Question

question 1 of 5
here are the travel times in minutes for 15 workers in north carolina, chosen at random by the census bureau.
30 20 10 40 25 20 10 60 15 40 5 30 12 10 10
identify the boundaries for outliers.
low outliers < minutes high outliers >
which of the following values are outliers, if any?
□ no outliers □ 30
□ 20 □ 10
□ 40 □ 25
□ 20 □ 10
□ 60 □ 15
□ 40 □ 5
□ 30 □ 12
□ 10 □ 10

Explanation:

Step1: Sort the data

First, we sort the travel times in ascending order: \( 5, 10, 10, 10, 10, 12, 15, 20, 20, 25, 30, 30, 40, 40, 60 \)

Step2: Find the median (Q2)

Since there are 15 data points (odd number), the median (Q2) is the 8th value. So, \( Q2 = 20 \)

Step3: Find Q1 (median of lower half)

The lower half of the data (excluding the median) is \( 5, 10, 10, 10, 10, 12, 15 \). There are 7 data points, so the median (Q1) is the 4th value. Thus, \( Q1 = 10 \)

Step4: Find Q3 (median of upper half)

The upper half of the data (excluding the median) is \( 25, 30, 30, 40, 40, 60 \). Wait, no, original upper half (after median) is \( 25, 30, 30, 40, 40, 60 \)? Wait, no, total data points 15, median at 8th (20). So lower half: first 7 (positions 1 - 7): \( 5, 10, 10, 10, 10, 12, 15 \) (7 points, median at 4th: 10). Upper half: positions 9 - 15: \( 25, 30, 30, 40, 40, 60 \)? Wait, no, position 9 is 20? Wait no, sorted data:

Wait let's re - sort:

Wait the data is: 5, 10, 10, 10, 10, 12, 15, 20, 20, 25, 30, 30, 40, 40, 60.

So positions:

1:5, 2:10, 3:10, 4:10, 5:10, 6:12, 7:15, 8:20 (Q2), 9:20, 10:25, 11:30, 12:30, 13:40, 14:40, 15:60.

So lower half (positions 1 - 7): 5, 10, 10, 10, 10, 12, 15. Median (Q1) is the 4th value: 10.

Upper half (positions 9 - 15): 20, 25, 30, 30, 40, 40, 60. Median (Q3) is the 4th value: 30.

Step5: Calculate IQR (Inter - Quartile Range)

\( IQR=Q3 - Q1=30 - 10 = 20 \)

Step6: Calculate boundaries for outliers

Lower boundary for outliers: \( Q1-1.5\times IQR=10 - 1.5\times20=10 - 30=- 20 \)

Upper boundary for outliers: \( Q3 + 1.5\times IQR=30+1.5\times20=30 + 30 = 60 \)

Now, we check each data point:

• For low outliers: We check if any data point is less than - 20. The minimum data point is 5, which is greater than - 20, so no low outliers.

• For high outliers: We check if any data point is greater than 60. The maximum data point is 60, which is equal to the upper boundary, so 60 is not an outlier (since the boundary is \(>60\), and 60 is not greater than 60). Wait, wait, the formula for upper outlier boundary is \( Q3 + 1.5IQR \), and a data point is an outlier if it is \(>Q3 + 1.5IQR\) or \(

Wait our \( Q3 + 1.5IQR=30 + 30 = 60 \). So a data point is a high outlier if it is greater than 60. Our data points are all less than or equal to 60. The minimum data point is 5, and \( Q1-1.5IQR = 10-30=-20 \), and 5 is greater than - 20. So there are no outliers.

Answer:

Low Outliers \(< - 20\) minutes, High outliers \(>60\) minutes. And the answer for outliers is "No outliers".