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The five-number summary of a data set is given below.

Minimum: 3 Q1: 12 Median: 15 Q3: 16 Maximum: 20

Which of the following is a true statement about outliers for this distribution?

The distribution’s maximum is an outlier.
The distribution’s minimum is an outlier.
There are no outliers for this distribution.
Distribution outliers cannot be determined.

User NYCeyes
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1 Answer

5 votes

To determine if there are any outliers in a distribution, we can use the interquartile range (IQR) method. The IQR is calculated by subtracting the first quartile (Q1) from the third quartile (Q3).

IQR = Q3 - Q1

In this case, the first quartile (Q1) is 12 and the third quartile (Q3) is 16. Therefore, the IQR is:

IQR = 16 - 12

= 4

To identify outliers, we can use the following rule: any value that is more than 1.5 times the IQR below Q1 or above Q3 is considered an outlier.

In this case, the lower bound for outliers is Q1 - 1.5 * IQR and the upper bound for outliers is Q3 + 1.5 * IQR.

Lower bound = 12 - 1.5 * 4

= 12 - 6

= 6

Upper bound = 16 + 1.5 * 4

= 16 + 6

= 22

Since the minimum value is 3 and the maximum value is 20, neither of them fall outside the lower and upper bounds. Therefore, there are no outliers for this distribution.

Hence, the correct statement is: There are no outliers for this distribution.

User Kjaquier
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