Boxplot and five-number summary practice problems

By Jude Wallis · Published

This set has 8 problems on quartiles (median-excluded convention), the interquartile range, the 1.5 IQR outlier rule, and reading and comparing boxplots. Solve each on paper, then check your work against the full solution steps.

AP Statistics: Unit 1 (topics 1.7 Summary Statistics for One Quantitative Variable, 1.8 Graphical Representations of Summary Statistics for One Quantitative Variable, 1.9 Comparisons of the Distributions for One Quantitative Variable). These problems cover Unit 1 topics 1.7 (quartiles, IQR, and the 1.5 IQR outlier rule), 1.8 (five-number summary and boxplots), and 1.9 (comparing boxplots) in the Fall 2026 AP Statistics course. Unit 1 is 20 to 30% of the multiple-choice section.

What these problems build

These problems build four core Unit 1 skills: locating the quartiles Q1Q_1 and Q3Q_3, finding the interquartile range IQR=Q3−Q1IQR = Q_3 - Q_1, applying the 1.5 IQR rule to flag outliers, and reading and comparing boxplots. Every problem uses the median-excluded (TI-84) convention: find the median first, then take Q1Q_1 as the median of the values below it and Q3Q_3 as the median of the values above it, leaving the median value itself out of both halves when the number of data values is odd.

Work each problem on paper before opening the steps, and check your five-number summary with the five-number summary calculator. For the underlying methods, see how to make a boxplot and how to find outliers with the 1.5 IQR rule.

Problem 1

A downtown bike-share station logs the number of bikes rented each day for one week: 20, 8, 26, 14, 11, 23, 17. Find the five-number summary and the interquartile range.

Show the worked solution

  1. Sort the values from smallest to largest: 8, 11, 14, 17, 20, 23, 26. There are n=7n = 7 values.

  2. Find the median. With 7 values the median is the 4th value, so the median =17= 17. Because the count is odd, leave this value out of both halves.

  3. Split the data around the median: the lower half is 8, 11, 14 and the upper half is 20, 23, 26.

  4. Find Q1Q_1, the median of the lower half 8, 11, 14. The middle value is 11, so Q1=11Q_1 = 11.

  5. Find Q3Q_3, the median of the upper half 20, 23, 26. The middle value is 23, so Q3=23Q_3 = 23.

  6. Read off the minimum (8) and the maximum (26), then compute IQR=Q3−Q1=23−11=12IQR = Q_3 - Q_1 = 23 - 11 = 12.

Five-number summary: minimum 8, Q1=11Q_1 = 11, median 17, Q3=23Q_3 = 23, maximum 26; IQR=12IQR = 12 bikes.

Problem 2

A ceramics studio records the number of mugs fired in the kiln during each of 8 sessions: 14, 3, 8, 19, 5, 16, 7, 12. Using the median-excluded convention, find Q1Q_1, the median, Q3Q_3, and the IQR.

Show the worked solution

  1. Sort the values: 3, 5, 7, 8, 12, 14, 16, 19. There are n=8n = 8 values.

  2. Find the median. With an even count it is the average of the 4th and 5th values: (8+12)/2=20/2=10(8 + 12)/2 = 20/2 = 10.

  3. The median falls between two values, so the lower half is the first four values 3, 5, 7, 8 and the upper half is the last four values 12, 14, 16, 19.

  4. Find Q1Q_1, the median of 3, 5, 7, 8: (5+7)/2=12/2=6(5 + 7)/2 = 12/2 = 6.

  5. Find Q3Q_3, the median of 12, 14, 16, 19: (14+16)/2=30/2=15(14 + 16)/2 = 30/2 = 15.

  6. Compute the spread: IQR=Q3−Q1=15−6=9IQR = Q_3 - Q_1 = 15 - 6 = 9.

Q1=6Q_1 = 6, median =10= 10, Q3=15Q_3 = 15, and IQR=9IQR = 9 mugs.

Problem 3

During a lunch rush a food truck records the number of minutes to serve each of 9 orders: 10, 30, 6, 9, 5, 12, 8, 11, 7. Use the 1.5 IQR rule to decide whether any order time is an outlier.

