Quartiles, IQR, and outliers practice problems

By Jude Wallis · Published July 30, 2026

These 8 problems build the five-number summary, the interquartile range, and the 1.5 IQR outlier rule, using the median-excluded (TI-84) quartile convention. Every solution writes out the sorted list first, then shows each arithmetic step.

AP Statistics: Unit 1 (topics 1.7 Summary Statistics for One Quantitative Variable, 1.8 Graphical Representations of Summary Statistics for One Quantitative Variable). These problems cover Unit 1 topics 1.7 (quartiles, the interquartile range, and other measures of center and variability) and 1.8 (the five-number summary and boxplots) in the Fall 2026 AP Statistics course, where Unit 1 carries 20 to 30% of the multiple-choice section.

What these problems build

These 8 problems build three linked Unit 1 skills: the five-number summary (minimum, $Q_1$ , median, $Q_3$ , maximum), the interquartile range $IQR = Q_3 - Q_1$ , and the 1.5 IQR rule for flagging outliers. Problems 1 and 2 are routine summaries for an odd and an even number of values. Problems 3 and 4 apply the outlier rule, problem 5 works backwards from a stated median or quartile, and problem 6 shows what one extreme value does to the mean and does not do to the median. Problems 7 and 8 pull the pieces together into a modified boxplot and an AP-style multi-part question.

Two conventions run through every solution. Quartiles follow the median-excluded (TI-84) rule: find the median first, then take $Q_1$ as the median of the values strictly below it and $Q_3$ as the median of the values strictly above it, leaving the median value out of both halves when the count is odd. A value is an outlier when it falls more than $1.5 \times IQR$ below $Q_1$ or more than $1.5 \times IQR$ above $Q_3$ , and a value sitting exactly on a fence is not flagged. Every solution writes the sorted list out first, because almost every error in this topic starts with an unsorted list.

For the methods behind these, see how to find quartiles by hand, the 1.5 IQR outlier rule, and how to make a boxplot. Check your work with the five-number summary calculator, and move values around to watch the quartiles shift in the descriptive statistics sandbox. The matching course pages are topic 1.7 and topic 1.8.

Problem 1

A climbing gym logs the number of routes each of 9 members completed during one Tuesday session: 12, 5, 9, 14, 7, 11, 6, 10, 8. Give the five-number summary and the interquartile range.

Show the worked solution

Sort the values from smallest to largest: 5, 6, 7, 8, 9, 10, 11, 12, 14. There are $n = 9$ members.
Find the median. With 9 values the median is the 5th value in the sorted list, so the median is 9 routes. The count is odd, so this value is left out of both halves.
Lower half: 5, 6, 7, 8. Its median is $Q_1 = (6 + 7)/2 = 13/2 = 6.5$ routes.
Upper half: 10, 11, 12, 14. Its median is $Q_3 = (11 + 12)/2 = 23/2 = 11.5$ routes.
Read the extremes off the sorted list: the minimum is 5 routes and the maximum is 14 routes.
Compute the spread: $IQR = Q_3 - Q_1 = 11.5 - 6.5 = 5$ routes. For comparison the range is $14 - 5 = 9$ routes.

Five-number summary: 5, 6.5, 9, 11.5, 14 routes; $IQR = 5$ routes.

Problem 2

A beekeeper weighs the honey harvested from each of 12 hives, in kilograms: 31, 18, 27, 22, 35, 20, 29, 24, 33, 26, 21, 28. Give the five-number summary and the interquartile range.

Show the worked solution

Sort the values: 18, 20, 21, 22, 24, 26, 27, 28, 29, 31, 33, 35. There are $n = 12$ hives.
Find the median. With an even count it is the average of the 6th and 7th values: $(26 + 27)/2 = 53/2 = 26.5$ kg.
The median falls between two data values, so no value is left out: the lower half is the first six values 18, 20, 21, 22, 24, 26 and the upper half is the last six values 27, 28, 29, 31, 33, 35.
$Q_1$ is the median of the lower half: $(21 + 22)/2 = 43/2 = 21.5$ kg.
$Q_3$ is the median of the upper half: $(29 + 31)/2 = 60/2 = 30$ kg.
The minimum is 18 kg and the maximum is 35 kg, so $IQR = Q_3 - Q_1 = 30 - 21.5 = 8.5$ kg.

Five-number summary: 18, 21.5, 26.5, 30, 35 kg; $IQR = 8.5$ kg.

Problem 3

A print shop records the number of minutes it took to finish each of 11 rush orders: 22, 19, 25, 21, 60, 23, 20, 26, 24, 18, 27. (a) Find $Q_1$ , $Q_3$ , and the IQR. (b) Compute the 1.5 IQR fences and decide which orders, if any, are outliers.

