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How do boxplots display the five-number summary, and how do we flag outliers formally?

Topic 1.8 Graphical Representations of Summary Statistics: construct and interpret boxplots from the five-number summary, and identify outliers using the 1.5 times IQR rule.

A focused answer to AP Statistics Topic 1.8, on building and reading boxplots from the five-number summary, the 1.5 times IQR rule for outliers, and what a boxplot does and does not reveal, with a worked construction.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The boxplot
The 1.5 times IQR rule
Reading shape from a boxplot
What a boxplot cannot show
Try this

What this topic is asking

The College Board (Topic 1.8) wants you to turn the five-number summary into a boxplot, read a boxplot, and apply the $1.5 \times \text{IQR}$ rule to flag outliers formally. It also wants you to know what a boxplot does and does not show.

The boxplot

The box itself spans the middle $50\%$ of the data (its width is the IQR), so the box shows where the bulk of the data sits and the line shows the center. A modified boxplot, which is what the AP course uses, applies the outlier rule so that the whiskers stop at the last non-outlier and outliers appear as dots; this is more informative than a plain boxplot whose whiskers always run to the min and max.

The 1.5 times IQR rule

This rule gives a reproducible, objective definition of "outlier," replacing the informal "looks far away" of Topic 1.6. Because the IQR is resistant, the fences are not themselves distorted by the very outliers they are detecting, which is the elegance of the rule.

Reading shape from a boxplot

A boxplot encodes skew through the relative lengths of its parts. If the median sits closer to $Q_1$ (left side of the box) and the right whisker is longer, the distribution is skewed right; the mirror image is skewed left; a centered median with balanced whiskers suggests symmetry. Comparing the whisker lengths and the position of the median inside the box is the standard way to read shape from a boxplot, and it is faster than computing anything. This makes boxplots superb for comparing several groups at once: drawing parallel boxplots on the same axis lets you compare centers (median lines), spreads (box widths), and skew (whisker lengths) across groups in a single glance, which is exactly the task of the next topic.

What a boxplot cannot show

The crucial limitation, and a favorite exam point, is that a boxplot is built only from five numbers, so it throws away the internal shape of the data. It cannot show how many modes a distribution has: a smoothly unimodal distribution and a sharply bimodal one can produce identical boxplots, because both can share the same five-number summary. It also hides gaps and clusters within the box or whiskers. For that reason, when a question asks whether a boxplot fully describes a distribution, the honest answer is no: to see modality, gaps, and clusters you need a dotplot, stemplot, or histogram. A strong exam answer states this limitation explicitly, often phrased as "a boxplot cannot reveal the number of peaks (modality) because it shows only the five-number summary." Knowing the strengths (comparison, skew, resistant outlier detection) and the weaknesses (no modality, no clusters) of each display is the meta-skill that Unit 1's display topics build toward.

Building a modified boxplot

A data set has ordered values $\{3, 7, 8, 12, 13, 15, 18, 21, 35\}$ ( $n = 9$ ). Find the five-number summary, apply the outlier rule, and describe the boxplot.

step 1 Median and quartiles

The median is the $5$ th value, $13$ . Lower half $\{3, 7, 8, 12\}$ gives $Q_1 = \frac{7 + 8}{2} = 7.5$ ; upper half $\{15, 18, 21, 35\}$ gives $Q_3 = \frac{18 + 21}{2} = 19.5$ .

step 2 Five-number summary and IQR

$\min = 3$ , $Q_1 = 7.5$ , median $= 13$ , $Q_3 = 19.5$ , $\max = 35$ . $\text{IQR} = 19.5 - 7.5 = 12$ .

step 3 Apply the fences

$1.5 \times \text{IQR} = 18$ . Lower fence $= 7.5 - 18 = -10.5$ ; upper fence $= 19.5 + 18 = 37.5$ .

step 4 Identify outliers

The minimum $3$ is above $-10.5$ , so no low outliers. The maximum $35$ is below $37.5$ , so $35$ is not an outlier under this rule (despite looking large). The right whisker therefore reaches all the way to $35$ .

step 5 Interpret

The box runs from $7.5$ to $19.5$ with the median at $13$ , and the right whisker (to $35$ ) is longer than the left (to $3$ ), suggesting a mild right skew. No values are flagged as outliers, which shows that "looks far" is not the same as "fails the rule."

Try this

Q1. A data set has $Q_1 = 10$ , $Q_3 = 22$ . Find the upper fence for outliers. [2 points]

Cue. $\text{IQR} = 12$ , so upper fence $= 22 + 1.5(12) = 22 + 18 = 40$ .

Q2. Give one feature of a distribution that two different data sets could hide while sharing the same boxplot. [1 point]

Cue. Modality (number of peaks), or gaps and clusters; a boxplot uses only the five-number summary.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2018 (style)1 marksSection I (multiple choice). A data set has

Q_1 = 20

and

Q_3 = 32

. Using the

1.5 \times \text{IQR}

rule, a value is an outlier if it is above which threshold? (A)

32

(B)

44

(C)

50

(D)

56

Show worked answer →

The correct answer is (C).

$\text{IQR} = 32 - 20 = 12$ , so $1.5 \times \text{IQR} = 18$ . The upper fence is $Q_3 + 1.5 \times \text{IQR} = 32 + 18 = 50$ . Any value above $50$ is a high outlier.

(A) is just $Q_3$ . (B) adds only $12$ (one IQR, not $1.5$ ). (D) adds $24$ (two IQR). The rule adds exactly $1.5 \times \text{IQR}$ to $Q_3$ for the upper fence.

AP 2023 (style)4 marksSection II (free response). A data set of daily download counts has five-number summary

\min = 5

Q_1 = 18

, median

= 24

Q_3 = 30

\max = 70

. (a) Apply the

1.5 \times \text{IQR}

rule to determine whether any values are outliers. (b) Describe what the boxplot of these data would look like and what it suggests about shape. (c) State one feature of the distribution a boxplot cannot reveal.

Show worked answer →

A 4-point question on boxplots and outliers.

(a) (2 points) $\text{IQR} = 30 - 18 = 12$ ; $1.5 \times \text{IQR} = 18$ . Lower fence $= 18 - 18 = 0$ ; upper fence $= 30 + 18 = 48$ (1 point). The minimum $5$ is above $0$ , so no low outliers; the maximum $70$ exceeds $48$ , so $70$ is a high outlier (1 point).
(b) (1 point) The box spans $Q_1 = 18$ to $Q_3 = 30$ with the median at $24$ ; the right whisker is long (or a point at $70$ is plotted as an outlier), suggesting the distribution is skewed right.
(c) (1 point) A boxplot cannot reveal the number of modes (peaks) or gaps and clusters within the data; for example a bimodal distribution and a uniform one can produce identical boxplots.

Markers reward correct fences and outlier identification, a description tying the long right whisker to right skew, and a correct limitation (modality or clusters).

Related dot points

Sources & how we know this

AP Statistics Course and Exam Description — College Board (2020)