Chapter 3.4, Problem 87E

(a)

To determine

The random numbers by summing twelve uniform variables and subtracting 6.

Answer to Problem 87E

Solution: The simulated distribution is shown in the below table.

0.543534	$–0.16771$	$–0.5377$	$–0.83375$	$\begin{array}{l}–1.35297\hfill \end{array}$
0.248992	$\begin{array}{l}–1.5897\hfill \end{array}$	1.493676	$\begin{array}{l}–0.32634\hfill \end{array}$	1.535747
0.742845	0.170866	0.394907	0.799959	2.122601
$\begin{array}{l}–0.18316\hfill \end{array}$	0.264532	1.407664	1.283963	$\begin{array}{l}–1.28957\hfill \end{array}$
0.442991	$\begin{array}{l}–0.81477\hfill \end{array}$	0.720228	2.229499	1.609478
$\begin{array}{l}–1.65737\hfill \end{array}$	0.708811	0.534153	0.903971	0.06244
1.238842	0.142653	$\begin{array}{l}-1.54798\hfill \end{array}$	0.134968	$\begin{array}{l}-0.27487\hfill \end{array}$
0.387513	$\begin{array}{l}-1.14603\hfill \end{array}$	0.525326	1.013777	0.665731
2.04422	$\begin{array}{l}-0.67698\hfill \end{array}$	$\begin{array}{l}-1.0171\hfill \end{array}$	$\begin{array}{l}-0.91308\hfill \end{array}$	$\begin{array}{l}-0.74691\hfill \end{array}$
0.415165	0.098935	2.275969	0.559715	0.925708
$\begin{array}{l}-0.53388\hfill \end{array}$	$\begin{array}{l}-0.50962\hfill \end{array}$	1.306352	$\begin{array}{l}-1.61856\hfill \end{array}$	$\begin{array}{l}-0.79513\hfill \end{array}$
0.199455	0.606878	2.31169	1.041178	$\begin{array}{l}-0.71081\hfill \end{array}$
$\begin{array}{l}-0.25882\hfill \end{array}$	1.371048	0.882886	0.45936	0.613565
1.145033	1.460973	$\begin{array}{l}-0.29857\hfill \end{array}$	$\begin{array}{l}-1.06949\hfill \end{array}$	$\begin{array}{l}-1.26279\hfill \end{array}$
$\begin{array}{l}-0.77\hfill \end{array}$	1.84498	$\begin{array}{l}-0.48357\hfill \end{array}$	1.512605	0.462096
1.207655	0.98702	0.661045	1.011263	0.805521
0.129918	0.287478	$\begin{array}{l}-0.06224\hfill \end{array}$	0.574382	$\begin{array}{l}-1.03787\hfill \end{array}$
0.203774	0.378354	1.203827	$\begin{array}{l}-1.42317\hfill \end{array}$	$\begin{array}{l}-0.56896\hfill \end{array}$
0.500351	$\begin{array}{l}-0.83593\hfill \end{array}$	1.292136	$\begin{array}{l}-0.09291\hfill \end{array}$	0.606476
$\begin{array}{l}-0.89808\hfill \end{array}$	1.330552	0.166358	$\begin{array}{l}-0.5962\hfill \end{array}$	$\begin{array}{l}-1.71095\hfill \end{array}$

Explanation of Solution

The below steps are followed in Minitab software to obtain the distribution for the variable.

Step 1: Open the Minitab worksheet. Go to Calc > Random Data> Uniform.

Step 2: Input 100 as the number of rows of data. Input C1-C12 in the “Store in column(s)” option and specify the Lower endpoint and the Upper endpoint as 0 and 1 respectively.

Step 3: Click OK.

Step 4: Go to Calc > Calculator.

Step 5: Enter C13 in “Store result in variable” and in “Expression”, write the expression as $\left(\text{C1+C2+C3+C4+C5+C6+C7+C8+C9+C10+C11+C12}\right)-6$.

Step 6: Click OK.

The generated random number is shown in the below table.

