Box-and-whisker plots — quartiles, IQR: Visual walkthrough

We will use one running dataset the whole way down so the pictures connect:

$8, 2, 15, 5, 6, 3, 9, 7, 30, 4$

Ten numbers, deliberately jumbled, with one suspiciously large value ( $30$ ). By the end you will know exactly where each one lands.

Step 1 — Line the numbers up (sort)

WHAT. We rearrange the list from smallest to largest:

$2, 3, 4, 5, 6, 7, 8, 9, 15, 30$

WHY. A quartile is a position — "the value one-quarter of the way along." That phrase is meaningless in a scrambled pile. The moment "one-quarter of the way" has to mean something, the data must be in order. Think of people in a queue by height: "the person 1/4 down the line" only makes sense once they've lined up.

PICTURE. Look at the figure: the jumbled numbers (top row) drop into their sorted slots (bottom row). Each slot has a position index $1, 2, \dots, 10$ written under it — those indices are the ruler we measure quartiles against.

Figure — Box-and-whisker plots — quartiles, IQR

Step 2 — Find the median $Q_{2}$ (the central cut)

WHAT. With $n = 10$ (an even count), there is no single middle value — the middle falls between positions $5$ and $6$ . We average those two:

$Q_{2} = \frac{6 position 5 + 7 position 6}{2} = 6.5$

$Q_{2}$ — read "Q-two", the median: the value with half the data below it, half above.
The $6$ and $7$ — the two innermost survivors when you close in from both ends (positions $5$ and $6$ of the sorted list).
Dividing by $2$ — takes the exact midpoint between them, because no data point sits there.

WHY average two? With an even count the queue has no lone centre person; the fair "middle" is the gap between the two central people, so we split the difference. (If $n$ were odd we'd just take the single middle value — see Step 6.)

PICTURE. Two arrows march inward from both ends and meet in the $6$ – $7$ gap. The dashed pink line marks $Q_{2} = 6.5$ sitting in the gap, touching neither data point.

Step 3 — Split into two halves

WHAT. The median cut divides the sorted list into a lower half and an upper half:

$lower half (5 values) 2, 3, 4, 5, 6 upper half (5 values) 7, 8, 9, 15, 30$

WHY. The plan is "median of each half," so first we must have two clean halves. Because $n = 10$ is even, the cut lands between the $6$ (position $5$ ) and the $7$ (position $6$ ), and each half gets exactly $n /2 = 5$ values — no value is orphaned. (When $n$ is odd the single median value belongs to neither half; Step 6 shows that case.)

PICTURE. A vertical pink bar drops through the $6$ – $7$ gap. Everything left of it is the blue lower half; everything right is the yellow upper half. Notice the counts printed under each: $5$ and $5$ , balanced.

Step 4 — Median of each half gives $Q_{1}$ and $Q_{3}$

WHAT. Repeat the "find the middle" trick inside each half. Each half has $5$ values (odd), so each has a genuine single middle:

$Q_{1} = median of {2, 3, middle 4, 5, 6} = 4$ $Q_{3} = median of {7, 8, middle 9, 15, 30} = 9$

$Q_{1}$ — "Q-one", the lower quartile: median of the lower half, so about a quarter of all data lies below it.
$Q_{3}$ — "Q-three", the upper quartile: median of the upper half, so about three-quarters lies below it.

WHY the median-of-half rule (not "position $n /4$ ")? As the parent note's steel-man warned: raw index counting lands between points and disagrees across textbooks. Taking a median of each half is unambiguous — it's the same reliable operation we already trust, just applied twice.

PICTURE. Two mini-versions of the Step 2 inward-arrows, one in each half. Blue arrows meet on $4$ ; yellow arrows meet on $9$ . Now three cuts $Q_{1} = 4$ , $Q_{2} = 6.5$ , $Q_{3} = 9$ carve the data into four equal-count blocks.

Step 5 — The IQR and the 1.5×IQR fences

WHAT. The interquartile range is the width from $Q_{1}$ to $Q_{3}$ :

$IQR = 9 Q_{3} - 4 Q_{1} = 5$

Then we build the fences — the boundary of "normal":

$Lower fence = 4 Q_{1} - yardsticks 1.5 \times 5 IQR = 4 - 7.5 = - 3.5$ $Upper fence = 9 Q_{3} + 1.5 \times 5 IQR = 9 + 7.5 = 16.5$

IQR — the width of the middle-50% box; our natural "step size."
$1.5$ — how many box-widths we step out before calling a point weird (a convention that flags ≈0.7% of normal data).
The subtraction/addition — walk left of $Q_{1}$ and right of $Q_{3}$ by that distance.

