Chapter 10.2, Problem 7IP

To determine

To find: The measure of variability and any outliers for data.

Answer to Problem 7IP

Range =15

Quartiles = 18,25

Interquartile range = 7

No outliers.

Explanation of Solution

Given Information:A data is given for measure of variability.

Data set $\{15,17,18,18,20,21,22,23,25,25,27,30\}$

Range: The range subtract the highest value from the lowest

Highest = 30

Lowest = 15

Range = 30-15 = 15

Quartiles:- The quartiles divide the set into two halves. Don’t include the median if there is an odd number of items.

  $\{15,17,18,18,20,21\},\{22,23,25,25,27,30\}$

The first quartile is the median of the lower half. First quartile = 18

The third quartile is the median of the upper half. Third quartile = 25

The interquartile range is the difference between the quartiles.

Interquartile range = 25-18 =7.

Outliners:- to find the outliners. The first multiply the interquartile range by 1.5 = $7\times 1.5=10.5$

Subtract 10.5 from the first quartile and add 10.5 to the third quartile

  $18-10.5=7.5,25+10.5=35.5$

Any value that is less than 7.5 or greater than 35.5 is an outlier

  $\{15,17,18,18,20,21,22,23,25,25,27,30\}$

There is no outliers.