Lesson 2: Describing Distributions — Shape, Center, SpreadAP

A graph shows the shape of a distribution, but the AP exam wants numbers too — and, more than that, it wants the numbers interpreted in context. This lesson covers the vocabulary of distributions: shape, centre, spread and unusual features. The CED expects you to describe and compare distributions using these features together; a response that reports only a mean is, at best, partially correct.

Describing shape

Look at the overall pattern first. A distribution is roughly symmetric if its two halves are approximate mirror images. It is skewed to the right if the tail stretches toward the large values (think incomes: a few very large values pull the tail right), and skewed to the left if the tail stretches toward small values (think easy exam scores bunched near 100). Count the peaks: unimodal (one peak), bimodal (two peaks — often a sign that two groups have been mixed), or roughly uniform. Note gaps and possible outliers — individual values that fall outside the overall pattern.

Measures of centre: mean and median

The mean x̄ is the arithmetic average: add all values and divide by n. The median M is the midpoint — half the values lie at or below it. The mean uses every value, so it is dragged toward outliers and toward the long tail of a skewed distribution; the median depends only on order, so it is resistant. For symmetric distributions mean ≈ median. For right-skewed data the mean usually exceeds the median — a relationship the exam tests constantly.

Worked example 1. The annual salaries (in thousands of dollars) of the 10 employees of a small firm are: 32, 35, 35, 38, 40, 42, 45, 48, 55, 130.

• Sum = 500, so the mean is 500/10 = $50k.
• Ordered as given, the median is the average of the 5th and 6th values: (40 + 42)/2 = $41k.
• The mean ($50k) exceeds the median ($41k) because the owner-level salary of $130k pulls the mean upward. For a "typical" employee's salary, the median is the more honest summary.

Measures of spread: range, IQR and standard deviation

The range (max − min) uses only two values and is easily distorted. The interquartile range IQR = Q3 − Q1 measures the spread of the middle 50% and pairs naturally with the median. The standard deviation s measures the typical distance of values from the mean and pairs with the mean. Together, min, Q1, median, Q3 and max form the five-number summary, displayed in a boxplot.

Continuing worked example 1:

• Lower half (32, 35, 35, 38, 40) has median Q1 = 35; upper half (42, 45, 48, 55, 130) has median Q3 = 48. IQR = 48 − 35 = 13 ($13k).
• Outlier rule: values below Q1 − 1.5×IQR or above Q3 + 1.5×IQR are flagged as outliers. Here the fences are 35 − 19.5 = 15.5 and 48 + 19.5 = 67.5, so the $130k salary is an outlier by the 1.5×IQR criterion; the five-number summary is (32, 35, 41, 48, 130).
• Standard deviation: the squared deviations from the mean of 50 sum to 7,536, so s² = 7,536/9 ≈ 837.3 and s ≈ $28.9k — huge relative to the salaries of most employees, again because of one extreme value. With outliers present, median and IQR are the summaries to report.

Worked example 2 (comparing distributions). Two grocery stores' delivery times (minutes) for online orders have these summaries. Store A: min 18, Q1 26, median 31, Q3 38, max 52. Store B: min 12, Q1 21, median 29, Q3 44, max 78. A full-credit comparison names the same features for both, in context, with direction: "Store B's median delivery time (29 min) is slightly lower than Store A's (31 min), so a typical order arrives a little faster from B. However, Store A is far more consistent: its IQR is 38 − 26 = 12 min versus 44 − 21 = 23 min for B, and its range is 34 min versus 66 min. Store B's distribution is right-skewed with deliveries stretching to 78 minutes; Store A's is roughly symmetric. A customer who needs a predictable arrival time should choose A; a customer who only cares about the typical time might choose B." Percentiles connect the two ideas: a value at the 80th percentile is greater than 80% of the data — being "in the top 20%" — not "80% correct".

Choosing summaries, and the real world

Government statisticians report median household income because income is right-skewed; the mean would overstate the typical household. Weather services report median rainfall in arid climates for the same reason. Whenever a headline quotes an "average", ask which average — the choice is a rhetorical act.

Common mistakes

• Describing shape with no context. "It is skewed right" must become "delivery times are skewed right, with a few orders taking much longer than typical."
• Comparing with centre alone. Two distributions can share a median and differ wildly in spread; readers require centre AND spread (and shape, if it differs).
• Reporting mean and SD for strongly skewed data with outliers without comment; median and IQR are the defensible pair.
• Percentile confusion. The 90th percentile is a position in the data, not a score of 90%.

AP Exam Experience

Question 1. The distribution of house prices in a town is strongly skewed to the right. Which pair of statistics best summarises the centre and spread of the distribution?

