Question

The following data represent the weights in pounds of a sample of 25 police officers:

Answer 1

Given:

The data values are:

164, 148, 137, 157, 173, 156, 177, 172, 169, 165, 145, 168, 163, 162, 174, 152, 156, 168, 154, 151, 174, 146, 134, 140, and 171.

To find;

a. Lower quartile.

b. Upper quartile.

c. Interquartile range.

Solution:

We have,

164, 148, 137, 157, 173, 156, 177, 172, 169, 165, 145, 168, 163, 162, 174, 152, 156, 168, 154, 151, 174, 146, 134, 140, 171.

Arrange the data values in ascending order.

134, 137, 140, 145, 146, 148, 151, 152, 154, 156, 156, 157, 162, 163, 164, 165, 168, 168, 169, 171, 172, 173, 174, 174, 177.

Divide the data values in two equal parts.

(134, 137, 140, 145, 146, 148, 151, 152, 154, 156, 156, 157), 162, (163, 164, 165, 168, 168, 169, 171, 172, 173, 174, 174, 177)

Divide each parentheses in two equal parts.

(134, 137, 140, 145, 146, 148), (151, 152, 154, 156, 156, 157), 162, (163, 164, 165, 168, 168, 169), (171, 172, 173, 174, 174, 177)

a. Location of lower quartile is:

$Q_1=\dfrac{1}{4}(n+1)\text{th term}$

$Q_1=\dfrac{1}{4}(25+1)\text{th term}$

$Q_1=\dfrac{26}{4}\text{th term}$

$Q_1=6.5\text{th term}$

The lower quartile of the weights is:

$Q_1=\dfrac{148+151}{2}$

$Q_1=\dfrac{299}{2}$

$Q_1=149.5$

Therefore, the location of the lower quartile of the weights is between 6th term and the 7th term. The value of the lower quartile is 149.5.

b. Location of upper quartile is:

$Q_3=\dfrac{3}{4}(n+1)\text{th term}$

$Q_3=\dfrac{3}{4}(25+1)\text{th term}$

$Q_3=\dfrac{3\cdot 26}{4}\text{th term}$

$Q_3=19.5\text{th term}$

The upper quartile of the weights is:

$Q_3=\dfrac{169+171}{2}$

$Q_3=\dfrac{340}{2}$

$Q_3=170$

Therefore, the location of the upper quartile of the weights is between 19th term and the 20th term. The value of the upper quartile is 170.

c. The interquartile range of the given data set is:

$IQR=Q_3-Q_1$

$IQR=170-149.5$

$IQR=20.5$

Therefore, the interquartile range of the weights is 20.5.