Show the worked solution

  1. Sort the values: 5, 6, 7, 8, 9, 10, 11, 12, 30. There are n=9n = 9 values.

  2. Find the median. With 9 values it is the 5th value, so the median =9= 9. Because the count is odd, leave this value out of both halves.

  3. Lower half: 5, 6, 7, 8. Then Q1=(6+7)/2=13/2=6.5Q_1 = (6 + 7)/2 = 13/2 = 6.5.

  4. Upper half: 10, 11, 12, 30. Then Q3=(11+12)/2=23/2=11.5Q_3 = (11 + 12)/2 = 23/2 = 11.5.

  5. Compute IQR=11.5−6.5=5IQR = 11.5 - 6.5 = 5, then 1.5×IQR=1.5×5=7.51.5 \times IQR = 1.5 \times 5 = 7.5.

  6. Set the fences: lower fence =6.5−7.5=−1= 6.5 - 7.5 = -1 and upper fence =11.5+7.5=19= 11.5 + 7.5 = 19.

  7. Compare each value. The 30-minute order is above the upper fence of 19, while every other value sits between −1-1 and 19.

The 30-minute order is an outlier; it lies above the upper fence of 19 minutes.

Problem 4

A greenhouse counts the ripe tomatoes picked from each of 10 plants: 20, 16, 1, 22, 18, 25, 14, 21, 17, 23. Use the 1.5 IQR rule to identify any outliers.

Show the worked solution

  1. Sort the values: 1, 14, 16, 17, 18, 20, 21, 22, 23, 25. There are n=10n = 10 values.

  2. Find the median: the average of the 5th and 6th values, (18+20)/2=38/2=19(18 + 20)/2 = 38/2 = 19.

  3. Lower half (first five values): 1, 14, 16, 17, 18. Then Q1Q_1 is the middle value, 16.

  4. Upper half (last five values): 20, 21, 22, 23, 25. Then Q3Q_3 is the middle value, 22.

  5. Compute IQR=22−16=6IQR = 22 - 16 = 6, then 1.5×IQR=1.5×6=91.5 \times IQR = 1.5 \times 6 = 9.

  6. Set the fences: lower fence =16−9=7= 16 - 9 = 7 and upper fence =22+9=31= 22 + 9 = 31.

  7. Compare each value. The plant with 1 tomato is below the lower fence of 7, while the largest count, 25, stays below the upper fence of 31.

The plant with 1 tomato is a low outlier; there are no high outliers (fences at 7 and 31).

Problem 5

A librarian draws a modified boxplot of the number of books checked out each hour. The box runs from Q1=20Q_1 = 20 to Q3=34Q_3 = 34 with the median line at 26, the lower whisker reaches the minimum of 12, the upper whisker ends at 40, and a single point is plotted at 58. (a) What is the IQR? (b) About what percent of hours had between 20 and 34 checkouts? (c) Confirm that the point at 58 is an outlier.

Show the worked solution

  1. Read the quartiles from the ends of the box: Q1=20Q_1 = 20 and Q3=34Q_3 = 34.

  2. Part (a): IQR=Q3−Q1=34−20=14IQR = Q_3 - Q_1 = 34 - 20 = 14.

  3. Part (b): the values 20 and 34 are Q1Q_1 and Q3Q_3, and the box always holds the middle 50% of the data, so about 50% of hours fall between 20 and 34 checkouts.

  4. Part (c): compute 1.5×IQR=1.5×14=211.5 \times IQR = 1.5 \times 14 = 21.

  5. Find the upper fence: Q3+21=34+21=55Q_3 + 21 = 34 + 21 = 55.

  6. Because 58>5558 > 55, the point at 58 lies beyond the upper fence, which is why it is drawn as a separate point rather than inside a whisker.

(a) IQR=14IQR = 14; (b) about 50% of hours; (c) 58>5558 > 55, so 58 is an outlier.

Problem 6

A robotics club records the number of code commits each of its 11 members made in one week: 16, 8, 21, 13, 45, 10, 19, 12, 22, 15, 18. (a) Give the five-number summary. (b) Use the 1.5 IQR rule to test for outliers. (c) Compare the mean and median to describe the shape.

Show the worked solution

  1. Sort the values: 8, 10, 12, 13, 15, 16, 18, 19, 21, 22, 45. There are n=11n = 11 values.

  2. The median is the 6th value, 16. Leave it out of both halves because the count is odd.

  3. Lower half: 8, 10, 12, 13, 15, so Q1Q_1 is the middle value, 12. Upper half: 18, 19, 21, 22, 45, so Q3Q_3 is the middle value, 21.

