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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: AP Statistics Unit 1 (2026-27) · Level: Grade 12 AP Statistics
Problem 1 Classifying Variables
(a) Blood type puts students in groups: categorical. (b) Height is a measured number: quantitative. (c) Number of siblings is a count: quantitative. (d) A zip code is a label; averaging zip codes is meaningless: categorical.
(a) categorical, (b) quantitative, (c) quantitative, (d) categorical
Problem 2 Relative Frequency Table
Divide each count by 50: Tacos 18/50 = 36%, BBQ 14/50 = 28%, Pizza 11/50 = 22%, Ramen 7/50 = 14%.
Tacos 36%, BBQ 28%, Pizza 22%, Ramen 14%
Problem 3 Reading a Bar Graph
(a) 142 + 96 + 38 + 24 = 300. (b) 96/300 = 32%. (c) 142 − 38 = 104.
(a) 300 students, (b) 32%, (c) 104 more students
Problem 4 Mean and Median of Scores
Sorted: 9, 11, 13, 14, 15, 17, 19, 22, 28. The median is the 5th value, 15. The sum is 148, so the mean is 148/9 = 16.44.
Mean ≈ 16.44 points, median = 15 points
Problem 5 Five-Number Summary
Sorted: 47, 49, 52, 55, 58, 59, 61, 64, 66, 70, 73. Median = 6th value = 59. Lower half (5 values): 47, 49, 52, 55, 58, median 52. Upper half: 61, 64, 66, 70, 73, median 66.
Min 47, Q1 52, median 59, Q3 66, max 73
Problem 6 Range and IQR
Range = 52 − 8 = 44. IQR = 29 − 15 = 14. Upper fence = 29 + 1.5(14) = 50. Since 52 > 50, the maximum is beyond the fence.
Range $44, IQR $14; yes, $52 is an outlier (fence $50)
Problem 7 Checking Plotted Outliers
IQR = 26 − 12 = 14, and 1.5 × 14 = 21. Upper fence = 26 + 21 = 47. Lower fence = 12 − 21 = −9, so nothing is low. Both 58 and 63 exceed 47, and the whisker end 41 is inside the fence.
Upper fence = 47 minutes; 58 and 63 are outliers
Problem 8 Standard Deviation by Hand
Mean = 135/5 = 27. Deviations: −4, 4, 0, 8, −8. Squares add to 160. Divide by n − 1 = 4: 40. Square root: 6.32.
s ≈ 6.32 cups
Problem 9 Percentile from a Dotplot
15 of the 20 students read fewer than 7 books, so the percentile is 15/20 = 75%. The right tail (9 and 12) is longer.
75th percentile; skewed right
Problem 10 Computing z-Scores
z = (95.0 − 86.4)/4.8 = 1.79. z = (79.2 − 86.4)/4.8 = −1.50.
z ≈ 1.79 and z = −1.50
Problem 11 Reading a Histogram
(a) Most bikes are cheap and the tail stretches right: skewed right. (b) Cumulative counts are 4, 13, 20; the 15th and 16th values are in $300–$400. (c) (5 + 3 + 2)/30 = 10/30 = 33.3%. (d) The long right tail pulls the mean up.
(a) skewed right, (b) $300–$400, (c) 33.3%, (d) the mean is larger
Problem 12 Effect of an Outlier
Before: mean 76/5 = 15.2, median 15, range 6. After: mean 116/5 = 23.2, median 15, range 46.
Mean 15.2 → 23.2; median stays 15; range 6 → 46
Problem 13 Comparing Two Distributions
Center: the medians are 70 (Section A) and 72 (Section B), so Section B is slightly higher. Spread: IQR is 78 − 61 = 17 for A and 77 − 66 = 11 for B; the ranges are 47 and 33. Section A scores are more spread out.
Section B has a slightly higher median (72 vs 70) and less variability (IQR 11 vs 17)
Problem 14 Using the Empirical Rule
(a) 226 and 274 are 2σ below and above the mean: 95%. (b) 238 and 262 are 1σ away: 68%. (c) 286 = 250 + 3(12); 99.7% lie within 3σ, so 0.3% lie outside and half of that, 0.15%, lie above.
(a) 95%, (b) 68%, (c) 0.15%
Problem 15 Normal Probability Between Values
z for 8.0: (8.0 − 9.6)/1.4 = −1.14. z for 11.0: (11.0 − 9.6)/1.4 = 1.00. Area = 0.8413 − 0.1265 = 0.7148.
About 0.715 (71.5%) of the tablets
Problem 16 Finding a Percentile Value
The 90th percentile has z = 1.282, so x = 48 + 1.282(7.5) = 57.6. By symmetry the 10th percentile has z = −1.282: x = 48 − 9.61 = 38.4.
10th percentile ≈ 38.4 minutes (finish faster than this); 90th percentile ≈ 57.6 minutes
Problem 17 Comparing Standardized Scores
Math: z = (83 − 74)/6 = 1.50. Science: z = (91 − 82)/11 = 0.82. A higher z-score means farther above the class mean.
Math: 1.50 standard deviations above the mean beats science at 0.82
Problem 18 Two Outlier Rules
(a) 18.4 + 2(4.1) = 26.6. Since 27 > 26.6, the member is an outlier by this rule. (b) IQR = 6, upper fence = 21 + 9 = 30. Since 27 < 30, the member is not an outlier by this rule.
Outlier by the 2σ rule only; not an outlier by the 1.5 × IQR rule
Problem 19 Changing Units of Measurement
Center values use the full formula: mean = 1.8(18.5) + 32 = 65.30, median = 1.8(19) + 32 = 66.2. Spread values ignore the added 32 and are only multiplied by 1.8: SD = 1.8(2.5) = 4.5, IQR = 1.8(3.5) = 6.3.
Mean 65.30°F, median 66.2°F, SD 4.5°F, IQR 6.3°F
Problem 20 Finding an Unknown Spread
The 15th percentile has z = −1.036. Solve −1.036 = (13.1 − 14.2)/σ, so σ = 1.1/1.036 = 1.06. For 16 pounds, z = (16 − 14.2)/1.06 = 1.70, and the area to the right is 0.0449.
σ ≈ 1.06 pounds; about 4.5% of bags weigh more than 16 pounds
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