Box-and-whisker plots — quartiles, IQR: Exercises

Before we start, one shared reminder of the whole toolkit, from the ground up — no symbol used before its meaning is fixed.

Definition The five numbers you always compute

Sort = write the data smallest-to-largest. Quartiles are positions, so an unsorted list is meaningless.
== $Q_{2}$ == = the median = the value with half the data below it.
== $Q_{1}$ == = median of the lower half.
== $Q_{3}$ == = median of the upper half.
== $IQR$ == = $Q_{3} - Q_{1}$ = the width of the middle 50%.

What exactly is "lower half" / "upper half"? After sorting a list of $n$ values:

If $n$ is odd: the median is one single value; the lower half is every value before it, the upper half is every value after it. The median itself is in neither half.
If $n$ is even: there is no single middle value, so we split the sorted list straight down the middle — the lower half is the first $n /2$ values, the upper half is the last $n /2$ values. (The two middle values, whose average is $Q_{2}$ , stay in their respective halves.)

See Box-and-whisker plots — quartiles, IQR for the full build-up of every idea used here.

Level 1 — Recognition

L1·Q1

The box plot below is drawn on a number line. Read off, straight from the picture: the median, $Q_{1}$ , $Q_{3}$ , the two whisker ends, and any outlier.

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

Recall Solution

Reading the marked positions on the axis:

Left whisker end (smallest non-outlier) $= 4$
Box left edge $= Q_{1} = 8$
Line inside box $= Q_{2} = 12$
Box right edge $= Q_{3} = 16$
Right whisker end (largest non-outlier) $= 22$
Lone dot beyond the right whisker $= 30 \Rightarrow$ outlier

That is the entire skill of L1: a box plot is a map, and each landmark has one meaning. From these we can also read $IQR = 16 - 8 = 8$ .

L1·Q2

For the sorted list $3, 5, 8, 11, 14$ ( $n = 5$ , odd), name the value that is $Q_{2}$ and say which values form the lower half.

Recall Solution

$n = 5$ is odd, so $Q_{2}$ is the single middle value at position $\frac{n + 1}{2} = \frac{6}{2} = 3$ , giving the 3rd value $= 8$ . The lower half = values below the median position = $3, 5$ (the median $8$ belongs to neither half because $n$ is odd).

Level 2 — Application

L2·Q1

Compute $Q_{1}, Q_{2}, Q_{3}$ and IQR for $12, 4, 7, 15, 9, 2, 20, 6, 11$ ( $n = 9$ ).

Recall Solution

Sort: $2, 4, 6, 7, 9, 11, 12, 15, 20$ . Median: position $\frac{9 + 1}{2} = 5$ gives value $9 = Q_{2}$ . Lower half ( $n$ odd, first 4, median excluded): $2, 4, 6, 7 \Rightarrow$ its median $= \frac{4 + 6}{2} = 5 = Q_{1}$ . Upper half (last 4): $11, 12, 15, 20 \Rightarrow$ median $= \frac{12 + 15}{2} = 13.5 = Q_{3}$ . IQR $= 13.5 - 5 = 8.5$ .

L2·Q2

Data (sorted): $2, 4, 4, 5, 7, 8, 10, 12$ ( $n = 8$ ). Find the five-number summary (min, $Q_{1}$ , $Q_{2}$ , $Q_{3}$ , max) and the IQR.

Recall Solution

$n = 8$ is even, so we split the list into the first $n /2 = 4$ values and the last $4$ values. Median: even $n$ , average of the 4th & 5th values $= \frac{5 + 7}{2} = 6 = Q_{2}$ . Lower half (first 4): $2, 4, 4, 5 \Rightarrow Q_{1} = \frac{4 + 4}{2} = 4$ . Upper half (last 4): $7, 8, 10, 12 \Rightarrow Q_{3} = \frac{8 + 10}{2} = 9$ . Five-number summary: min $= 2$ , $Q_{1} = 4$ , $Q_{2} = 6$ , $Q_{3} = 9$ , max $= 12$ . IQR $= 9 - 4 = 5$ .

L2·Q3

Find the outlier fences for the data in L2·Q2, then say where each whisker ends.

