_i=n\times p\\ =500\times 0.0796\\ =39.80\end{array}$

To obtain the probability less than 0.6, use Table A. Look for 0.6 in the row and 0.00 in the column and the intercept term is found to be 0.7257. Similarly, the probability for less than 0.1 is 0.5398. The required probability further can be calculated as given below:

$\begin{array}{c}P\left[0.1<X\le 0.6\right]=\left[X\le 0.6\right]-P\left[X\le 0.1\right]\\ =0.7257-0.5398\\ =0.1859\end{array}$

The expected number of frequency is calculated as

$\begin{array}{c}{E}_i=n\times p\\ =500\times 0.1859\\ =92.95\end{array}$

To obtain the probability less than 0.6, use Table A. Look for 0.6 in the row and 0.00 in the column and the intercept term is found to be 0.7257. The required probability further can be calculated as given below:

$\begin{array}{c}P\left[X<0.6\right]=0.7257\\ P\left[X\ge 0.6\right]=1-0.7257\\ =0.2743\end{array}$

The expected number of frequency is calculated as

$\begin{array}{c}E=n\times p\\ =500\times 0.2743\\ =137.15\end{array}$

The table that shows the expected counts is given below:

Expected counts	Observed Counts
137.15	139
92.95	102
39.80	41
92.95	78
137.15	140

Defining the null and alternative hypothesis for goodness of fit testing:

$H_0$: The random number generated follows the standard normal distribution.

$H_{{}_1}$: The random number generated does not follow the standard normal distribution.

Further, chi-square testing goodness of fit has been done using Minitab software by following steps:

Step 1: Input the data in the Minitab worksheet.

Step 2: Go to Stat and select “Chi-Square Goodness-of-Fit (One- variable)” under the option tables.

Step 3: Enter the frequency as “Observed Count.”

Step 4: Specify “proportions specified by historical counts” by choosing the “Input column” from the pop-up menu.

After implementing these 4 steps, a table will appear in the output box where chi-square value of the desired result is given, that is, 3.41 and the P-value is 0.492.

Interpretation: The ${\chi }^2$-statistic is 3.41 and the P-value is 0.492, which is greater than the significance value 0.05. Hence, the null hypothesis is fails to be rejected. This concludes that the generated random number follows the standard normal distribution.