Quartiles, IQR, and the box plot

Median-of-halves quartiles, the five-number summary, the 1.5·IQR fences, and what to do with a value beyond them.

Position, not distance

The standard deviation measures spread as a distance from the mean. The quartiles measure it by position: $Q_{1}$ has a quarter of the sorted data below it, the median half, $Q_{3}$ three-quarters. The interquartile range $I QR = Q_{3} - Q_{1}$ is the width of the middle half, and it is untouched by whatever is happening in the tails — which is why it is the spread to report beside a median.

The course rule for quartiles is median-of-halves. Sort. Find the median. $Q_{1}$ is the median of the lower half and $Q_{3}$ the median of the upper half; when $n$ is odd, the median itself belongs to neither half. Software uses other rules (L3.7); on the paper, this one.

Worked example — eleven overdue invoices

Maple Ledger now has eleven overdue invoices. Sorted days overdue: 1, 2, 4, 5, 7, 8, 10, 12, 15, 19, 41.

$n = 11$ , median position $(11 + 1) /2 = 6$ : median = 8.
Lower half, excluding the 8: 1, 2, 4, 5, 7 → $Q_{1} = 4$ .
Upper half: 10, 12, 15, 19, 41 → $Q_{3} = 15$ .
$I QR = 15 - 4 = 11$ .

Five-number summary: min 1 · $Q_{1}$ 4 · median 8 · $Q_{3}$ 15 · max 41. A box plot draws the box from $Q_{1}$ to $Q_{3}$ with a line at the median, and whiskers out to the most extreme values that are not flagged by the fences.

Fences. Lower fence $= Q_{1} - 1.5 \times I QR = 4 - 16.5 = - 12.5$ ; upper fence $= Q_{3} + 1.5 \times I QR = 15 + 16.5 = 31.5$ . The 41 is beyond the upper fence: a potential outlier, drawn as a separate point, with the whisker stopping at 19.

What to do with the 41

Investigate it. Is it a data-entry error (411 days? 4.1?), a disputed account, a genuine long-overdue invoice? If it is an error, correct it. If it is genuine, it stays in the data — and the report says so, giving the summary with and without it when the two differ materially. Deleting a true value because it is inconvenient is the one thing never to do.

The even case

With $n = 10$ there is no middle value to exclude: the lower half is simply the five smallest values and the upper half the five largest. For 6, 9, 11, 14, 16, 18, 19, 20, 24, 27: lower half 6, 9, 11, 14, 16 → $Q_{1} = 11$ ; upper half 18, 19, 20, 24, 27 → $Q_{3} = 20$ ; median $(16 + 18) /2 = 17$ .

The box plot is the paper's tool for comparing groups: two boxes side by side show at a glance which branch has the higher median, the wider middle half, and the longer tail. It cannot show modality — a bimodal histogram and a unimodal one can produce the same box.

Median-of-halves on the sorted values. The median (8) is excluded from both halves; each half’s own median is a quartile. The 41 lies beyond Q₃ + 1.5·IQR = 31.5.