Recall Solution

$IQR = 5$ . Lower fence $= Q_{1} - 1.5 IQR = 4 - 1.5 (5) = 4 - 7.5 = - 3.5$ . Upper fence $= Q_{3} + 1.5 IQR = 9 + 7.5 = 16.5$ . Every value lies between $- 3.5$ and $16.5$ , so no outliers. Whiskers reach the extreme data points: left whisker to $2$ , right whisker to $12$ (not to the fences $- 3.5$ or $16.5$ ).

Level 3 — Analysis

L3·Q1

Data (sorted): $5, 6, 7, 8, 9, 10, 11, 40$ ( $n = 8$ ). Show numerically that the IQR barely changes if you delete the $40$ , but the range collapses. Explain what this reveals.

Recall Solution

With the $40$ : median $= \frac{8 + 9}{2} = 8.5$ ; lower half $5, 6, 7, 8 \Rightarrow Q_{1} = 6.5$ ; upper half $9, 10, 11, 40 \Rightarrow Q_{3} = \frac{10 + 11}{2} = 10.5$ . IQR $= 10.5 - 6.5 = 4$ . Range $= 40 - 5 = 35$ . Delete the $40$ ( $n = 7$ ): $5, 6, 7, 8, 9, 10, 11$ ; median $= 8$ ; lower half $5, 6, 7 \Rightarrow Q_{1} = 6$ ; upper half $9, 10, 11 \Rightarrow Q_{3} = 10$ . IQR $= 10 - 6 = 4$ . Range $= 11 - 5 = 6$ . Reveal: the IQR stayed at $4$ (unchanged), while the range crashed from $35$ to $6$ . The range is built from the two most extreme points, so one wild value dominates it; the IQR uses only the inner quartiles, so it is robust. This is exactly the point of Outliers and Robust Statistics and Range and Spread.

L3·Q2

For $2, 4, 6, 7, 9, 11, 12, 15, 20$ (L2·Q1: $Q_{1} = 5, Q_{2} = 9, Q_{3} = 13.5$ ), compare the gap below the median ( $Q_{2} - Q_{1}$ ) to the gap above ( $Q_{3} - Q_{2}$ ). What does this asymmetry say about the shape of the data?

Recall Solution

Lower gap $= Q_{2} - Q_{1} = 9 - 5 = 4$ . Upper gap $= Q_{3} - Q_{2} = 13.5 - 9 = 4.5$ . The upper gap is larger, so the box's right half is wider than its left half. A longer stretch on the high side means values are more spread out above the median, so the distribution is right-skewed (a tail toward large values). This is how a box plot reveals Skewness without any formula.

Level 4 — Synthesis

L4·Q1

You are told a data set of $n = 10$ has $Q_{1} = 20$ , $Q_{3} = 32$ , and exactly one outlier on the high side. Its sorted values are $14, 20, 22, 25, 28, 30, 31, 32, 35, v$ Find the smallest integer $v$ that makes the last value a genuine upper outlier, and state where the right whisker then ends.

Recall Solution

$IQR = Q_{3} - Q_{1} = 32 - 20 = 12$ . Upper fence $= Q_{3} + 1.5 IQR = 32 + 1.5 (12) = 32 + 18 = 50$ . A point is an outlier when it is strictly greater than the fence: $v > 50$ . The smallest integer is $v = 51$ . With $v = 51$ , the largest value inside the fence is $35$ , so the right whisker ends at $35$ and $51$ is drawn as a lone dot. (Check: we must keep $Q_{1}, Q_{3}$ as given. Changing only the last value doesn't move $Q_{3}$ : upper half is $30, 31, 32, 35, v$ , whose median is the middle value $32$ regardless of how large $v$ is — consistent.)

L4·Q2

Construct a 7-value data set (integers) whose median is $10$ , whose IQR is $6$ , and which has no outliers. Then verify your set.

