How to find Q1 and Q3 by hand

By Jude Wallis · Published

Sort the data and find the median. Q1 (the first quartile) is the median of the lower half and Q3 (the third quartile) is the median of the upper half. When the number of values is odd, leave the median itself out of both halves. This site uses that median-excluded convention, matching the TI-84.

AP Statistics: Unit 1 (topics 1.7 Summary Statistics for One Quantitative Variable, 1.8 Graphical Representations of Summary Statistics for One Quantitative Variable). In the Fall 2026 AP Statistics course, quartiles are Unit 1 topic 1.7, where the CED defines Q1 and Q3 as the medians of the lower and upper halves, the IQR as Q3 minus Q1, and the 1.5 IQR outlier rule.

The rule used here: median-excluded

Q1Q_1, the first quartile, is the median of the lower half of the sorted data, and Q3Q_3, the third quartile, is the median of the upper half. About 25% of the values sit at or below Q1Q_1 and about 75% sit at or below Q3Q_3, which makes them the 25th and 75th percentiles. Together they bound the middle 50% of the data.

The one judgment call is what to do with the median itself when the number of values nn is odd. This site always uses the median-excluded convention: the median value is left out of both halves. That is the rule TI-83 and TI-84 calculators use, so their 1-Var Stats output will agree with your hand arithmetic.

The AP course describes Q1Q_1 as the median of the lower half of the ordered data, running from the minimum to the position of the median, and Q3Q_3 as the median of the upper half. Stating your convention costs one clause and removes any ambiguity from your work.

The four steps

  1. Sort the values from smallest to largest. Nothing below works on unsorted data.
  2. Find the median. With nn odd it is the single middle value. With nn even it is the average of the two middle values.
  3. Split into halves. With nn odd, the lower half is every value before the median and the upper half is every value after it, with the median itself in neither. With nn even, the lower half is the first n/2n/2 values and the upper half is the last n/2n/2 values.
  4. Take the median of each half. The median of the lower half is Q1Q_1; the median of the upper half is Q3Q_3.

Each half contains the same number of values, either (n1)/2(n-1)/2 when nn is odd or n/2n/2 when nn is even. If your two halves have different sizes, you split in the wrong place. A quartile can land between two data values, so Q1=7.5Q_1 = 7.5 is a perfectly good answer even when every value is a whole number.

Why odd and even n feel different

With an even nn the median falls in the gap between two data values, so the split point is already between observations and nothing is left over. Every value goes into one half or the other.

With an odd nn the median is an actual data value sitting at the split point, and it has to go somewhere. Median-excluded drops it from both halves. Median-included, the other common rule, keeps it in both.

That single choice is the whole difference between the two conventions. For even nn the two rules always agree, so the question only arises when nn is odd.

From quartiles to the IQR and the 1.5 IQR fences

Once you have the quartiles, two more results follow immediately. The interquartile range is

IQR=Q3Q1\text{IQR} = Q_3 - Q_1

and it measures the spread of the middle 50% of the data. Because it ignores the smallest and largest quarter of the values, the IQR is resistant: a single extreme value barely moves it, which is why the AP course pairs the IQR with the median for skewed data.

The fences for the 1.5 IQR outlier rule come next:

Q11.5×IQRandQ3+1.5×IQRQ_1 - 1.5 \times \text{IQR} \quad \text{and} \quad Q_3 + 1.5 \times \text{IQR}

Any value below the lower fence or above the upper fence is an outlier. Keep full precision through this calculation; rounding Q1Q_1 or the IQR early can flip a borderline value from inside the fence to outside it. The 1.5 IQR rule covers the fences in more depth, and how to make a boxplot shows where they end up on a graph.

Other conventions give slightly different quartiles

There is no single universal definition of a quartile, and honest work says which rule it used. Take the 9 sorted values 3, 7, 8, 12, 14, 19, 21, 26, 60 from the first worked example below.

MethodQ1Q_1Q3Q_3IQR
Median-excluded (TI-84, used here)7.523.516
Median-included (Tukey's hinges)82113
Linear interpolation (Excel QUARTILE.INC, R's default quantile)82113

All three are defensible; they answer slightly different questions about where the 25th percentile sits in a finite data set. Software usually interpolates between the two data values nearest the quartile position, which is why a spreadsheet can hand you a quartile that is not in your data at all. On the 10 protein values in the second worked example, the median-excluded rule gives Q1=7Q_1 = 7 and Q3=14Q_3 = 14, while Excel's QUARTILE.INC gives Q1=7.25Q_1 = 7.25 and Q3=13.5Q_3 = 13.5.