Show the worked solution

Sort the values: 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 60. There are $n = 11$ orders.
The median is the 6th value, 23 minutes. Because the count is odd, leave it out of both halves.
(a) Lower half: 18, 19, 20, 21, 22. Its middle value is $Q_1 = 20$ minutes. Upper half: 24, 25, 26, 27, 60. Its middle value is $Q_3 = 26$ minutes.
(a) $IQR = Q_3 - Q_1 = 26 - 20 = 6$ minutes.
(b) $1.5 \times IQR = 1.5 \times 6 = 9$ minutes.
(b) Lower fence: $Q_1 - 9 = 20 - 9 = 11$ minutes. Upper fence: $Q_3 + 9 = 26 + 9 = 35$ minutes.
(b) Compare every value with the fences. The fastest order, 18 minutes, sits above the lower fence of 11, and every order from 18 to 27 minutes sits inside both fences. Only the 60-minute order exceeds the upper fence of 35.

(a) $Q_1 = 20$ minutes, $Q_3 = 26$ minutes, $IQR = 6$ minutes. (b) Fences at 11 and 35 minutes; the 60-minute order is the only outlier.

Problem 4

A trailhead counter at a state park records the number of hikers who start the loop each day for 12 days: 45, 40, 2, 47, 44, 90, 43, 50, 41, 46, 52, 48. An ice storm closed the access road on one of those days, and another was the day of the trail festival. Use the 1.5 IQR rule to find every outlier.

Show the worked solution

Sort the values: 2, 40, 41, 43, 44, 45, 46, 47, 48, 50, 52, 90. There are $n = 12$ days.
The median is the average of the 6th and 7th values: $(45 + 46)/2 = 91/2 = 45.5$ hikers.
The lower half is the first six values 2, 40, 41, 43, 44, 45, so $Q_1 = (41 + 43)/2 = 84/2 = 42$ hikers.
The upper half is the last six values 46, 47, 48, 50, 52, 90, so $Q_3 = (48 + 50)/2 = 98/2 = 49$ hikers.
$IQR = 49 - 42 = 7$ hikers, so $1.5 \times IQR = 1.5 \times 7 = 10.5$ hikers.
Lower fence: $42 - 10.5 = 31.5$ hikers. Upper fence: $49 + 10.5 = 59.5$ hikers.
Check both tails. At the low end the ice-storm day at 2 hikers is below 31.5, so it is a low outlier, while the next smallest day at 40 hikers is well above the fence. At the high end the festival day at 90 hikers is above 59.5, so it is a high outlier, while the next largest day at 52 hikers is below the fence.
Notice what having one outlier on each end does to the mean. The 12 values sum to 548, so $\bar{x} = 548/12 = 45.67$ hikers (x-bar is the sample mean), only 0.17 above the median of 45.5. The two outliers pull the mean in opposite directions and nearly cancel, so a mean close to the median is no evidence that a data set is free of outliers.

Fences at 31.5 and 59.5 hikers: the 2-hiker ice-storm day is a low outlier and the 90-hiker festival day is a high outlier. ( $Q_1 = 42$ , median 45.5, $Q_3 = 49$ , $IQR = 7$ hikers.)

Problem 5

A wildlife rehabilitation center logs how many animals it admits each week, but two entries in the logbook have been smudged.

(a) In one 7-week stretch the readable entries are 12, 15, 9, 18, 11, and 16, and the center's report states that the median for those 7 weeks was 14 animals. Find the smudged value.

(b) In a different 8-week stretch the readable entries are 5, 8, 10, 13, 16, 19, and 24, and the report states that $Q_3$ for those 8 weeks was 18 animals. Find the smudged value.