Recall Solution

Design plan: $n = 7$ odd means drop the median from the halves. Need $Q_{2} = 10$ , and $Q_{3} - Q_{1} = 6$ . Pick $Q_{1} = 8, Q_{3} = 14$ (difference $6$ ). Lower half (3 values) must have median $8$ ; upper half (3 values) median $14$ . A valid set: $6, 8, 9, 10, 12, 14, 16$ . Verify: sorted ✓. Median position $4$ gives $10 = Q_{2}$ ✓. Lower half $6, 8, 9$ has median $8 = Q_{1}$ ✓. Upper half $12, 14, 16$ has median $14 = Q_{3}$ ✓. IQR $= 14 - 8 = 6$ ✓. Fences: lower $= 8 - 1.5 (6) = 8 - 9 = - 1$ ; upper $= 14 + 9 = 23$ . All data lie in $[- 1, 23]$ , so no outliers ✓. (Many answers exist; any set matching the three targets is correct.)

Level 5 — Mastery

L5·Q1 (degenerate / edge case)

Consider the data $7, 7, 7, 7, 7, 7$ ( $n = 6$ , all identical). Compute the five-number summary, the IQR, and the fences. What does the box plot look like?

Recall Solution

Sorted (already): all $7$ . Median $= \frac{7 + 7}{2} = 7 = Q_{2}$ . Lower half $7, 7, 7 \Rightarrow Q_{1} = 7$ . Upper half $7, 7, 7 \Rightarrow Q_{3} = 7$ . $IQR = 7 - 7 = 0$ . Fences: lower $= 7 - 1.5 (0) = 7$ ; upper $= 7 + 1.5 (0) = 7$ . Both fences sit exactly at $7$ . The plot: the box has zero width (a single vertical line at $7$ ), median line on top of it, whiskers of length zero. Every point equals $7$ , which is on (not beyond) the fence, so nothing is flagged. A zero IQR is the honest picture of "no spread at all."

L5·Q2 (four-value halves & tie-breaking)

Data (sorted): $1, 1, 2, 2, 2, 8, 9, 9, 10, 10$ ( $n = 10$ ). Find the five-number summary, IQR, fences, list any outliers, and state both whisker ends.

Recall Solution

Median: even $n$ , average of 5th & 6th $= \frac{2 + 8}{2} = 5 = Q_{2}$ . Lower half (first 5): $1, 1, 2, 2, 2 \Rightarrow$ median (3rd) $= 2 = Q_{1}$ . Upper half (last 5): $8, 9, 9, 10, 10 \Rightarrow$ median (3rd) $= 9 = Q_{3}$ . IQR $= 9 - 2 = 7$ . Fences: lower $= 2 - 1.5 (7) = 2 - 10.5 = - 8.5$ ; upper $= 9 + 1.5 (7) = 9 + 10.5 = 19.5$ . All values lie within $[- 8.5, 19.5]$ , so no outliers. Whiskers: left to $1$ (min), right to $10$ (max). Notice the median $5$ is a value that doesn't even appear in the data — that's fine, a median is a position-based cut point, not required to be an observation. This distribution is clearly bimodal (a low clump and a high clump), yet the summary handles it cleanly. See Percentiles and Quantiles.

L5·Q3 (reverse-engineering from a summary)

A report states: min $= 3$ , $Q_{1} = 10$ , median $= 15$ , $Q_{3} = 22$ , max $= 60$ , and there is at least one outlier. Prove an outlier must exist and identify which extreme is the offender.

Recall Solution

$IQR = Q_{3} - Q_{1} = 22 - 10 = 12$ . Upper fence $= 22 + 1.5 (12) = 22 + 18 = 40$ . Lower fence $= 10 - 1.5 (12) = 10 - 18 = - 8$ . The max $= 60 > 40$ , so it lies beyond the upper fence and is an outlier (the offender is on the high side). The min $= 3$ satisfies $3 > - 8$ , so the low end is fine. Thus an outlier must exist, and it is the large value near $60$ . We can't recover the exact data, but the summary alone forces this conclusion — the power of the five-number summary as a compact fingerprint.

Recall checklist

Connections

Box-and-whisker plots — quartiles, IQR — the parent recipe these drill.
Median and Measures of Central Tendency — every quartile is a median.
Range and Spread — contrasted with IQR in L3·Q1.
Outliers and Robust Statistics — the fence rule throughout L2–L5.
Percentiles and Quantiles — quartiles as the 25/50/75 cut points.
Skewness — read from half-widths in L3·Q2.
Normal Distribution — origin of the $1.5$ multiplier.