Chapter 8, Problem 30CE

The Appliance Center has six sales representatives at its North Jacksonville outlet. The following table lists the number of refrigerators sold by each representative last month.

1. a. How many samples of size 2 are possible?
2. b. Select all possible samples of size 2 and compute the mean number sold.
3. c. Organize the sample means into a frequency distribution.
4. d. What is the mean of the population? What is the mean of the sample means?
5. e. What is the shape of the population distribution?
6. f. What is the shape of the distribution of the sample mean?

To determine

Find the possible number of different samples of size 2.

Answer to Problem 30CE

The possible number of different samples of size 2 is 15.

Explanation of Solution

From the given information, there are six representatives.

The possible number of different samples of size two is obtained by using the following formula:

${}_N{C}_n=\frac{N!}{\left(N-n\right)!n!}$

Substitute 6 for the N and 2 for the n.

Then,

$\begin{array}{c}{}_6{C}_2=\frac{6!}{\left(6-2\right)!2!}\\ =\frac{6\times 5\times 4!}{4!2!}\\ =\frac{6\times 5}{2}\\ =15\end{array}$

Thus, the possible number of different samples of size 2 is 15.

To determine

Give the all possible samples of size 2.

Find the mean of each sample.

Answer to Problem 30CE

All possible samples of size 2 are $\left(54,50\right)$, $\left(54,52\right)$, $\left(54,48\right)$, $\left(54,50\right)$, $\left(54,52\right)$, $\left(50,52\right)$, $\left(50,48\right)$, $\left(50,50\right)$, $\left(50,52\right)$, $\left(52,48\right)$, $\left(52,50\right)$, $\left(52,52\right)$, $\left(48,50\right)$, $\left(48,52\right)$ and $\left(50,52\right)$.

The mean of each sample is 52, 53, 51, 52, 53, 51, 49, 50, 51, 50, 51, 52, 49, 50 and 51.

Explanation of Solution

The mean is calculated by using the following formula:

$\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{the}\text{observations}}{\text{Number}\text{of}\text{observations}}$

Sample	Mean
$\left(54,50\right)$	$\frac{54+50}{2}=52$
$\left(54,52\right)$	$\frac{54+52}{2}=53$
$\left(54,48\right)$	$\frac{54+48}{2}=51$
$\left(54,50\right)$	$\frac{54+50}{2}=52$
$\left(54,52\right)$	$\frac{54+52}{2}=53$
$\left(50,52\right)$	$\frac{50+52}{2}=51$
$\left(50,48\right)$	$\frac{50+48}{2}=49$
$\left(50,50\right)$	$\frac{50+50}{2}=50$
$\left(50,52\right)$	$\frac{50+52}{2}=51$
$\left(52,48\right)$	$\frac{52+48}{2}=50$
$\left(52,50\right)$	$\frac{52+50}{2}=51$
$\left(52,52\right)$	$\frac{52+52}{2}=52$
$\left(48,50\right)$	$\frac{48+50}{2}=49$
$\left(48,52\right)$	$\frac{48+52}{2}=50$
$\left(50,52\right)$	$\frac{50+52}{2}=51$

Thus, all possible samples of size 2 are $\left(54,50\right)$, $\left(54,52\right)$, $\left(54,48\right)$, $\left(54,50\right)$, $\left(54,52\right)$, $\left(50,52\right)$, $\left(50,48\right)$, $\left(50,50\right)$, $\left(50,52\right)$, $\left(52,48\right)$, $\left(52,50\right)$, $\left(52,52\right)$, $\left(48,50\right)$, $\left(48,52\right)$ and $\left(50,52\right)$.

Thus, the mean of each sample is 52, 53, 51, 52, 53, 51, 49, 50, 51, 50, 51, 52, 49, 50 and 51.

To determine

Create a frequency distribution for the sample means.

Answer to Problem 30CE

A frequency distribution for the sample means is

Sample mean	Probability
49	$\frac{2}{15}=0.13$
50	$\frac{3}{15}=0.2$
51	$\frac{5}{15}=0.33$
52	$\frac{3}{15}=0.2$
53	$\frac{2}{15}=0.13$

Explanation of Solution

A frequency distribution for the sample means is obtained as follows:

Let $\overline{x}$ is the sample mean and f is the frequency.

Sample mean	f	Probability
49	2	$\frac{2}{15}=0.13$
50	3	$\frac{3}{15}=0.2$
51	5	$\frac{5}{15}=0.33$
52	3	$\frac{3}{15}=0.2$
53	2	$\frac{2}{15}=0.13$
	$N=15$

To determine

Find the population mean.

Find the mean of the sample means.

Answer to Problem 30CE

The population mean is 51.

The mean of the distribution of the sample means is 51.

Explanation of Solution

Population mean is calculated as follows:

From the given information, the number of refrigerators sold by each representative is 54, 50, 52, 48, 50 and 52.

$\begin{array}{c}\text{Mean}=\frac{54+50+52+48+50+52}{6}\\ =\frac{306}{6}\\ =51\end{array}$

The mean of the sample means is calculated as follows:

$\begin{array}{c}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{the}\text{sample}\text{means}}{\text{Number}\text{of}\text{samples}}\\ =\frac{\begin{array}{l}52+53+51+52+53+51+49+50+51+\\ 50+51+52+49+50+51.\end{array}}{15}\\ =\frac{765}{15}\\ =51\end{array}$

Comparison:

The mean of the distribution of the sample mean is 51 and the population mean is 51. The two means are exactly same.

Thus, the mean of the distribution of the sample means is equal to the population mean.

To determine

Find the shape of the population distribution.

Answer to Problem 30CE

The shape of the population distribution is normal.

Explanation of Solution

A frequency distribution for the number sold is obtained as follows:

Number sold	f	Probability
48	1	$\frac{1}{6}=0.17$
50	2	$\frac{2}{6}=0.33$
52	2	$\frac{2}{6}=0.33$
54	1	$\frac{1}{6}=0.17$
	$N=6$

Step-by-step procedure to obtain the bar chart using MINITAB:

• Choose Stat > Graph > Bar chart.
• Under Bars represent, enter select Values from a table.
• Under One column of values select Simple.
• Click on OK.
• Under Graph variables enter probability and under categorical variable enter Number sold.
• Click OK.

Output using MINITAB software is given below:

From the bar chart it can be observed that the shape of the population distribution is normal.

To determine

Find the shape of the distribution of the sample mean.

Answer to Problem 30CE

The shape of the distribution of the sample means is normal.

Explanation of Solution

Step-by-step procedure to obtain the bar chart using MINITAB:

• Choose Stat > Graph > Bar chart.
• Under Bars represent, enter select Values from a table.
• Under One column of values select Simple.
• Click on OK.
• Under Graph variables enter probability and under categorical variable enter sample mean.
• Click OK.

Output using MINITAB software is given below:

From the bar chart it can be observed that the shape of the distribution of the sample means is normal.

Thus, the shape of the distribution of the sample means is normal.