Show the worked solution

(a) Sort the six readable values: 9, 11, 12, 15, 16, 18. Call the smudged value $x$ . The full stretch has $n = 7$ values, so the median is the 4th value in sorted order.
(a) Test where $x$ could sit. If $x \le 12$ , the four smallest values are $x$ along with 9, 11, and 12, so the 4th value is 12 and the median would be 12, not 14. If $x \ge 15$ , the three smallest are 9, 11, 12 and the 4th value is 15, so the median would be 15, not 14.
(a) That leaves $12 \le x \le 15$ , where the sorted list is 9, 11, 12, $x$ , 15, 16, 18 and the 4th value is $x$ itself. Setting $x = 14$ produces the stated median.
(a) Check it. With $x = 14$ the sorted list is 9, 11, 12, 14, 15, 16, 18, and the 4th of 7 values is 14 animals, as reported.
(b) Sort the seven readable values: 5, 8, 10, 13, 16, 19, 24. Call the smudged value $y$ . With it included there are $n = 8$ weeks, so the upper half is the last four values and $Q_3$ is the average of the 6th and 7th values in sorted order.
(b) Test where $y$ could sit. If $y \le 16$ , the 6th and 7th values are 16 and 19, giving $Q_3 = (16 + 19)/2 = 17.5$ , not 18. If $y \ge 19$ , the 6th value is 19 and the 7th is at least 19, so $Q_3$ would be at least $(19 + 19)/2 = 19$ , above 18.
(b) That leaves $16 \le y \le 19$ , where the sorted list is 5, 8, 10, 13, 16, $y$ , 19, 24. Then $Q_3 = (y + 19)/2 = 18$ , so $y + 19 = 36$ and $y = 17$ animals, which does fall between 16 and 19.
(b) Check it. With $y = 17$ the sorted list is 5, 8, 10, 13, 16, 17, 19, 24. The upper half is 16, 17, 19, 24, whose median is $(17 + 19)/2 = 18$ animals, as reported.

(a) 14 animals. (b) 17 animals.

Problem 6

A used bookstore records the number of books brought in at each of 10 trade-ins one week: 3, 5, 6, 8, 9, 11, 12, 14, 16, 16. (a) Find the mean and the median. (b) An eleventh trade-in arrives from an estate sale with 210 books. Recompute the mean, the median, and the IQR using all 11 trade-ins. (c) Explain what the two sets of numbers show about the mean and the median.

Show the worked solution

(a) Sort the first 10 values: 3, 5, 6, 8, 9, 11, 12, 14, 16, 16. The sum is $3 + 5 + 6 + 8 + 9 + 11 + 12 + 14 + 16 + 16 = 100$ books, so $\bar{x} = 100/10 = 10$ books, where $\bar{x}$ (x-bar) is the sample mean.
(a) With 10 values the median is the average of the 5th and 6th: $(9 + 11)/2 = 20/2 = 10$ books. Here the mean and the median agree exactly.
(b) Add the estate sale and sort again: 3, 5, 6, 8, 9, 11, 12, 14, 16, 16, 210. There are now $n = 11$ trade-ins.
(b) New mean: the sum is $100 + 210 = 310$ books, so $\bar{x} = 310/11 = 28.18$ books to two decimal places.
(b) New median: with 11 values it is the 6th value in sorted order, which is 11 books.
(b) New quartiles. The count is odd, so leave the median out. The lower half is 3, 5, 6, 8, 9, giving $Q_1 = 6$ , and the upper half is 12, 14, 16, 16, 210, giving $Q_3 = 16$ . Then $IQR = 16 - 6 = 10$ books. Checking the rule, $1.5 \times 10 = 15$ , so the upper fence is $16 + 15 = 31$ books and the 210-book trade-in is an outlier.
(c) Line the two sets up. The mean went from 10 books to 28.18 books, nearly tripling, because it adds in every value and 210 is more than 13 times the next largest trade-in. The median went from 10 books to 11 books, a move of one book, because it depends only on which value lands in the middle position.
(c) The mean of 28.18 books is larger than 10 of the 11 trade-ins, so it describes none of them. That is the general point: the median is resistant to an extreme value and the mean is not, so report the median and the IQR when a distribution carries an outlier.

(a) Mean 10 books, median 10 books. (b) Mean 28.18 books, median 11 books, $IQR = 10$ books, and the 210-book trade-in is an outlier above the 31-book fence. (c) One extreme value nearly tripled the mean but moved the median by 1 book, because the mean uses every value while the median uses only position.

Problem 7

A library branch records the number of new library cards issued each day for 15 days: 12, 5, 34, 9, 15, 2, 19, 11, 41, 8, 14, 6, 17, 10, 13. (a) Give the five-number summary. (b) Identify any outliers with the 1.5 IQR rule. (c) Describe exactly what a modified boxplot of these data looks like: where the box ends, where the median line sits, where each whisker stops, and what is plotted separately. (d) How would a boxplot drawn without the outlier rule differ?