That gap is not cosmetic. With the median-excluded quartiles the IQR is 7 and the upper fence is 14+1.5×7=24.514 + 1.5 \times 7 = 24.5, so the value 24 is not an outlier. With the interpolated quartiles the IQR is 13.57.25=6.2513.5 - 7.25 = 6.25 and the upper fence is 13.5+1.5×6.25=22.87513.5 + 1.5 \times 6.25 = 22.875, so the same value 24 gets flagged. When you report an outlier, say which quartile rule produced it.

Checking your work on a calculator

A graphing calculator with statistical capability is expected on the AP exam, and its one-variable statistics output lists the five-number summary directly: minX, Q1Q_1, Med, Q3Q_3, and maxX. Because TI calculators use the median-excluded rule, their Q1Q_1 and Q3Q_3 should match your hand arithmetic exactly.

If they disagree by a small amount, check three things before assuming the calculator is right. Did you sort the data. Did you leave the median out of both halves for odd nn. Did you enter every value, since one missing entry changes nn and can move both quartiles.

You can also paste a data set into the five-number summary calculator, which uses the same median-excluded convention, or drag values around in the descriptive statistics sandbox to watch Q1Q_1 and Q3Q_3 move.

Common mistakes

  • Not sorting first. Quartiles are positional, so unsorted data produces meaningless answers.
  • **Keeping the median in both halves when nn is odd.** That is a different convention, and it will disagree with a TI calculator.
  • **Splitting at the wrong place when nn is even.** There is no middle value to remove; the halves are simply the first n/2n/2 and the last n/2n/2 values.
  • **Using position 25% of the way through as Q1Q_1.** For n=10n = 10, the 2.5th value is not the median-excluded Q1Q_1.
  • **Averaging Q1Q_1 and Q3Q_3 to get the median.** That gives the midhinge, which usually is not the median.
  • Rounding too early. Carry the exact quartiles into the IQR and the fences.
  • Reporting a quartile with no units. Q1=7.5Q_1 = 7.5 means nothing without "paint cans" or "minutes" after it.

Where this sits in the AP course

Quartiles are topic 1.7 in Unit 1 of the Fall 2026 AP Statistics course, which covers measures of center and position, and the course defines the IQR as Q3Q1Q_3 - Q_1 and lists the 1.5 IQR outlier rule in the same topic. They reappear in topic 1.8 as part of the five-number summary and the boxplot, and in topic 1.9 when you compare two distributions by IQR.

Unit 1 carries 20 to 30% of the multiple-choice section, and quartile arithmetic is the kind of small calculation that appears inside larger free-response parts. Once you have the quartiles, use them to describe the distribution in context rather than leaving them as bare numbers. The official course description is at AP Central.

Odd number of values: 9 days of paint can sales

A hardware store records how many cans of paint it sold on each of 9 days: 14, 3, 21, 8, 60, 12, 26, 7, 19. Find Q1Q_1, Q3Q_3, the IQR, and the 1.5 IQR fences.

  1. Sort the values: 3, 7, 8, 12, 14, 19, 21, 26, 60. There are n=9n = 9 values.

  2. Find the median. With n=9n = 9 the median is the 5th value, so the median is 14 cans.

  3. Split the data, leaving the median out of both halves. Lower half: 3, 7, 8, 12. Upper half: 19, 21, 26, 60. Each half has (91)/2=4(9 - 1)/2 = 4 values.

  4. Find Q1Q_1, the median of 3, 7, 8, 12. With 4 values it is the average of the middle two: Q1=(7+8)/2=15/2=7.5Q_1 = (7 + 8)/2 = 15/2 = 7.5 cans.

  5. Find Q3Q_3, the median of 19, 21, 26, 60. Again average the middle two: Q3=(21+26)/2=47/2=23.5Q_3 = (21 + 26)/2 = 47/2 = 23.5 cans.

