91Ó°ÊÓ

The data consists of the blood alcohol concentrations of drivers.

a.

The 5-number summary of a dataset consists of the following five points:

Minimum value

First quartile

Median/Second quartile

Third quartile

Maximum value

The following table shows the data in ascending order:

The minimum value of the dataset is equal to 0.09g/dL.

The first quartile is equal to the 25^th percentile.

$Q_1=P_{25}$

The formula to calculate the percentile is equal to:

$L=\frac{k}{100}×n$

Here,

L is the locator of the value.

k is the percentile of the value.

n is the total number of values.

So, for k=25 and n=12, the value of L is equal to:

$$\begin{gathered} L=\frac{k}{100}×n \\ =\frac{25}{100}×12 \\ =3 \end{gathered}$$

Since L is a whole number, compute the sum of the 3^rd and the 4^th observation and divide by 2.

$$\begin{gathered} P_{25}=\frac{3^{rd}\text{obs}+4^{th}\text{obs}}{2} \\ =\frac{0.11+0.13}{2} \\ =0.12 \end{gathered}$$

Thus, the first quartile is 0.12 g/dL.

Similarly, the second quartile is the 50^th percentile.

$Q_2=P_{50}$

So, for k=50 and n=12, the value of L is equal to:

$$\begin{gathered} L=\frac{k}{100}×n \\ =\frac{50}{100}×12 \\ =6 \end{gathered}$$

Since L is a whole number, compute the sum of the 6^thand the 7^th observation and divide by 2.

$$\begin{gathered} P_{50}=\frac{6^{th}\text{obs}+7^{th}\text{obs}}{2} \\ =\frac{0.15+0.17}{2} \\ =0.16 \end{gathered}$$

Thus, the second quartile/median is 0.16 g/dL.

Similarly, the third quartile is the 75^th percentile.

$Q_3=P_{75}$

So, for k=75 and n=12, the value of L is equal to:

$$\begin{gathered} L=\frac{k}{100}×n \\ =\frac{75}{100}×12 \\ =9 \end{gathered}$$

Since L is a whole number, compute the sum of the 9^thand the 10^th observation and divide by 2.

$$\begin{gathered} P_{75}=\frac{9^{th}\text{obs}+{10}^{th}\text{obs}}{2} \\ =\frac{0.18+0.18}{2} \\ =0.18 \end{gathered}$$

Thus, the third quartile is 0.18 g/dL.

The maximum value of the data is 0.35g/dL.

Thus, the 5-number summary is as follows:

Minimum Value: 0.09 g/dL

First Quartile: 0.12 g/dL

Median: 0.16g/dL

Third Quartile: 0.18g/dL

Maximum Value: 0.35g/dL

An outlier is an observation that is below 1.5 times the interquartile range from the first quartile or greater than 1.5 times the interquartile range from the third quartile.

The interquartile range (IQR) for the given data is calculated as shown below:

$$\begin{gathered} IQR=Q_3-Q_1 \\ =0.18-0.12 \\ =0.06 \end{gathered}$$

The value of 1.5 times IQR is calculated below:

$$\begin{gathered} 1.5×IQR=1.5×0.06 \\ =0.09 \end{gathered}$$

The limits are calculated as:

$$\begin{gathered} Q_1-0.09=0.12-0.09 \\ =0.03 \\ {Q}_3+0.09=0.18+0.09 \\ =0.27 \end{gathered}$$

Thus, values below 0.03 and above 0.27 are outliers.

The value equal to 0.35 is the only outlier present in the data.

b.

Using the 5-number summary, the boxplot is constructed by plotting the five points on the graph and constructing a box-shaped plot. The constructed boxplot is shown below

c.

The stemplot is constructed using the following steps:

• Consider the whole number and the tenth part of the decimal as the stems, and the hundredth place as the leaves.
• Draw a vertical line in the middle.
• Write the stems on the left side of the vertical bar and the leaves on the right side of the vertical bar.
• Write the key.

The following stemplot is constructed:

Key: 0.3|5 represents 0.35 g/dL blood alcohol concentration.