Box-and-whisker plots — quartiles, IQR: Worked examples

The scenario matrix

Before working anything, let us list every case class this topic can throw at you. Think of it like a checklist of "shapes of data" — each row is a situation where the procedure branches (does something different).

#	Case class	What is tricky about it	Example that covers it
A	$n$ odd	Median is a real data point; drop it from both halves	Ex 1
B	$n$ even	Median is an average of two points; halves split cleanly	Ex 2
C	Half-size odd ( $n$ leaves 3-per-half etc.)	$Q_{1}, Q_{3}$ themselves are single middle values	Ex 1, Ex 3
D	Half-size even ( $Q_{1}, Q_{3}$ are averages)	$Q_{1}, Q_{3}$ need their own averaging step	Ex 3
E	Repeated values	Repeats still count as separate positions	Ex 4
F	Outlier on the high side	Upper fence catches it; whisker stops short	Ex 2, Ex 5
G	Outlier on the low side	Lower fence catches it (people forget this side!)	Ex 5
H	All values identical (degenerate)	$IQR = 0$ ; fences collapse onto the value	Ex 6
I	Smallest possible set ( $n = 4$ )	Boundary case for the halving rule	Ex 3
J	Real-world word problem (units, meaning)	Must interpret, not just compute	Ex 7
K	Exam twist — find a missing value given $Q_{1}$	Work the rule backwards	Ex 8

Every cell above is hit by at least one worked example below. Let us go.

Ex 1 — Case A + C: odd $n$ , single-value quartiles

Step 1 — Sort ascending. $6, 8, 9, 11, 14, 15, 22$ . Why this step? Quartiles are defined by position in a sorted queue; positions are meaningless until we sort.

Step 2 — Median $Q_{2}$ . Position $\frac{n + 1}{2} = \frac{7 + 1}{2} = 4$ → the 4th value $= 11$ . Why this step? With 7 items there is exactly one item in the dead centre — three below, three above.

Step 3 — Split, dropping the median. Lower half $= 6, 8, 9$ ; upper half $= 14, 15, 22$ . Why this step? $n$ is odd, so the median $11$ belongs to neither side — including it would make the halves unequal.

Step 4 — $Q_{1}, Q_{3}$ = medians of the halves. Each half has 3 values → middle one. $Q_{1} = 8$ , $Q_{3} = 15$ . Why this step? $Q_{1}$ is "the median of the lower half" — that is the median-of-half rule from the parent note.

Step 5 — IQR. $IQR = Q_{3} - Q_{1} = 15 - 8 = 7$ .

Verify: Below $Q_{1} = 8$ sits only ${6}$ (1 of 7 ≈ 14%, near 25% for tiny $n$ ✓). Below the median $11$ sit ${6, 8, 9}$ = 3 of 7 ≈ 43% ≈ 50% ✓. All quartiles are actual data points, as expected for odd $n$ . ✓

Ex 2 — Case B + F: even $n$ , high-side outlier

Step 1 — Already sorted. $0, 2, 2, 3, 4, 5, 6, 40$ . Confirm ascending ✓. Why this step? Never assume — a hidden unsorted value breaks everything.

Step 2 — Median $Q_{2}$ . Even $n$ : average positions 4 and 5 → $\frac{3 + 4}{2} = 3.5$ . Why this step? No single middle item exists for even $n$ ; the fair centre sits between the two middle values.

Step 3 — Split into two halves of 4. Lower $= 0, 2, 2, 3$ ; upper $= 4, 5, 6, 40$ . Why this step? Even $n$ divides cleanly into two equal groups; the median is not a data point so nothing is dropped.

Step 4 — $Q_{1}, Q_{3}$ (each half is even → average). Lower middle pair $2, 2$ → $Q_{1} = \frac{2 + 2}{2} = 2$ . Upper middle pair $5, 6$ → $Q_{3} = \frac{5 + 6}{2} = 5.5$ . Why this step? Each half has 4 items — even — so its median is itself an average of two middles.

Step 5 — IQR and fences. $IQR = 5.5 - 2 = 3.5$ $Upper fence = Q_{3} + 1.5 IQR = 5.5 + 1.5 (3.5) = 5.5 + 5.25 = 10.75$ $Lower fence = Q_{1} - 1.5 IQR = 2 - 5.25 = - 3.25$ Why this step? The fences are the "1.5 boxes out" yardstick — anything beyond them is unusual.

Step 6 — Flag outliers. $40 > 10.75$ → outlier. The right whisker stops at the largest value $\leq 10.75$ , which is $6$ . Why this step? Whiskers reach the last non-outlier, not the fence.

