Median vs Quartile

Sort the values from smallest to largest and the median is whatever sits at position $\frac{n+1}{2}$ (n plus 1, over 2), where $n$ is how many values there are. When $n$ is odd that position lands on an actual observation. When $n$ is even it lands between two of them and the median is their average, so the median of an even-sized set need not be a number that appears in the data at all. It is commonly written $M$.

Take the seven quiz scores 12, 15, 15, 18, 22, 24, 91. They are already sorted and $n = 7$, so the median is the value in position 4: $M = 18$. The mean of the same seven scores is 28.14. Now change the 91 to 27. The median is still 18, while the mean drops to 19. That is resistance: the median only counts how many values sit on each side of it and ignores how far away they are, so moving one extreme value cannot shift it by more than one position.

With an even count the averaging step kicks in. For 2, 6, 8, 10 the median is $(6 + 8) / 2 = 7$, a number that appears nowhere in the list.

The most common error is reporting the position instead of the value. For those seven scores the answer is 18, not 4; position 4 only tells you where to look. Two related slips are taking the middle of the list before sorting it, and averaging the smallest and largest values, which gives the midrange rather than the median.

The median is also the 50th percentile and the center of the five-number summary. The quartiles are the medians of the halves on either side of it, with the median itself left out of each half under the TI-84 convention this site uses: for the seven scores, 12, 15, 15 gives $Q_1 = 15$ and 22, 24, 91 gives $Q_3 = 24$. Comparing the two centers is also a quick read on shape: here the mean of 28.14 sits far above the median of 18, which usually signals either a right skew or a single high value dragging the mean up. In this set it is the second of those, since the 1.5 IQR rule puts the upper fence at $24 + 1.5(9) = 37.5$ and flags 91 as an outlier.

$Q_1$, the first quartile, is the median of the lower half of the sorted data, and $Q_3$, the third quartile, is the median of the upper half, with the overall median between them as the second quartile. Everything turns on what you do with that median when $n$ is odd. This site uses the median-excluded convention, the one TI-83 and TI-84 calculators use: the median is left out of both halves before you take the median of each.

Take the seven values 4, 6, 9, 11, 15, 18, 20. The median is the fourth value, 11. Median-excluded leaves a lower half of 4, 6, 9 and an upper half of 15, 18, 20, so $Q_1 = 6$, $Q_3 = 18$, and the IQR is 12. The median-included convention keeps the 11 in both halves: 4, 6, 9, 11 gives $Q_1 = (6 + 9)/2 = 7.5$, and 11, 15, 18, 20 gives $Q_3 = 16.5$, for an IQR of 9. Same seven numbers, different quartiles, neither an arithmetic error.

So "my calculator says $Q_1 = 6$ but the spreadsheet says 7.5, so one of them is wrong" is itself the mistake. There is no universal definition of a quartile in a finite data set, software disagrees on purpose, and naming the convention you used ends the argument. The two rules always agree when $n$ is even, since there is no middle value to place, so the question arises only for odd $n$.

A quartile is a value, not a group. "The first quartile is the bottom 25 percent of the scores" describes a set of observations; $Q_1 = 6$ is one number in the units of the data. The four groups are only roughly equal too: with $n = 7$ no cut gives four of the same size.

The choice propagates: the IQR is $Q_3 - Q_1$ and the 1.5 IQR fences are built from it, so switching conventions can move a borderline value from inside a fence to outside. In the Fall 2026 course this sits in Unit 1, whose topic 1.7 is titled Summary Statistics for One Quantitative Variable.