• (A) Mean and standard deviation
• (B) Mean and range
• (C) Median and IQR
• (D) Median and standard deviation
• (E) Mode and range

Answer: (C). For skewed data with extreme values, the resistant pair — median for centre, IQR for spread — is not pulled around by the few mansions in the tail. The mean and standard deviation are both non-resistant.

Question 2. A data set of 15 values has Q1 = 20 and Q3 = 44. Which of the following values would be flagged as an outlier by the 1.5×IQR rule?

• (A) −10
• (B) 0
• (C) 78
• (D) 82
• (E) Both (C) and (D)

Answer: (D). IQR = 44 − 20 = 24, so 1.5×IQR = 36. The fences are 20 − 36 = −16 and 44 + 36 = 80. Only 82 lies outside the fences: 78 is inside (just barely), and −10 and 0 are both above −16. Moral: compute the fences before judging — "far away" by eye is not the criterion, and choice (E) is the trap for students who skip the arithmetic.

Question 3. Two sections of a course took the same test. Section 1: mean 74, median 78, SD 6. Section 2: mean 74, median 70, SD 11. Which statement is best supported?

• (A) Section 1's scores are skewed left and were more consistent than Section 2's.
• (B) Section 2's scores are symmetric.
• (C) The sections have identical distributions.
• (D) Section 1 had the higher typical score by every measure.
• (E) Section 2's highest score must be an outlier.

Answer: (A). In Section 1 the mean (74) is below the median (78), the signature of left skew, and its SD of 6 versus 11 shows less variability — more consistency. Equal means do not imply identical distributions, which is precisely the trap in (C).

Free-response question. A bakery records the number of loaves sold each day for 25 days. The five-number summary is: min 84, Q1 102, median 118, Q3 131, max 166. (a) Calculate the IQR and use the 1.5×IQR rule to determine whether the maximum, 166, is an outlier. (b) A second bakery has median 121 and IQR 41. Compare the two bakeries' daily sales distributions in context. (c) The first bakery's mean is 121.4 loaves. What does comparing the mean with the median suggest about the shape of the distribution?

Scoring guideline.

• Part (a): Essentially correct if IQR = 131 − 102 = 29, fences are 102 − 43.5 = 58.5 and 131 + 43.5 = 174.5, and the conclusion states 166 < 174.5 so the maximum is NOT an outlier. Partially correct if the IQR and fences are right but the conclusion is missing or reversed.
• Part (b): Essentially correct if the comparison addresses centre AND spread with direction and context: the second bakery sells slightly more on a typical day (median 121 vs 118 loaves) but is less consistent (IQR 41 vs 29 loaves). Partially correct if only centre or only spread is compared. Incorrect if it claims one bakery is "better" with no feature compared.
• Part (c): Essentially correct if it states that the mean (121.4) being greater than the median (118) suggests a slight right skew — a few high-sales days pulling the mean up. Incorrect if it claims the data must be symmetric or that the difference proves outliers exist.

Deeper dive: how outliers move the mean but not the median

Return to the salary data and watch the two centres react when the largest value changes. If the owner's salary doubles from 130 to 260, the new sum is 630, so the mean jumps from 50 to 63 — yet the median stays exactly 41, because the ordered middle values did not move. Delete the owner entirely and the mean of the remaining nine salaries is 370/9 ≈ 41.1, suddenly agreeing with the median (40). This is the operational meaning of *resistant*: the median and IQR barely notice extreme values, while the mean, range and standard deviation chase them. The exam exploits this in reverse: given that a mean exceeds a median, you should *infer* right skew; given a request to summarise income-like data, you should *choose* the resistant pair unprompted. Both directions — reading shape from summaries and choosing summaries from shape — are scored skills.

Case file: two phones, two boxplots

Battery life (hours) for two phone models, from samples of 40 phones each. Model A: min 9.2, Q1 12.8, median 14.1, Q3 15.6, max 18.9. Model B: min 6.5, Q1 10.9, median 13.8, Q3 17.9, max 24.6. Work the machinery: Model A's IQR = 15.6 − 12.8 = 2.8 h, fences at 12.8 − 4.2 = 8.6 and 15.6 + 4.2 = 19.8 — neither the min nor the max crosses a fence, so no outliers are flagged. Model B's IQR = 17.9 − 10.9 = 7.0 h, fences 10.9 − 10.5 = 0.4 and 17.9 + 10.5 = 28.4 — again no flagged outliers, despite the much wider range. The comparison a reader wants: Model A is slightly better on a typical day (median 14.1 vs 13.8 h) and dramatically more *consistent* (IQR 2.8 vs 7.0 h); Model B has the higher ceiling — its top quarter outlasts almost every Model A phone — but also the floor of 6.5 h. "Which phone is better?" is not a statistical question until the criterion (typical value? consistency? best case?) is named; distributions, unlike single averages, force that honesty.