Problem: Ten students recorded their commute times in minutes: 5, 8, 10, 12, 12, 14, 15, 18, 25, 41. Describe the distribution.
Step 1: Center. The mean is
xˉ=160/10=16.0 minutes. With 10 values, the median is the average of the 5th and 6th:
(12+14)/2=13 minutes.
Step 2: Quartiles and IQR. Q1 is the median of the lower five values (5, 8, 10, 12, 12), so
Q1=10.
Q3 is the median of the upper five (14, 15, 18, 25, 41), so
Q3=18.
IQR=18−10=8 minutes.
Step 3: Outliers. 1.5×8=12. The fences are
10−12=−2 and
18+12=30. The value 41 is above 30, so it is an outlier.
Step 4: Other spread. sx≈10.37 minutes; range
=41−5=36 minutes.
Step 5: Describe. The distribution of commute times is skewed right with a high outlier at 41 minutes. Because of the skew and outlier, use resistant measures: median 13 minutes, IQR 8 minutes. The mean (16.0) is above the median, as the right tail predicts.
Compare in the simulator: set Skew to 1.00 for the same pattern on a larger data set: mode below median below mean (45.28, 50.00, 53.02), and SD (10.00) above IQR (9.58).