Chapter 12.4, Problem 18P

Make box-and-whisker plots to compare the data sets

Test scores;

Andrew’s: 79 80 87 87 99 94 77 86

Dipak s: 93 79 78 82 91 87 80 99

To determine

Draw the box-and-whisker plots to compare the given data sets.

Answer to Problem 18P

  

Explanation of Solution

Given:

Test scores:

A’s : 79 , 80 , 87 , 87 , 99 , 94 , 77 , 86

D’s : 93 , 79 , 78 , 82 , 91 , 87 , 80 , 99

Calculation: ,

A’s test scores:

Arrange the data in ascending order:

77 , 79 , 80 , 86 , 87 , 87 , 94 , 99

The minimum is 77.

The first quartile is the median of the lower half of the data set.

Since the total entries are 8, which is even , So, the lower half is the first $\frac{8}{2}=4$ entries.

So , the first quartile is $\frac{79+80}{2}=79.5$

Median = The mean of $\frac{8}{2}=4^{th}\&5^{th}$ data values = $\frac{86+87}{2}=86.5$

The third quartile is the median of the upper half of the data set.

Since the total entries are 8, which is even , So, the upper half is the last $\frac{8}{2}=4$ entries.

So , the third quartile is $\frac{87+94}{2}=90.5$

The maximum is 99.

D’s test scores:

Arrange the data in ascending order:

78 , 79 , 80 , 82 , 87 , 91 , 93 , 99

The minimum is 78.

The first quartile is the median of the lower half of the data set.

Since the total entries are 8, which is even , So, the lower half is the first $\frac{8}{2}=4$ entries.

So , the first quartile is $\frac{79+80}{2}=79.5$

Median = The mean of $\frac{8}{2}=4^{th}\&5^{th}$ data values = $\frac{82+87}{2}=84.5$

The third quartile is the median of the upper half of the data set.

Since the total entries are 8, which is even , So, the upper half is the last $\frac{8}{2}=4$ entries.

So , the third quartile is $\frac{91+93}{2}=92$

The maximum is 99.

Represent the above details on the number line. Draw the box through the first and third quartile and make a vertical line through the median , and whisker from the minimum to the maximum data entry.

  

Since the box of D’s test score is longer than that of A’s test score . So, D has more interquartile range.That is the median of D’s test score varies more widely than the median or middle test score of A.