Chapter R, Problem 1.1RE

a.

To determine

Construct appropriate graph for the data.

Answer to Problem 1.1RE

The appropriate graph for the data is given below:

Explanation of Solution

Given info:

The data represents the sample of 15 bananas price surveyed by the U.S Department of agriculture.

Calculation:

The Histogram using MINITAB software is obtained below.

Software procedure:

Step by step procedure to draw the frequency histogram chart by using MINITAB software.

• Choose Graph > Histogram.
• Choose Simple, and then click OK.
• In Graph variables, enter the corresponding column of data.
• To modify the interval settings, double click on the horizontal axis of the graph. Then, select Labels > Specified. In this box, enter the values for the cut points of the bin intervals (42, 46, 50, 54)
• In scale on y-axis, make click on frequency.
• Click on ok.

Skew ness:

The data is said to be skewed if there is lack of symmetry and values fall on one side that is, either left or right of the distribution.

Left skewed:

If the tail on the distribution is elongated towards the left then it is a left skewed distribution. It is also called negatively skewed.

The mean is less than the median when the distribution of the data is left skewed. That is, $\text{Mean}<\text{Median}<\text{Mode}$.

In this case, the given data approximately follows left skewed distribution.

b.

To determine

Determine the appropriate summary statistics for the given data.

Answer to Problem 1.1RE

Median and IQR is the appropriate summary statistics for the given data.

Explanation of Solution

Calculation:

The summary statistics for the given data is obtained below.

Software procedure:

Step-by-step procedure to obtain summary statistics is given below.

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables, enter the columns Bananas
• Choose Statistics and select the median and IQR.
• Click OK.

Output obtained from the MINITAB is given below:

From the output, the median is 49 cents per pound, and the interquartile range is 6 cents per pound.

Relationship between mean and median:

• The value of mean is equal to the median when the distribution of the data is symmetric. $\text{Mean}=\text{Median}=\text{Mode}$.
• The mean is greater than the median when the distribution of the data is right skewed. $\text{Mean}>\text{Median}>\text{Mode}$.
• The mean is less than the median when the distribution of the data is left skewed. That is, $\text{Mean}<\text{Median}<\text{Mode}$.

From part a., the distribution of the data is left skewed. This indicates that median will be greater than the mean.

Inter quartile range:

In contrast to the range, which measures only differences between the extremes, the inter quartile range (also called mid spread) is the difference between the third quartile and the first quartile. Thus, it measures the variation in the middle 50 percent of the data, and, unlike the range, is not affected by extreme values. $IQR=Q_3-Q_1$

Thus, for a skewed distribution median and IQR are the better summary statistics.

c.

To determine

Describe the distribution with few sentences.

Explanation of Solution

Justification:

Histogram:

The graphical representation of classes of the quantitative data in the horizontal axis and class frequency in the vertical axis displayed in terms of bar is termed as histogram of the data.

Based on the number of peaks, the modality is classified into three types namely unimodal, bi-modal and multimodal.

Unimodal: The number of peak is one.

Bi-modal: The number of peak is two.

Multi-modal: The number of peak is three or more than three.

Here, the histogram has only one peak and it is elongated to left side. Hence it is unimodal and approximately left skewed.

Moreover, the center of the distribution lies approximately about 50 cents and it also contains of lowest price near 42 cents per pound and the highest price near 53 cents per pound.