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

Verify: All fences are consistent (lower < $Q_{1}$ < median < $Q_{3}$ < upper ✓). $40$ is above the upper fence; nothing is below $- 3.25$ , so no low outlier. Right whisker end $= 6$ . ✓

Ex 3 — Case D + I: smallest set ( $n = 4$ ), averaged quartiles

Step 1 — Sort. $8, 12, 16, 20$ . Why this step? Same reason as always — positions need order.

Step 2 — Median $Q_{2}$ . Even $n = 4$ : average of the 2nd and 3rd values $= \frac{12 + 16}{2} = 14$ . Why this step? No single middle; the centre lies between $12$ and $16$ .

Step 3 — Split. Lower $= 8, 12$ ; upper $= 16, 20$ . Why this step? $n$ even → clean 2/2 split, no value dropped.

Step 4 — Quartiles as averages. $Q_{1} = \frac{8 + 12}{2} = 10$ , $Q_{3} = \frac{16 + 20}{2} = 18$ . Why this step? Each half has 2 values (even) → median is their average. This is Case D: the quartiles themselves need averaging.

Step 5 — IQR. $18 - 10 = 8$ .

Verify: $Q_{1} < Q_{2} < Q_{3}$ → $10 < 14 < 18$ ✓, spacing symmetric because the data is symmetric ✓. This is the boundary case: $n = 4$ is the fewest points for which "median of each half" still produces two-element halves. ✓

Ex 4 — Case E: repeated values

Step 1 — Already sorted, $n = 9$ (odd). Confirm ✓. Why this step? Order first — repeats stay in place.

Step 2 — Median $Q_{2}$ . Position $\frac{9 + 1}{2} = 5$ → 5th value $= 8$ . Why this step? Nine items → the 5th is the exact centre. We count positions, not distinct values — the three 7s occupy positions 2, 3, 4 separately.

Step 3 — Split, drop median. Lower $= 6, 7, 7, 7$ ; upper $= 8, 9, 10, 10$ . Why this step? Odd $n$ → median (position 5) belongs to neither half.

Step 4 — Quartiles. Lower half has 4 values (even) → $Q_{1} = \frac{7 + 7}{2} = 7$ . Upper half → $Q_{3} = \frac{9 + 10}{2} = 9.5$ . Why this step? Repeats do not collapse: the lower half's middle pair is the 2nd and 3rd 7s, and $\frac{7 + 7}{2} = 7$ .

Step 5 — IQR. $9.5 - 7 = 2.5$ .

Verify: $Q_{1} = 7$ (repeated value can legitimately equal a quartile), $Q_{3} = 9.5$ , IQR $= 2.5$ ; ordering $7 \leq 8 \leq 9.5$ holds ✓.

Ex 5 — Case F + G: outliers on BOTH sides

Step 1 — Sort. $- 50, 10, 11, 12, 13, 14, 15, 60$ — already ascending, $n = 8$ . Why this step? Confirm order; the negative must sit first.

Step 2 — Median $Q_{2}$ . Average of 4th & 5th $= \frac{12 + 13}{2} = 12.5$ . Why this step? Even $n$ → averaged centre.

Step 3 — Split into 4/4. Lower $= - 50, 10, 11, 12$ ; upper $= 13, 14, 15, 60$ .

Step 4 — Quartiles. $Q_{1} = \frac{10 + 11}{2} = 10.5$ , $Q_{3} = \frac{14 + 15}{2} = 14.5$ . Why this step? Each half even → average the middle pair. Notice $- 50$ and $60$ sit at the edges of the halves, so they barely affect $Q_{1}, Q_{3}$ — that is robustness in action.

Step 5 — Fences. $IQR = 14.5 - 10.5 = 4$ $Lower fence = 10.5 - 1.5 (4) = 10.5 - 6 = 4.5$ $Upper fence = 14.5 + 1.5 (4) = 14.5 + 6 = 20.5$

Step 6 — Flag BOTH sides. $- 50 < 4.5$ → low outlier; $60 > 20.5$ → high outlier. Left whisker stops at $10$ (smallest value $\geq 4.5$ ); right whisker stops at $15$ . Why this step? The fence rule is symmetric — you must test both directions every time.

Verify: Two outliers, one each side; IQR $= 4$ unchanged by the wild edge values (compare: plain range $= 60 - (- 50) = 110$ !) — a vivid demo that IQR is robust while range is not. (Range and Spread, Outliers and Robust Statistics.) ✓

Ex 6 — Case H: degenerate, all values identical

Step 1 — Sorted (trivially), $n = 6$ . All equal. Why this step? Even identical data must be treated by the same rule — no shortcuts.

