Chapter 3, Problem 15RE

a.

To determine

Find the first and the third quartiles.

Answer to Problem 15RE

The first and the third quartiles are 2.1 and 15.7 respectively.

Explanation of Solution

Calculation:

The profit of 15 largest U.S. Corporation is given.

Three quartiles:

• The first quartile separates the lowest 25% of the observations from the other 75% of the observations. The first quartile is denoted by $Q_1$.
• The second quartile separates the lower 50% of the observations from the other 50% of the observations. The second quartile is denoted by $Q_2$. The second quartile is same as median.
• The third quartile separates the lowest 75% of the observations from the other 25% of the observations. The third quartile is denoted by $Q_3$.
• Procedure for finding the first and the third quartile:
• Step 1: The observations should be arranged in increasing order.
• Step 2: The size of the data is n.
•                                     For finding first quartile, $L=0.25n$
•                                     For finding third quartile, $L=0.75n$
• Step 3: The quartile will be the average of the observation of the position L and the observation in position $L+1$, if L is a whole number. If the L value is not a whole number, the next higher whole number will be considered. The quartile is the observation in the position of rounded-up value.
• The observations are arranged in increasing order:
• $\left\{\begin{array}{l}-16.9,1.2,1.5,2.1,2.4,3.9,7.1,9.2,\\ 10.2,12.4,14.2,15.7,20.2,26.9,41.1\end{array}\right\}$
• The size of the data is $n=15$.
• For finding first quartile,
•  $\begin{array}{c}L=0.25\times 15\\ =3.75\end{array}$
• Here, 3.75 is not a whole number, hence the 1^st quartile will be the observation in the 4^th position.
• From the arranged observations the first quartile is 2.1.
• For finding third quartile,
• $\begin{array}{c}L=0.75\times 15\\ =11.25\end{array}$
• Here, 11.25 is not a whole number, hence the 3^rd quartile will be the observation in the 12^th position.
• From the arranged observations the third quartile is 15.7.

Hence, the first and the third quartiles are 2.1 and 15.7 respectively.

b.

To determine

Find the median of the data.

Answer to Problem 15RE

The median amount spent on advertising is 9.2.

Explanation of Solution

Calculation:

Median:

Let $x_1,x_2\mathrm{...}{x}_n$ be n values.

The steps for finding the median:

• The all data values should be arranged in ascending order.
• If the total number of data values, n is odd, then the median will be the middle value or if n is even, then the median will be the average of middle two values.
• The observations are arranged in increasing order:
• $\left\{\begin{array}{l}-16.9,1.2,1.5,2.1,2.4,3.9,7.1,9.2,\\ 10.2,12.4,14.2,15.7,20.2,26.9,41.1\end{array}\right\}$
• The size of the data is $n=15$.
• Hence, the sample size is odd. Therefore, the median is the middle value that is 8th observation.
• From the arranged observations the median is 9.2.

Thus, the median of the data is 9.2.

c.

To determine

Find the upper and lower outlier boundaries.

Answer to Problem 15RE

The upper and lower outlier boundaries are –18.3 and 36.1 respectively.

Explanation of Solution

Calculation:

Interquartile range:

• The interquartile range is the difference between the third quartile and first quartile. For detecting outlier this measure can be used.
• Interquartile range can be found as, $IQR=Q_3-Q_1$. Where, the first quartile is denoted by $Q_1$ and third quartile is denoted by $Q_3$.
• From part (a), the first and the third quartiles are 2.1 and 15.7 respectively.
• Substitute these values in the interquartile range formula,
• $\begin{array}{c}IQR=15.7-2.1\\ =13.6\end{array}$
• Outlier boundaries:
• Lower outlier boundary is $Q_1-1.5\times IQR$.
• Upper outlier boundary is $Q_3+1.5\times IQR$.
• Where, the first quartile is denoted by $Q_1$ and third quartile is denoted by $Q_3$.
• Substitute these values in the formulae,
• $\begin{array}{c}\text{Lower outlier boundary}=2.1-1.5\times 13.6\\ =2.1-20.4\\ =-18.3\end{array}$.
•  $\begin{array}{c}\text{Upper outlier boundary}=15.7+\left(1.5\times 13.7\right)\\ =15.7+20.4\\ =36.1\end{array}$

Thus, the upper and lower outlier boundaries are –18.3 and 36.1 respectively.

d.

To determine

Find the outliers.

Answer to Problem 15RE

There is one outliers in the data and is 41.1.

Explanation of Solution

Calculation:

• Condition for outlier:
• If any observation is less than the lower outlier boundary, the observation will be outlier.
• If any observation is greater than the upper outlier boundary, the observation will be outlier.

From part (c), the upper and lower outlier boundaries are –18.3 and 36.1 respectively.

There is one observation which is greater than the upper outlier boundary that is, $41.1>36.1$.

Other observations are greater than the lower outlier boundary and less than the upper outlier boundary.

Hence, there is one outliers in the data.

e.

To determine

Draw a boxplot of the data.

Answer to Problem 15RE

The boxplot is given below,

Explanation of Solution

Calculation:

• Boxplot:

Software procedure:

• Step-by-step procedure to draw a boxplot using the MINITAB software:
• Choose Graph > Boxplot.
• Choose Simple. Click OK.
• In Graph variables, enter the data of profit..
• Click OK.

Output using the MINITAB software is given below:

• From the MINITAB output, it is clear that there is one outlier in the data.