Chapter 4, Problem 14E

Kevin Horn is the national sales manager for National Textbooks Inc. He has a sales staff of 40 who visit college professors all over the United States. Each Saturday morning he requires his sales staff to send him a report. This report includes, among other things, the number of professors visited during the previous week. Listed below, ordered from smallest to largest, are the number of visits last week.

1. a. Determine the median number of calls.
2. b. Determine the first and third quartiles.
3. c. Determine the first decile and the ninth decile.
4. d. Determine the 33rd percentile.

To determine

Find the first quartile.

Find the third quartile.

Answer to Problem 14E

The first quartile is 33.25.

The third quartile is 50.25.

Explanation of Solution

Calculation:

Location of percentile:

The formula for percentile is,

$L_P=\left(n+1\right)\frac{P}{100}$

In the formula, P denotes the location of the percentile and n denotes the total number of observations.

The median is the midpoint of the dataset which is same as the 50^th percentile of the data set.

There are 40 observations in the dataset. The observations are arranged in numerical order as, 38, 40, 41, 45, 48, 48, 50, 50, 51, 51, 52, 52, 53, 54, 55, 55, 55, 56, 56, 57, 59, 59, 59, 62, 62, 62, 63, 64, 65, 66, 66, 67, 67, 69, 69, 71, 77, 78, 79, and 79.

Substitute $P=50,n=40$ in the percentile formula.

$\begin{array}{c}{L}_{50}=\left(40+1\right)\frac{50}{100}\\ =\frac{41\times 50}{100}\\ =\frac{2,050}{100}\\ =20.5\end{array}$

The position of median is the 20.5^th number in the dataset. The average of 20^th and 21^st value in the dataset has to be taken.

$\begin{array}{c}\text{Median}=\frac{{\text{20}}^{\text{th}}\text{value}+{\text{21}}^{\text{st}}\text{value}}{2}\\ =\frac{57+59}{2}\\ =\frac{116}{2}\\ =58\end{array}$

Hence, the median is 58.

To determine

Find the first quartile.

Find the third quartile.

Answer to Problem 14E

The first quartile is 51.25.

The third quartile is 66.0.

Explanation of Solution

Calculation:

First quartile:

The first quartile represents the 25% of observation lies below first quartile. That is $P=25$. There are 40 observations in the dataset arranged in ascending order.

Substitute, $P=25,n=40$ in the percentile formula.

$\begin{array}{c}{L}_{25}=\left(40+1\right)\frac{25}{100}\\ =\frac{41\times 25}{100}\\ =\frac{1,025}{100}\\ =10.25\end{array}$

The position of first quartile is 10.25^th value in the dataset.

$\begin{array}{c}\text{First quartile}={10}^{\text{th}}\text{value}+\left({11}^{\text{th}}\text{value}-{10}^{\text{th}}\text{value}\right)0.25\\ =51+\left(52-51\right)0.25\\ =51+0.25\\ =51.25\end{array}$

Hence, the first quartile is 51.25.

Third quartile:

The third quartile represents the 75% of observation lies above third quartile. That is $P=75$.

Substitute, $P=75,n=40$ in the percentile formula.

$\begin{array}{c}{L}_{75}=\left(40+1\right)\frac{75}{100}\\ =\frac{41\times 75}{100}\\ =\frac{3,075}{100}\\ =30.75\end{array}$

The position of third quartile is 30.75^th value in the dataset.

$\begin{array}{c}\text{Third quartile}={30}^{\text{th}}\text{value}+\left({31}^{\text{st}}\text{value}-{30}^{\text{th}}\text{value}\right)0.75\\ =66+\left(66-66\right)0.75\\ =66+0\\ =66.0\end{array}$

Hence, the third quartile is 66.0.

To determine

Find the first decile.

Find the ninth decile.

Answer to Problem 14E

The first decile is 45.3.

The ninth decile is 76.4.

Explanation of Solution

Calculation:

Location of decile:

The deciles divide the total observation into 10 equal parts. The formula for decile is,

$D_i=\left(n+1\right)\frac{i}{10}$

In the formula, i denotes the location of the decile and n denotes the total number of observations.

First decile:

Substitute, $i=1,n=40$ in the percentile formula.

$\begin{array}{c}{D}_2=\left(40+1\right)\frac{1}{10}\\ =\frac{41\times 1}{10}\\ =\frac{41}{10}\\ =4.1\end{array}$

The position of first decile is 4.1^th value in the dataset.

$\begin{array}{c}{D}_1=4^{\text{th}}\text{value}+\left(5^{\text{th}}\text{value}-4^{\text{th}}\text{value}\right)0.10\\ =45+\left(48-45\right)0.10\\ =45+0.30\\ =45.30\end{array}$

Hence, the first decile is 45.30.

Ninth decile:

Substitute, $i=9,n=40$ in the percentile formula.

$\begin{array}{c}{D}_9=\left(40+1\right)\frac{9}{10}\\ =\frac{41\times 9}{10}\\ =\frac{369}{10}\\ =36.9\end{array}$

The position of ninth decile is 36.9^th value in the dataset.

$\begin{array}{c}{D}_8={36}^{\text{th}}\text{value}+\left({37}^{\text{th}}\text{value}-{36}^{\text{th}}\text{value}\right)0.90\\ =71+\left(77-71\right)0.90\\ =71+5.4\\ =76.4\end{array}$

Hence, the ninth decile is 76.4.

To determine

Find the 33^rd percentile.

Answer to Problem 14E

The 33^rd percentile is 53.53.

Explanation of Solution

Calculation:

Substitute, $P=33,n=40$ in the percentile formula.

$\begin{array}{c}{L}_{67}=\left(40+1\right)\frac{33}{100}\\ =\frac{41\times 33}{100}\\ =\frac{1,353}{100}\\ =13.53\end{array}$

The position of 33^rd percentile is 13.53^th value in the dataset.

$\begin{array}{c}{\text{33}}^{\text{rd}}\text{percentile}={13}^{\text{th}}\text{value}+\left({14}^{\text{th}}\text{value}-{13}^{\text{th}}\text{value}\right)0.53\\ =53+\left(54-53\right)0.53\\ =53+0.53\\ =53.53\end{array}$

Hence, the 33^rd percentile is 53.53.