  4. Part (a) five-number summary: minimum 8, Q1=12Q_1 = 12, median 16, Q3=21Q_3 = 21, maximum 45.

  5. Part (b): IQR=21−12=9IQR = 21 - 12 = 9 and 1.5×IQR=1.5×9=13.51.5 \times IQR = 1.5 \times 9 = 13.5, so the upper fence is 21+13.5=34.521 + 13.5 = 34.5 and the lower fence is 12−13.5=−1.512 - 13.5 = -1.5. The value 45 is above 34.5, so it is an outlier.

  6. Part (c): the sum of all values is 8+10+12+13+15+16+18+19+21+22+45=1998 + 10 + 12 + 13 + 15 + 16 + 18 + 19 + 21 + 22 + 45 = 199, so the mean is xˉ=199/11=18.09\bar{x} = 199/11 = 18.09 (to 2 decimal places), while the median is 16.

  7. Because the mean 18.09 is larger than the median 16 and one large value sits far to the right, the distribution is skewed to the right.

(a) 8, 12, 16, 21, 45; (b) 45 is an outlier (fences at −1.5-1.5 and 34.5); (c) mean 18.09 > median 16, so skewed right.

Problem 7

A cafe records the number of oat-milk lattes sold each hour over an 8-hour day: 14, 6, 17, 10, 19, 9, 15, 12. What is the smallest whole number of lattes in an hour that the 1.5 IQR rule would flag as a high outlier?

Show the worked solution

  1. Sort the values: 6, 9, 10, 12, 14, 15, 17, 19. There are n=8n = 8 values.

  2. The median is (12+14)/2=26/2=13(12 + 14)/2 = 26/2 = 13, so the lower half is 6, 9, 10, 12 and the upper half is 14, 15, 17, 19.

  3. Find the quartiles: Q1=(9+10)/2=19/2=9.5Q_1 = (9 + 10)/2 = 19/2 = 9.5 and Q3=(15+17)/2=32/2=16Q_3 = (15 + 17)/2 = 32/2 = 16.

  4. Compute IQR=16−9.5=6.5IQR = 16 - 9.5 = 6.5, then 1.5×IQR=1.5×6.5=9.751.5 \times IQR = 1.5 \times 6.5 = 9.75.

  5. Find the upper fence: Q3+9.75=16+9.75=25.75Q_3 + 9.75 = 16 + 9.75 = 25.75. A high outlier must exceed this value.

  6. The smallest whole number greater than 25.75 is 26, so an hour would need at least 26 lattes to be flagged.

26 lattes; the upper fence is 25.75, and 26 is the smallest whole number above it.

Problem 8

Boxplots compare the travel times (in minutes) for two bus routes to campus. Route A: minimum 8, Q1=15Q_1 = 15, median 18, Q3=22Q_3 = 22, maximum 30. Route B: minimum 10, Q1=16Q_1 = 16, median 24, Q3=28Q_3 = 28, upper whisker ending at 38, with a single point plotted at 52. (a) Which route is typically faster, and by how much? (b) Which route is more consistent? (c) Confirm whether the 52-minute day on Route B is an outlier.

Show the worked solution

  1. Part (a): compare the medians. Route A has median 18 and Route B has median 24, so Route A is typically faster by 24−18=624 - 18 = 6 minutes.

  2. Part (b): compare the IQRs. Route A has IQR=22−15=7IQR = 22 - 15 = 7 minutes.

  3. Route B has IQR=28−16=12IQR = 28 - 16 = 12 minutes.

  4. Because 7<127 < 12, Route A has the smaller spread across its middle 50%, so Route A is more consistent.

  5. Part (c): for Route B, 1.5×IQR=1.5×12=181.5 \times IQR = 1.5 \times 12 = 18, so the upper fence is 28+18=4628 + 18 = 46.

  6. Because 52>4652 > 46, the 52-minute day lies beyond the upper fence and is an outlier.

(a) Route A, by 6 minutes; (b) Route A (IQR 7 vs 12); (c) 52>4652 > 46, so the 52-minute day is an outlier.