← Back to the Binomial And Normal worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: AP Statistics Unit 2 (2026-27) · Level: Grade 12 AP Statistics
Problem 1 Is It Binomial?
(a) Two outcomes, independent trials, fixed n = 15, constant p: binomial. (b) The number of trials is not fixed (cards are dealt until the first heart): not binomial. (c) The sample is 5 of 28, more than 10% of the class, and sampling without replacement changes the probability: independence fails, so not binomial.
(a) binomial; (b) not binomial, number of trials not fixed; (c) not binomial, independence fails
Problem 2 A Binomial Probability by Formula
P(X = 2) = C(6, 2)(0.4)2(0.6)4 = 15 × 0.16 × 0.1296 = 0.3110.
≈ 0.3110
Problem 3 A Cumulative Binomial Probability
P(0) = 0.758 = 0.1001. P(1) = 8(0.25)(0.75)7 = 0.2670. The sum is 0.3671.
≈ 0.3671
Problem 4 Mean and Standard Deviation of a Count
μ = 60(0.35) = 21. σ = √(60 × 0.35 × 0.65) = √13.65 = 3.695.
μ = 21, σ ≈ 3.69
Problem 5 Reading a Binomial Histogram
The tallest bar is at 5 wins (0.2787). P(X ≥ 7) = P(7) + P(8) = 0.0896 + 0.0168 = 0.1064. The mean is 8(0.6) = 4.8.
5 wins; 0.1064; mean 4.8
Problem 6 At Least One Success
Let a failure to sprout have probability 0.15. P(at least one fails) = 1 − P(all sprout) = 1 − 0.855 = 1 − 0.4437 = 0.5563.
≈ 0.5563
Problem 7 Binomial with a Calculator
P(X ≥ 7) = 1 − P(X ≤ 6) = 1 − binomcdf(15, 0.3, 6) = 1 − 0.8689 = 0.1311.
≈ 0.1311
Problem 8 Is the Result Unusual?
P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.9568 = 0.0432. A probability of about 4.3% is small but not extremely small, so the result is somewhat unusual yet could occur by chance.
0.0432; fairly unlikely, though not impossible
Problem 9 Normal Area to the Left
z = (12 − 14.5)/2.1 = −1.19. The area to the left is 0.1169.
≈ 0.1169
Problem 10 Normal Area to the Right
z = (18 − 14.5)/2.1 = 1.67. The area to the right is 1 − 0.9522 = 0.0478.
≈ 0.0478
Problem 11 Normal Area Between Two Values
z = (66 − 71)/4.5 = −1.11 and z = (77 − 71)/4.5 = 1.33. Area = 0.9088 − 0.1333 = 0.7755.
≈ 0.7755
Problem 12 Finding a Cutoff
The cutoff has 90% of scores below it, so z = 1.282 and x = 78 + 1.282(8) = 88.25.
about 88.3 points
Problem 13 Quartiles of a Normal Model
For the 25th and 75th percentiles, z = ±0.6745. Q1 = 24 − 0.6745(6.5) = 19.62; Q3 = 24 + 0.6745(6.5) = 28.38.
Q1 ≈ $19.62, Q3 ≈ $28.38
Problem 14 Comparing Two Normal Models
Reading: z = (672 − 640)/24 = 1.33, percentile 90.9. Math: z = (58 − 52)/5 = 1.20, percentile 88.5.
Reading: about the 91st percentile, higher than math at about the 88th
Problem 15 Binomial Probability in Context
P(X ≤ 2) = 0.6127 + 0.3064 + 0.0702 = 0.9893. The expected number is 12(0.04) = 0.48.
0.9893; 0.48 defective batteries per box
Problem 16 Surprising Counts
μ = 50(0.3) = 15, σ = √(50 × 0.3 × 0.7) = 3.240. z = (25 − 15)/3.240 = 3.09. Exact: P(X ≥ 25) = 1 − binomcdf(50, 0.3, 24) = 0.0024. A result this far above the mean is very unlikely if the claim is true.
μ = 15, σ ≈ 3.24; z ≈ 3.09; P ≈ 0.0024
Problem 17 Normal Approximation Setup
Check: np = 60 ≥ 10 and n(1 − p) = 90 ≥ 10. μ = 60, σ = √(150 × 0.4 × 0.6) = 6.000. z = (70 − 60)/6.000 = 1.67, so P(X ≥ 70) ≈ 0.0478. (The exact binomial probability is 0.0575.)
≈ 0.048
Problem 18 Approximating a Lower Tail
np = 48 ≥ 10 and n(1 − p) = 352 ≥ 10. μ = 48, σ = √(400 × 0.12 × 0.88) = 6.499. z = (40 − 48)/6.499 = −1.23, so the area to the left is 0.1092.
≈ 0.109
Problem 19 Normal Model Then Binomial
(a) z = (12 − 12.1)/0.2 = −0.50, so p = 0.3085. (b) P(none underfilled) = (1 − 0.3085)8 = 0.0523, so P(at least one) = 0.9477.
(a) 0.3085; (b) 0.9477
Problem 20 Finding μ and σ from Two Percentiles
For 10% below, z = −1.282, so 40 = μ − 1.282σ. For 5% above, z = 1.645, so 70 = μ + 1.645σ. Subtract: 30 = 2.9264σ, so σ = 10.25. Then μ = 40 + 1.282(10.25) = 53.14.
μ ≈ 53.1 minutes, σ ≈ 10.3 minutes
Mis à jour le