Chapter 3, Problem 62SE

a.

To determine

Find the mean and median.

Answer to Problem 62SE

The mean and median are 2.95 and 3, respectively.

Explanation of Solution

Calculation:

The data represent the number of times a sample of 20 families dined out in the previous week.

Store the variable, “Number of times the families dined outside in the previous week” in Column C11 of a MINITAB worksheet.

Mean and Median:

Software Procedure:

Step-by-step procedure to obtain the mean and median using MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns C11.
• In Statistics select Mean and Median.
• Click OK.

The output obtained using MINITAB software is given below:

Thus, mean and median are 2.95 and 3, respectively.

b.

To determine

Find the first quartile and third quartile.

Answer to Problem 62SE

The first quartile is 1 and the third quartile is 4.75.

Explanation of Solution

Calculation:

First quartile and Third quartile:

Software Procedure:

Step-by-step procedure to obtain the first and third quartiles using MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns C11.
• In Statistics select First quartile and Third quartile.
• Click OK.

The output obtained using MINITAB software is given below:

Thus, the first and third quartiles are 1 and 4.75, respectively.

c.

To determine

Find the range and interquartile range.

Answer to Problem 62SE

The range and interquartile range are 7 and 3.75, respectively.

Explanation of Solution

Calculation:

Range, and Interquartile range:

Software Procedure:

Step-by-step procedure to obtain the range and interquartile range using the MINITAB software:

• Choose Stat > Basic Statistics >Display Descriptive Statistics.
• Click OK.
• In Variables, enter the column of C11.
• In Statistics select Range and Interquartile range.
• Click OK.

The output obtained using MINITAB software is given below:

Thus, the range of the data is 7 and the interquartile range is 3.75.

d.

To determine

Find the variance and standard deviation.

Answer to Problem 62SE

The variance and standard deviation are 4.366 and 2.089, respectively.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the variance and standard deviation using MINITAB software:

• Choose Stat > Basic Statistics >Display Descriptive Statistics.
• Click OK.
• In Variables, enter the column of C11.
• In Statistics select Variance and Standard deviation.
• Click OK.

The output obtained using MINITAB software is given below:

Thus, the variance of the data is 4.366 and the standard deviation is 2.089.

e.

To determine

Explain about the shape of the distribution and also explain the reason for the expected shape.

Explanation of Solution

The given information is that the skewness measure for the data is 0.34.

Skewness:

• If the value of skewness is equal to zero, then the distribution is symmetric.
• If the skewness value is less than zero, then the distribution is negatively skewed.
• If the skewness value is greater than zero, then the distribution is positively skewed.

Here, the skewness is 0.34, which is greater than zero. Hence, the shape of the distribution is positively skewed.

f.

To determine

Check whether the data contain any outliers.

Answer to Problem 62SE

The data has no outliers.

Explanation of Solution

Calculation:

The formula for the lower limit to detect outliers is as follows:

$\text{Lower limit}=Q_1-1.5\text{IQR}$,

where $Q_1$ is the first quartile and IQR is the interquartile range.

Substitute $IQR=3.75$ and $Q_1=1$.

$\begin{array}{c}\text{Lower limit}=1-1.5\left(3.75\right)\\ =1-5.625\\ =-4.625\end{array}$

Thus, the lower limit is –4.625.

The formula for the upper limit to detect outliers is as follows:

$\text{Upper limit}=Q_3+1.5\text{IQR}$,

where $Q_3$ is the third quartile and IQR is the interquartile range.

Substitute $IQR=3.75$ and $Q_3=4.75$.

$\begin{array}{c}\text{Upper limit}=4.75+1.5\left(3.75\right)\\ =4.75+5.625\\ =10.375\end{array}$

Thus, the upper limit is 10.375.

Outlier:

The observation that falls outside of the overall pattern of the data is called an outlier.

The data values less than –4.625 or greater than 10.375 are considered as the outliers.

Here, all data values lie within the limits. Thus, the data have no outliers.