Chapter 6, Problem 70DE

Refer to the Real Estate data, which report information on homes sold in the Goodyear, Arizona, area last year.

1. a. Create a probability distribution for the number of bedrooms. Compute the mean and the standard deviation of this distribution.
2. b. Create a probability distribution for the number of bathrooms. Compute the mean and the standard deviation of this distribution.

To determine

Construct the probability distribution for the number of bedrooms.

Calculate the mean and standard deviation of this distribution.

Answer to Problem 70DE

The probability distribution for the number of bedrooms is as follows:

Number of Bedrooms	Frequency	Probability $P\left(x\right)$
2	24	0.2286
3	26	0.2476
4	26	0.2476
5	11	0.1048
6	14	0.1333
7	2	0.0190
8	2	0.0190
Total	105	1

The mean of the distribution is 3.8.

The standard deviation of the distribution is 1.4954.

Explanation of Solution

Step-by-step procedure to obtain the frequency table for the number of bedrooms using EXCEL is given below:

• Enter the data in an EXCEL sheet.
• Go to Insert > PivotTable > PivotTable.
• In Table/Range, select the column Bedrooms and click OK.
• In PivotTable Field List, drag Bedrooms to Row Labels and to ∑ values.
• Click on it from ∑ values.
• Choose Value Field Settings.
• In Summarize value field by, choose Count and click OK.

The output obtained is as follows:

The probability distributions for the number of bedrooms are calculated as follows:

Number of Bedrooms	Frequency	Probability $P\left(x\right)$
2	24	$\frac{24}{105}=0.2286$
3	26	$\frac{26}{105}=0.2476$
4	26	$\frac{26}{105}=0.2476$
5	11	$\frac{11}{105}=0.1048$
6	14	$\frac{14}{105}=0.1333$
7	2	$\frac{2}{105}=0.0190$
8	2	$\frac{2}{105}=0.0190$
Total	105	1

The mean of the distribution is calculated as follows:

$\begin{array}{c}\mu ={\displaystyle \sum xP\left(x\right)}\\ =\left(2\times 0.2286\right)+\left(3\times 0.2476\right)+\left(4\times 0.2476\right)+\left(5\times 0.1048\right)+\left(6\times 0.1333\right)+\left(7\times 0.0190\right)+\left(8\times 0.0190\right)\\ =0.4571+0.7429+0.9905+0.5238+0.8+0.1333+0.1524\\ =3.8\end{array}$

Therefore, the mean of the distribution is 3.8.

Consider the following table that shows preliminary calculations to compute the standard deviation:

Bedrooms	Frequency	Probability $P\left(x\right)$	${\left(x-\mu \right)}^2$	${\left(x-\mu \right)}^2\cdot P\left(x\right)$
2	24	0.2286	${\left(2-3.8\right)}^2=3.24$	$3.24\times 0.2286=0.7406$
3	26	0.2476	${\left(3-3.8\right)}^2=0.64$	$0.64\times 0.2476=0.1585$
4	26	0.2476	${\left(4-3.8\right)}^2=0.04$	$0.04\times 0.2476=0.0099$
5	11	0.1048	${\left(5-3.8\right)}^2=1.44$	$1.44\times 0.1048=0.1509$
6	14	0.1333	${\left(6-3.8\right)}^2=4.84$	$4.84\times 0.1333=0.6453$
7	2	0.0190	${\left(7-3.8\right)}^2=10.24$	$10.24\times 0.0190=0.1950$
8	2	0.0190	${\left(8-3.8\right)}^2=17.64$	$17.64\times 0.0190=0.3360$
	105			2.2362

The standard deviation of the distribution is calculated as follows:

$\begin{array}{c}\sigma =\sqrt{{\displaystyle \sum \left[{\left(x-\mu \right)}^{2}P\left(x\right)\right]}}\\ =\sqrt{2.2362}\\ =1.4954\end{array}$

Therefore, the standard deviation of the distribution is 1.4954.

To determine

Construct the probability distribution for the number of bathrooms.

Find the mean and standard deviation for this distribution.

Answer to Problem 70DE

The probability distributions for the number of bathrooms are as follows:

Number of Bathrooms	Frequency	Probability $P\left(x\right)$
1.5	16	0.1524
2	65	0.6191
2.5	15	0.1429
3	9	0.0857
	105

The mean of the distribution is 2.0812.

The standard deviation of the distribution is 0.3912.

Explanation of Solution

Step-by-step procedure to obtain the frequency table for bedrooms using EXCEL:

• Enter the data in an EXCEL sheet.
• Go to Insert > PivotTable > PivotTable.
• In Table/Range, select the column Bathrooms and click OK.
• In PivotTable Field List, drag Bathrooms to Row Labels and to ∑ values.
• Click on it from ∑ values.
• Choose Value Field Settings.
• In Summarize value field by, choose Count and click OK.

The output obtained is as follows:

The probability distributions for the number of bathrooms are calculated as follows:

Number of Bathrooms	Frequency	Probability $P\left(x\right)$
1.5	16	$\frac{16}{105}=0.1524$
2	65	$\frac{65}{105}=0.6191$
2.5	15	$\frac{15}{105}=0.1429$
3	9	$\frac{9}{105}=0.0857$
Total	105	1

The mean of the distribution is calculated as follows:

$\begin{array}{c}\mu ={\displaystyle \sum xP\left(x\right)}\\ =\left(1.5\times 0.1524\right)+\left(2\times 0.6191\right)+\left(2.5\times 0.1429\right)+\left(3\times 0.0857\right)\\ =0.2286+1.2382+0.3573+0.2571\\ =2.0812\end{array}$

Therefore, the mean of the distribution is 2.0812.

Consider the following table that shows preliminary calculations to compute the standard deviation:

Number of Bathrooms	Frequency	Probability $P\left(x\right)$	${\left(x-\mu \right)}^2$	${\left(x-\mu \right)}^2\cdot P\left(x\right)$
1.5	16	0.1524	$\begin{array}{l}{\left(1.5-2.0812\right)}^2\\ =0.3378\end{array}$	$\begin{array}{l}0.3374\times 0.1524\\ =0.0515\end{array}$
2	65	0.6191	$\begin{array}{l}{\left(2-2.0812\right)}^2\\ =0.0066\end{array}$	$\begin{array}{l}0.0066\times 0.6191\\ =0.0041\end{array}$
2.5	15	0.1429	$\begin{array}{l}{\left(2.5-2.0812\right)}^2\\ =0.1754\end{array}$	$\begin{array}{l}0.1754\times 0.1429\\ =0.0251\end{array}$
3	9	0.0857	$\begin{array}{l}{\left(3-2.0812\right)}^2\\ =0.8442\end{array}$	$\begin{array}{l}0.8442\times 0.0857\\ =0.0723\end{array}$
Total	105			$0.1530$

The standard deviation of the distribution is calculated as follows:

$\begin{array}{c}\sigma =\sqrt{{\displaystyle \sum \left[{\left(x-\mu \right)}^{2}P\left(x\right)\right]}}\\ =\sqrt{0.1530}\\ =0.3912\end{array}$

Therefore, the standard deviation of the distribution is 0.3912.