Show the worked solution

Sort the values: 2, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 17, 19, 34, 41. There are $n = 15$ days.
(a) The median is the 8th value, 12 cards. The count is odd, so leave it out of both halves.
(a) Lower half: 2, 5, 6, 8, 9, 10, 11. Its middle value is $Q_1 = 8$ cards. Upper half: 13, 14, 15, 17, 19, 34, 41. Its middle value is $Q_3 = 17$ cards. The five-number summary is 2, 8, 12, 17, 41 cards.
(b) $IQR = 17 - 8 = 9$ cards, so $1.5 \times IQR = 1.5 \times 9 = 13.5$ cards. The fences are $8 - 13.5 = -5.5$ cards and $17 + 13.5 = 30.5$ cards.
(b) No day can fall below $-5.5$ cards, so there are no low outliers. Two days, 34 cards and 41 cards, sit above the upper fence of 30.5, so both are high outliers.
(c) Build the plot piece by piece. The box runs from $Q_1 = 8$ cards to $Q_3 = 17$ cards, with a line drawn inside it at the median of 12 cards.
(c) The lower whisker runs from the edge of the box at 8 cards down to the smallest value that is not an outlier. The minimum of 2 cards is inside the fences, so that whisker reaches 2 cards.
(c) The upper whisker runs from the edge of the box at 17 cards out to the largest value that is not an outlier. Skipping 34 and 41, that value is 19 cards, so the whisker stops at 19. The two outliers are then drawn as separate marks at 34 cards and at 41 cards.
(d) A boxplot drawn without the rule keeps the same box and the same median line, because the quartiles do not change. Its upper whisker runs all the way to the maximum of 41 cards and nothing is plotted separately, so the two busiest days are buried inside a long whisker instead of flagged.

(a) 2, 8, 12, 17, 41 cards. (b) Fences at $-5.5$ and 30.5 cards, so the 34-card and 41-card days are outliers. (c) Box from 8 to 17 with the median line at 12, lower whisker to 2, upper whisker to 19, and separate points at 34 and 41. (d) Same box and median line, but the upper whisker would stretch to 41 and no points would be plotted separately.

Problem 8

A drop-in tutoring center records the length, in minutes, of each of the 12 sessions on one afternoon: 41, 38, 44, 40, 36, 42, 39, 43, 37, 45, 96, 40. (a) Give the five-number summary and the IQR. (b) Use the 1.5 IQR rule to test for outliers. (c) Find the mean with and without the longest session, and the median with and without it. (d) The 96-minute entry was logged in error and the real length is unknown. What is the largest whole number of minutes it could have been without being flagged as an outlier by the same rule?

Show the worked solution

Sort the values: 36, 37, 38, 39, 40, 40, 41, 42, 43, 44, 45, 96. There are $n = 12$ sessions.
(a) The median is the average of the 6th and 7th values: $(40 + 41)/2 = 81/2 = 40.5$ minutes.
(a) The lower half is 36, 37, 38, 39, 40, 40, so $Q_1 = (38 + 39)/2 = 77/2 = 38.5$ minutes. The upper half is 41, 42, 43, 44, 45, 96, so $Q_3 = (43 + 44)/2 = 87/2 = 43.5$ minutes. The five-number summary is 36, 38.5, 40.5, 43.5, 96 minutes, and $IQR = 43.5 - 38.5 = 5$ minutes.
(b) $1.5 \times IQR = 1.5 \times 5 = 7.5$ minutes, so the fences are $38.5 - 7.5 = 31$ minutes and $43.5 + 7.5 = 51$ minutes. The shortest session at 36 minutes is above 31, and only the 96-minute session exceeds 51, so it is the one outlier.
(c) With all 12 sessions the sum is $36 + 37 + 38 + 39 + 40 + 40 + 41 + 42 + 43 + 44 + 45 + 96 = 541$ minutes, so $\bar{x} = 541/12 = 45.08$ minutes to two decimal places, and the median is 40.5 minutes.
(c) Drop the 96-minute session. The remaining 11 sessions sum to $541 - 96 = 445$ minutes, so $\bar{x} = 445/11 = 40.45$ minutes, and with 11 values the median is the 6th, which is 40 minutes.
(c) Compare the two moves. The mean fell 4.63 minutes, from 45.08 to 40.45, while the median fell 0.5 minutes, from 40.5 to 40. With the outlier in, the mean of 45.08 minutes was longer than 11 of the 12 sessions.
(d) As long as the corrected value stays at or above 45 minutes it is still the largest value in the list, so the sorted order of the other 11 values does not change and $Q_1$ , $Q_3$ , and the IQR stay at 38.5, 43.5, and 5 minutes.
(d) The upper fence therefore stays at $43.5 + 7.5 = 51$ minutes, and the rule flags a value only when it lies beyond a fence. A 51-minute session is not beyond it, so 51 minutes is the largest whole number that would not be flagged, and 52 minutes would be.

(a) 36, 38.5, 40.5, 43.5, 96 minutes; $IQR = 5$ minutes. (b) Fences at 31 and 51 minutes, so the 96-minute session is the only outlier. (c) Mean 45.08 minutes with it and 40.45 without (a drop of 4.63); median 40.5 with it and 40 without (a drop of 0.5). (d) 51 minutes.