the interquartile range of the data set is 4.\n2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12\nwhich explains whether or…

the interquartile range of the data set is 4.\n2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12\nwhich explains whether or not 12 is an outlier?\n○ twelve is an outlier because it is greater than the sum of 7 and 4.\n○ twelve is an outlier because it is less than the sum of 7 and 6.\n○ twelve is not an outlier because it is greater than the sum of 7 and 4.\n○ twelve is not an outlier because it is less than the sum of 7 and 6.

the interquartile range of the data set is 4.\n2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12\nwhich explains whether or not 12 is an outlier?\n○ twelve is an outlier because it is greater than the sum of 7 and 4.\n○ twelve is an outlier because it is less than the sum of 7 and 6.\n○ twelve is not an outlier because it is greater than the sum of 7 and 4.\n○ twelve is not an outlier because it is less than the sum of 7 and 6.

Answer

Explanation:

Step1: Recall the outlier formula

To determine if a value is an outlier, we use the formula for upper outliers: A value is an outlier if it is greater than ( Q_3 + 1.5 \times \text{IQR} ). We know ( \text{IQR} = 4 ), and we need to find ( Q_3 ). First, find the median of the data set. The data set is ( 2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12 ) (11 values). The median (second quartile ( Q_2 )) is the 6th value, which is 5. The upper half of the data (above ( Q_2 )) is ( 5, 6, 7, 9, 12 ) (5 values). The median of this upper half is ( Q_3 ), which is the 3rd value, so ( Q_3 = 7 ).

Step2: Calculate the upper bound for outliers

Using the formula ( \text{Upper Bound} = Q_3 + 1.5 \times \text{IQR} ). Substitute ( Q_3 = 7 ) and ( \text{IQR} = 4 ): ( \text{Upper Bound} = 7 + 1.5\times4 = 7 + 6 = 13 ). Wait, but the options mention sums with 4 or 6. Wait, maybe a simpler way: the formula can also be thought of as ( Q_3 + 1.5\times\text{IQR} ), but ( 1.5\times\text{IQR}=1.5\times4 = 6 ), so ( Q_3 + 1.5\times\text{IQR}=7 + 6 = 13 ). Now, check the value 12. Since ( 12 < 13 ) (the upper bound), we check the options. The options talk about sums: let's see the options. The last option says "Twelve is not an outlier because it is less than the sum of 7 and 6." Since ( 7 + 6 = 13 ), and 12 < 13, so 12 is not an outlier because it is less than ( Q_3 + 1.5\times\text{IQR} = 7 + 6 = 13 ).

Answer:

D. Twelve is not an outlier because it is less than the sum of 7 and 6.