The Tutoring Center's "10-Point Gain" Claim
A tutoring center's flyer reads: "Our students improve by an average of 10 points." The director pulls the practice-test records for the five students who finished the program this term (before → after): Student A 60 → 66; Student B 72 → 78; Student C 85 → 90; Student D 55 → 85; Student E 90 → 94. A parent touring the center asks what gain a typical new student should expect. The director also has a prospective sixth student who, based on the intake assessment, is projected to gain 6 points.
1. Verify the flyer's claim numerically and assess whether it is a fair summary of what happened.
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Individual gains: A 6, B 6, C 5, D 30, E 4. Sum = 51.
Mean gain: 51 ÷ 5 = 10.2 points, so the flyer's "average of 10 points" is arithmetically true.
Ordered gains: 4, 5, 6, 6, 30 → median = 6 points.
Assessment: four of the five students gained 4–6 points; the mean is pulled up by a single 30-point outlier (Student D, who started lowest at 55). The flyer reports a real statistic but is misleading: it describes no actual student well and would overstate the typical result by roughly 4 points. A fair flyer would say "median gain 6 points, range 4–30."
Score means for context: before = 362 ÷ 5 = 72.4; after = 413 ÷ 5 = 82.6.
2. What should the director tell the parent who asks about a "typical" gain, and about how much of the 10-point average is driven by one student?
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Recommendation to the parent: cite the median of 6 points, with a typical range of 4–6 points for four of five students, noting that one student with a low starting score gained 30.
Robustness check: without Student D, the mean gain is (6 + 6 + 5 + 4) ÷ 4 = 5.25 points - barely half the advertised figure.
Sensitivity: Student D's gain alone contributes 30 ÷ 5 = 6.0 points of the 10.2-point mean, i.e. about 59% of the advertised "average."
Why median is better here: with n = 5 and one extreme value, the median resists distortion and matches what the parent is actually asking ("what will happen to my child"). Quote the mean only alongside the median and the spread, never alone.
3. The projected sixth student joins with a 6-point gain. Recompute the mean and median and state whether the flyer's claim still holds.
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New gains: 6, 6, 5, 30, 4, 6 → sum = 57 over 6 students.
New mean: 57 ÷ 6 = 9.5 points → 9.5 < 10, so the "average of 10 points" claim now fails on its own terms.
New median: ordered 4, 5, 6, 6, 6, 30 → average of the 3rd and 4th values = (6 + 6) ÷ 2 = 6 points (unchanged).
Interpretation: adding one ordinary student moved the mean by 0.7 points and broke the advertising claim, while the median did not move at all - evidence the claim was always resting on a single outlier. The center should stop quoting a rounded mean as a headline and report median gain with the range.