Step 2 — Median. Average of 3rd & 4th $= \frac{20 + 20}{2} = 20$ .

Step 3 — Halves. Lower $= 20, 20, 20$ , upper $= 20, 20, 20$ → $Q_{1} = 20$ , $Q_{3} = 20$ .

Step 4 — IQR and fences. $IQR = 20 - 20 = 0, fences = 20 \pm 1.5 (0) = 20.$ Why this step? With zero spread the box has zero width — it collapses to a single vertical line at 20, whiskers vanish, and no point can be an outlier (nothing is beyond $20$ ).

Verify: IQR $= 0$ , both fences $= 20$ , no outliers — the degenerate case behaves sensibly. A box plot here is just a line. ✓

Ex 7 — Case J: real-world word problem (with units)

Step 1 — Sort. Already ascending, $n = 12$ . Why this step? Confirm; the $25$ sits last.

Step 2 — Median $Q_{2}$ . Average of 6th & 7th $= \frac{5 + 5}{2} = 5$ minutes. Why this step? Even $n = 12$ → centre between positions 6 and 7.

Step 3 — Split into 6/6. Lower $= 2, 3, 3, 4, 4, 5$ ; upper $= 5, 6, 7, 8, 9, 25$ .

Step 4 — Quartiles. Lower middle pair (3rd,4th) $= 3, 4$ → $Q_{1} = 3.5$ min. Upper middle pair (3rd,4th of upper $= 7, 8$ ) → $Q_{3} = 7.5$ min. Why this step? Each half has 6 (even) → average middle pair.

Step 5 — Interpret IQR + fences. $IQR = 7.5 - 3.5 = 4 min$ $Upper fence = 7.5 + 1.5 (4) = 13.5 min, Lower fence = 3.5 - 6 = - 2.5 min .$ Since a wait can't be negative, the lower fence is effectively "no lower outliers possible." $25 > 13.5$ → the 25-minute wait is an outlier.

Step 6 — Answer in words. "The middle half of customers waited within a 4-minute band (from 3.5 to 7.5 min), with a typical wait of 5 min. One customer's 25-minute wait was an outlier — flag it for investigation." Why this step? A word problem demands interpretation with units, not just a number.

Verify: $Q_{1} = 3.5$ , $Q_{3} = 7.5$ , IQR $= 4$ min, upper fence $= 13.5$ min, $25 > 13.5$ → outlier ✓. Units carried throughout (minutes). ✓

Ex 8 — Case K: exam twist, work the rule BACKWARDS

Step 1 — Identify $n$ and structure. $n = 8$ (even) → lower half is the first 4 values $= 4, 6, x, 10$ . Why this step? $Q_{1}$ is the median of the lower half; only those 4 values matter.

Step 2 — Write $Q_{1}$ in terms of $x$ . Lower half has 4 values (even) → $Q_{1} = \frac{2nd + 3rd}{2} = \frac{6 + x}{2}$ . Why this step? We reverse the usual computation: express the quantity we know as a formula in the unknown.

Step 3 — Solve. $\frac{6 + x}{2} = 7 \Rightarrow 6 + x = 14 \Rightarrow x = 8$ . Why this step? Simple algebra once the relationship is set up.

Step 4 — Check the constraint. $x = 8$ satisfies $6 \leq 8 \leq 10$ ✓ and keeps the list sorted ( $6, 8, 10$ ) ✓. Why this step? An exam wants you to confirm the answer is admissible, not just algebraically valid.

Verify: With $x = 8$ the lower half is $4, 6, 8, 10$ , whose median is $\frac{6 + 8}{2} = 7 = Q_{1}$ ✓. The dataset stays sorted. ✓

Recall Which cell did each example hit?

Ex 1 → A,C · Ex 2 → B,F · Ex 3 → D,I · Ex 4 → E · Ex 5 → F,G · Ex 6 → H · Ex 7 → J · Ex 8 → K. Every matrix cell is covered.

Connections

Box-and-whisker plots — quartiles, IQR — parent: the rules these examples exercise.
Median and Measures of Central Tendency — $Q_{2}$ throughout.
Range and Spread — Ex 5 contrasts IQR ( $4$ ) vs range ( $110$ ).
Outliers and Robust Statistics — Ex 2, 5, 7 apply the fences.
Percentiles and Quantiles — quartiles as 25/50/75 percentiles.
Skewness — Ex 2 and Ex 7 are right-skewed (long high whisker).
Normal Distribution — why the 1.5 multiplier flags ≈0.7%.