WHY use IQR as the yardstick? Because IQR is robust: it ignores the extreme outer quarters, so the fence can't be blown outward by the very outlier it's meant to catch. Using the plain range would let the giant $30$ define its own "normal."

PICTURE. The box (from $Q_{1}$ to $Q_{3}$ ) is drawn, then two fence-flags planted at $- 3.5$ and $16.5$ by stepping $1.5$ box-widths outward. The value $30$ sits clearly past the upper flag.

Step 6 — Edge case: odd $n$ (where does the median go?)

WHAT. Drop the $30$ so $n = 9$ (odd): $2, 3, 4, 5, 6, 7, 8, 9, 15$ . Now the middle is a single value at position $\frac{n + 1}{2} = 5$ , namely $Q_{2} = 6$ . Crucially, that $6$ is removed before halving:

$lower half 2, 3, 4, 5 [6] upper half 7, 8, 9, 15$

$Q_{1} = \frac{3 + 4}{2} = 3.5, Q_{3} = \frac{8 + 9}{2} = 8.5$

WHY exclude the median? With odd $n$ the middle value belongs to neither side — it's the pivot itself. If we glued it to one half, that half would have one extra member and the four blocks would no longer be equal-count. Leaving it out keeps both halves at $4$ values each.

PICTURE. The single median value $6$ is boxed and greyed out, with arrows showing it stepping aside so the two equal halves of size $4$ form on either side.

Step 7 — Assemble the finished plot

WHAT. Put it all together for the original $n = 10$ data. The whisker does not reach the fence — it stops at the last real data point still inside it:

Left whisker → smallest value $\geq$ lower fence $= 2$ .
Box: left edge $Q_{1} = 4$ , line $Q_{2} = 6.5$ , right edge $Q_{3} = 9$ .
Right whisker → largest value $\leq$ upper fence $= 16.5$ . That is $15$ (since $16.5 \geq 15$ ), not $30$ .
$30 > 16.5$ → drawn as a lone dot: an outlier.

WHY the whisker stops at $15$ , not $16.5$ . A whisker marks a real observation; the fence is an imaginary threshold. The last observed value that clears the fence is $15$ , so the arm reaches exactly there and $30$ is exiled to a dot.

PICTURE. The complete box-and-whisker: whisker $\to$ box $\to$ whisker, with the fence lines faint behind it and $30$ hovering as a pink dot beyond the right whisker.

The one-picture summary

Every earlier step, stacked into a single ladder: sort $\to$ cut at $Q_{2}$ $\to$ cut each half at $Q_{1}, Q_{3}$ $\to$ measure IQR $\to$ step out $1.5 \times$ IQR $\to$ draw box, whiskers, outlier.

Recall Feynman retelling — the whole walkthrough in plain words

First I made everyone stand in a line, shortest to tallest, because "one-quarter along" only means something once they're lined up. I found the person exactly in the middle — with an even crowd that's the gap between the two middle people, so I picked the point halfway between them: that's $Q_{2}$ . That middle cut split the line into a shorter half and a taller half. I did the same middle-finding trick inside each half, giving me two more cuts, $Q_{1}$ and $Q_{3}$ — now the line is in four equal groups. The distance between those two inner cuts is the IQR, the width of the "middle bunch." To spot weirdos, I stepped one-and-a-half of those widths past each inner cut and planted a fence. Anyone past the fence is a suspicious loner (that was $30$ ). Finally I drew a box from $Q_{1}$ to $Q_{3}$ with a line at the middle, stretched arms out to the last normal person on each side, and drew the loner as a lonely dot. When the crowd is odd, the single middle person steps aside and doesn't join either half, so the halves stay equal.

Connections

Box-and-whisker plots — quartiles, IQR — the parent this walkthrough builds.
Median and Measures of Central Tendency — $Q_{2}$ is the median, applied three times here.
Range and Spread — IQR vs plain range in Step 5.
Outliers and Robust Statistics — the $1.5 \times$ IQR fence logic.
Percentiles and Quantiles — quartiles as the 25/50/75 cut points.
Skewness — an off-centre median line hints at skew.
Normal Distribution — why $1.5$ flags ≈0.7%.