0.543534	$–0.16771$	$–0.5377$	$–0.83375$	$\begin{array}{l}–1.35297\hfill \end{array}$
0.248992	$\begin{array}{l}–1.5897\hfill \end{array}$	1.493676	$\begin{array}{l}–0.32634\hfill \end{array}$	1.535747
0.742845	0.170866	0.394907	0.799959	2.122601
$\begin{array}{l}–0.18316\hfill \end{array}$	0.264532	1.407664	1.283963	$\begin{array}{l}–1.28957\hfill \end{array}$
0.442991	$\begin{array}{l}–0.81477\hfill \end{array}$	0.720228	2.229499	1.609478
$\begin{array}{l}–1.65737\hfill \end{array}$	0.708811	0.534153	0.903971	0.06244
1.238842	0.142653	$\begin{array}{l}-1.54798\hfill \end{array}$	0.134968	$\begin{array}{l}-0.27487\hfill \end{array}$
0.387513	$\begin{array}{l}-1.14603\hfill \end{array}$	0.525326	1.013777	0.665731
2.04422	$\begin{array}{l}-0.67698\hfill \end{array}$	$\begin{array}{l}-1.0171\hfill \end{array}$	$\begin{array}{l}-0.91308\hfill \end{array}$	$\begin{array}{l}-0.74691\hfill \end{array}$
0.415165	0.098935	2.275969	0.559715	0.925708
$\begin{array}{l}-0.53388\hfill \end{array}$	$\begin{array}{l}-0.50962\hfill \end{array}$	1.306352	$\begin{array}{l}-1.61856\hfill \end{array}$	$\begin{array}{l}-0.79513\hfill \end{array}$
0.199455	0.606878	2.31169	1.041178	$\begin{array}{l}-0.71081\hfill \end{array}$
$\begin{array}{l}-0.25882\hfill \end{array}$	1.371048	0.882886	0.45936	0.613565
1.145033	1.460973	$\begin{array}{l}-0.29857\hfill \end{array}$	$\begin{array}{l}-1.06949\hfill \end{array}$	$\begin{array}{l}-1.26279\hfill \end{array}$
$\begin{array}{l}-0.77\hfill \end{array}$	1.84498	$\begin{array}{l}-0.48357\hfill \end{array}$	1.512605	0.462096
1.207655	0.98702	0.661045	1.011263	0.805521
0.129918	0.287478	$\begin{array}{l}-0.06224\hfill \end{array}$	0.574382	$\begin{array}{l}-1.03787\hfill \end{array}$
0.203774	0.378354	1.203827	$\begin{array}{l}-1.42317\hfill \end{array}$	$\begin{array}{l}-0.56896\hfill \end{array}$
0.500351	$\begin{array}{l}-0.83593\hfill \end{array}$	1.292136	$\begin{array}{l}-0.09291\hfill \end{array}$	0.606476
$\begin{array}{l}-0.89808\hfill \end{array}$	1.330552	0.166358	$\begin{array}{l}-0.5962\hfill \end{array}$	$\begin{array}{l}-1.71095\hfill \end{array}$

(b)

To determine

To find: The numerical and graphical summary of the obtained distribution.

Answer to Problem 87E

Solution: The shape of the distribution is more or less symmetric and the centre of the curve is near the mean value, 0.2460. The standard deviation of the distribution is almost 1.

Explanation of Solution

Calculation: To obtain the numerical summary of the data, the below steps are followed in the Minitab software.

Step 1: Go to Stat $\to$ Basic Statistics $\to$ Display Descriptive Statistics.

Step 2: Select ‘C13’ in the variable option.

Step 3: Click on the Statistics button.

Step 4:Select minimum, maximum, mean and standard deviation.

Step 5: Click OK.

The mean, standard deviation, minimum, and maximum values are obtained as 0.2460, 0.994, $-1.711$, and 2.312.

Graph: The below steps are followed in Minitab software to obtain the histogram for the distribution.

Step 1: Insert all the observations in the worksheet.

Step 2: Go to Graph $\to$ Histogram

Step 3: Select ‘C13’ in the ‘Graph variable’ option.

Step 4: Click OK.

The obtained histogram is,

Interpretation: The histogram shows that the shape of the distribution is more or less symmetric. Moreover, the centre of the curve is near the mean value, 0.2460. The standard deviation of the distribution is almost 1. The distribution can be considered as approximately normal.

(c)

To determine

The summary of the findings of the provided simulation.

Answer to Problem 87E

Solution: The simulated distribution is approximated by the standard normal distribution as its mean and standard deviation are 0.2460 and 0.994.

Explanation of Solution

The 100 samples are generated by summing 12 uniform variables and subtracting 6 from the sum. The mean of the distribution is 0.2460 and standard deviation is 1. The obtained distribution can be approximated by the standard normal distribution whose mean is 0 and standard deviation is 1.