Question

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

Explanation:

Step1: Recall the outlier formula for upper bound

To determine if a value is an outlier, we use the formula for the upper fence: \( Q_3 + 1.5 \times \text{IQR} \). We know \( \text{IQR} = 4 \), and from the data set, we can find \( Q_3 \). First, let's find the quartiles. The data set is \( 2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12 \). The median (second quartile \( Q_2 \)) of 11 data points is the 6th value, which is 5. The upper half (for \( Q_3 \)) is \( 5, 6, 7, 9, 12 \), so the median of the upper half ( \( Q_3 \)) is the 3rd value, which is 7.

Step2: Calculate the upper fence

Now, calculate the upper fence: \( Q_3 + 1.5 \times \text{IQR} = 7 + 1.5\times4 = 7 + 6 = 13 \).

Step3: Compare 12 with the upper fence

We check if 12 is greater than the upper fence. Since \( 12 < 13 \) (and \( 13 = 7 + 6 \), because \( 1.5\times4 = 6 \)), we see that 12 is less than the sum of \( Q_3 = 7 \) and \( 1.5\times\text{IQR}=6 \). So, 12 is not an outlier because it is less than the sum of 7 and 6.

Answer:

Twelve is not an outlier because it is less than the sum of 7 and 6. (The corresponding option, e.g., if it's the fourth option: D. Twelve is not an outlier because it is less than the sum of 7 and 6)