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13th Edition

Anderson

Publisher: CENGAGE L

ISBN: 9781305881884

Chapter 3, Problem 62SE

**a.**

To determine

Find the mean and median.

Expert Solution

The mean and median are 2.95 and 3, respectively.

**Calculation:**

The given information is that the data represent the number of times a sample of 20 families dined out last week.

**Software Procedure:**

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

- Choose
**Stat > Basic Statistics > Display Descriptive Statistics**. - In
**Variables**enter the columns**Variable**. - In
**Statistics**select**mean**and**median.** - Click
**OK**.

Output using the MINITAB software is given below:

Thus, mean and median is 2.95 and 3.

**b.**

To determine

Find the first and third quartiles.

Expert Solution

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

**Calculation:**

**Software Procedure:**

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

- Choose
**Stat > Basic Statistics > Display Descriptive Statistics**. - In
**Variables**enter the columns**Variable**. - In
**Statistics**select**first quartile**and**third quartile.** - Click
**OK**.

Output using the 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.

Expert Solution

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

**Calculation:**

**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**Variable.** - In
**Statistics**select**range**and**interquartile range.** - Click
**OK**.

Output using the MINITAB software is given below:

From the MINITAB output, it is clear that the range of the data is 7 and the interquartile range is 3.75.

**d.**

To determine

Find the variance and standard deviation.

Expert Solution

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

**Calculation:**

**Software Procedure:**

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

- Choose
**Stat**>**Basic Statistics >.Display Descriptive Statistics.** - Click
**OK**. - In
**Variables**, enter the column of**Variable.** - In
**Statistics**select**variance**and**standard deviation.** - Click
**OK**.

Output using the MINITAB software is given below:

From the MINITAB output, it is clear that the variance of the data is 4.366 and the standard deviation is 2.089.

**e.**

To determine

Explain the shape of the distribution and the shape one can expect and also explain the reason for that.

Expert 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 normal and 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. This measure is greater than zero. Hence, the shape of the distribution is positively skewed.

**f.**

To determine

Check whether the data contain any outliers.

Expert Solution

There are no outliers.

**Calculation:**

The formula for lower limit is,

Where,

Substitute

Thus, the lower limit is –4.625.

The formula for upper limit is,

Substitute

Thus, the upper limit is 10.375.

*Outlier:*

The observation that fall 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 are lies within the limits. Thus, there are no outliers.

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