  6. Compute the IQR: IQR=Q3Q1=23.57.5=16\text{IQR} = Q_3 - Q_1 = 23.5 - 7.5 = 16 cans.

  7. Compute 1.5×IQR=1.5×16=241.5 \times \text{IQR} = 1.5 \times 16 = 24 cans.

  8. Lower fence: Q124=7.524=16.5Q_1 - 24 = 7.5 - 24 = -16.5 cans. Upper fence: Q3+24=23.5+24=47.5Q_3 + 24 = 23.5 + 24 = 47.5 cans.

  9. Compare the extremes to the fences. The minimum, 3, is above -16.5, so there are no low outliers. The maximum, 60, is above 47.5, so 60 is a high outlier.

Q1=7.5Q_1 = 7.5 cans and Q3=23.5Q_3 = 23.5 cans, so IQR=16\text{IQR} = 16 cans. The fences are -16.5 and 47.5 cans, and the day with 60 cans is a high outlier. The full five-number summary is 3, 7.5, 14, 23.5, 60 cans.

Even number of values: 10 snack bars

A shopper records the grams of protein per serving in 10 snack bars: 9, 4, 12, 7, 24, 5, 16, 11, 8, 14. Find Q1Q_1, Q3Q_3, the IQR, and the 1.5 IQR fences.

  1. Sort the values: 4, 5, 7, 8, 9, 11, 12, 14, 16, 24. There are n=10n = 10 values.

  2. Find the median. With n=10n = 10 it is the average of the 5th and 6th values: (9+11)/2=20/2=10(9 + 11)/2 = 20/2 = 10 grams. No data value sits at the median, so nothing needs to be excluded.

  3. Split the data at that point. Lower half (the first 5 values): 4, 5, 7, 8, 9. Upper half (the last 5 values): 11, 12, 14, 16, 24.

  4. Find Q1Q_1, the median of 4, 5, 7, 8, 9. With 5 values it is the 3rd, so Q1=7Q_1 = 7 grams.

  5. Find Q3Q_3, the median of 11, 12, 14, 16, 24. The 3rd of those is Q3=14Q_3 = 14 grams.

  6. Compute the IQR: IQR=147=7\text{IQR} = 14 - 7 = 7 grams.

  7. Compute 1.5×IQR=1.5×7=10.51.5 \times \text{IQR} = 1.5 \times 7 = 10.5 grams.

  8. Lower fence: 710.5=3.57 - 10.5 = -3.5 grams. Upper fence: 14+10.5=24.514 + 10.5 = 24.5 grams.

  9. Compare the extremes. The maximum is 24 grams, which is below the upper fence of 24.5 grams, so it is not an outlier even though it looks far from the rest. The minimum, 4 grams, is above -3.5 grams.

Q1=7Q_1 = 7 grams and Q3=14Q_3 = 14 grams, so IQR=7\text{IQR} = 7 grams. The fences are -3.5 and 24.5 grams, and no value falls outside them, so this data set has no outliers. The 24-gram bar misses the upper fence by 0.5 grams, which is why you compute the fence instead of judging by eye.

Frequently asked questions

Do you include the median when finding Q1 and Q3?

Not on this site, and not on a TI-83 or TI-84. When the number of values is odd, the median is left out of both halves, and Q1Q_1 and Q3Q_3 are the medians of the values strictly below and strictly above it. When the number of values is even there is no data value at the median, so the question does not arise.

Why does my calculator or spreadsheet give a different Q1?

Because it uses a different quartile convention. Excel's QUARTILE.INC and the default method in R interpolate between data values, so they can return a quartile that is not one of your observations. For the 9 values 3, 7, 8, 12, 14, 19, 21, 26, 60 the median-excluded rule gives Q1=7.5Q_1 = 7.5 while interpolation gives 8.

Can a quartile be a value that is not in the data set?

Yes. Whenever the relevant half has an even number of values you average the middle two, which often lands between observations. In the paint can example Q1=7.5Q_1 = 7.5 cans even though no day had 7.5 cans.

What is Q2?

Q2Q_2 is another name for the median, the 50th percentile. It splits the data in half, while Q1Q_1 and Q3Q_3 mark the ends of the middle 50%. A five-number summary lists the minimum, Q1Q_1, Q2Q_2, Q3Q_3, and the maximum.

Does the quartile convention change my answer on the AP exam?

Rarely, because exam data sets are usually built so the quartiles are unambiguous, and the median-excluded rule matches the calculator output students are expected to use. Show your split and your two half-medians so your reasoning is visible either way.