Medians are nice, but sometimes we need more than just a middle marker. We’ll add to our repertoire quartiles, which sound like quarters, because that’s what they do: they break the data in quarters.
There are three of them (fencepost problem!). The first quartile $Q_1$ marks the end of the first 25% of data. $Q_2$ marks where 50% of the data falls, but this is the same thing as the median. $Q_3$ marks where 75% of the data falls.
Finding quartiles begins by finding the median $Q_2$. Put your data in order, then find the middle.
\[\begin{array}{ccccccc} 2 & 4 & 4 & 5 & 6 & 7 & 8 \\ & & & Q_2 & & & \end{array}\]Now you have split data. With each half, repeat the what you did before. Make sure you don’t include the median when finding your other quartiles.
\[\begin{array}{ccccccc} 2 & 4 & 4 & 5 & 6 & 7 & 8 \\ & Q_1 & & Q_2 & & Q_3 & \end{array}\]With an even number of entries, the process is the same. We still don’t include the median, but that’s not a problem since the median won’t technically be an entry.
\[\begin{array}{ccccc|ccccc} 1 & 3 & 3 & 4 & 5 & 6 & 6 & 7 & 8 & 8 \\ \end{array}\]Our median is 5.5, and again we look at our two halves and find the middle.
\[\begin{array}{ccccc|ccccc} 1 & 3 & 3 & 4 & 5 & 6 & 6 & 7 & 8 & 8 \\ & & Q_1 & & & & & Q_3 & & \end{array}\]If you recall, range was the difference between the maximum and minimum data values. Interquartile range is the difference between $Q_1$ and $Q_3$, or the spread of the middle 50% of your data.
In our first examples, our interquartile range was $7-4 = 3$. In the second, it was $7-3 = 4$.