Sampling Distributions Worksheet Answer Key – AP Statistics

Problem 1 Parameter or Statistic

(a) It describes the whole population: a parameter, μ. (b) It describes a sample: a statistic, p̂. (c) It describes a sample of batteries: a statistic, x̄.

(a) parameter μ; (b) statistic p̂; (c) statistic x̄

Problem 2 Center and Spread of p̂

μ = p = 0.40. σ = √(0.40 × 0.60/100) = √0.0024 = 0.0490.

mean 0.40, standard deviation ≈ 0.0490

Problem 3 Checking Conditions for p̂

The sample is under 10% of 5,000 (120 < 500). For n = 120: np = 9.6 < 10, so the normal approximation is not appropriate. For n = 250: np = 20 ≥ 10 and n(1 − p) = 230 ≥ 10, so it is appropriate.

Not normal for n = 120 (np = 9.6); approximately normal for n = 250

Problem 4 Probability for a Sample Proportion

σ = √(0.35 × 0.65/200) = 0.0337. Conditions: np = 70 and n(1 − p) = 130 are both at least 10. z = (0.40 − 0.35)/0.0337 = 1.48. P(p̂ ≥ 0.40) = 0.0691.

≈ 0.0691

Problem 5 Reading a Simulated Sampling Distribution

The distribution is roughly symmetric and mound-shaped. The mean of the 40 values is 0.283, close to 0.30, which suggests p̂ is unbiased. The standard deviation of the simulated values is 0.059, near the theoretical √(0.30 × 0.70/50) = 0.065.

Symmetric, center ≈ 0.28, spread ≈ 0.059 (theory 0.065)

Problem 6 Sample Size and Spread

n = 100: √(0.5 × 0.5/100) = 0.050. n = 400: √(0.5 × 0.5/400) = 0.025. The standard deviation is cut in half.

0.050 versus 0.025: quadrupling n halves the standard deviation

Problem 7 Matching Bias and Variability

(a) A: centered on the bullseye with little scatter. (b) C: dots are tightly grouped but away from the target. (c) D. (d) Method B is unbiased but scattered; taking larger samples reduces variability.

(a) A; (b) C; (c) D; (d) increase the sample size

Problem 8 Mean and SD of a Sample Mean

μ_x̄ = 72. σ_x̄ = 9/√36 = 9/6 = 1.5.

mean 72, standard deviation 1.5

Problem 9 Normal Population, Sample Mean

The population is normal, so x̄ is normal with mean 160 and σ = 14/√25 = 2.8. z = (164 − 160)/2.8 = 1.43. P(x̄ > 164) = 0.0766.

≈ 0.0766

Problem 10 Central Limit Theorem Application

The population is skewed, but n = 45 ≥ 30, so by the Central Limit Theorem x̄ is approximately normal with mean 15.2 and σ = 11/√45 = 1.640. z = (18 − 15.2)/1.640 = 1.71, and the probability is 0.0439.

≈ 0.0439

Problem 11 Individual Versus Mean

(a) z = (150 − 160)/14 = −0.71, probability 0.2375. (b) σ_x̄ = 14/4 = 3.5, z = −2.86, probability 0.0021. (c) Averages vary much less than individuals, so a mean far from 160 is rarer than an individual far from 160.

(a) 0.2375; (b) 0.0021

Problem 12 Sample Size for a Target Standard Deviation

Solve 14/√n ≤ 2: √n ≥ 7, so n ≥ 49.

n = 49

Problem 13 Difference of Two Proportions

Mean = 0.55 − 0.48 = 0.07. σ = √(0.55 × 0.45/120 + 0.48 × 0.52/150) = √(0.002063 + 0.001664) = 0.0610.

mean 0.07, standard deviation ≈ 0.0610

Problem 14 Probability for a Difference of Proportions

Conditions: 120(0.55) = 66, 120(0.45) = 54, 150(0.48) = 72, 150(0.52) = 78, all at least 10. z = (0 − 0.07)/0.0610 = −1.15. P(difference < 0) = 0.1258.

≈ 0.1258

Problem 15 Difference of Two Sample Means

Mean = 82 − 76 = 6. σ = √(12²/40 + 10²/35) = √(3.600 + 2.857) = 2.541.

mean 6, standard deviation ≈ 2.541

Problem 16 Probability for a Difference of Means

z = (10 − 6)/2.541 = 1.57. P(x̄_A − x̄_B > 10) = 0.0577.

≈ 0.0577

Problem 17 A Complete Sampling Distribution

The population mean is (3 + 5 + 9 + 11)/4 = 7.

The mean of the 16 values of x̄ is 7, equal to μ, so x̄ is an unbiased estimator of μ.

Mean of x̄ = 7 = μ

Problem 18 Quality Control Proportion

Conditions: 500(0.04) = 20 ≥ 10, 500(0.96) = 480 ≥ 10, and the shipment is very large. σ = √(0.04 × 0.96/500) = 0.00876. z = (0.06 − 0.04)/0.00876 = 2.28. P(p̂ > 0.06) = 0.0112.

≈ 0.0112

Problem 19 Is the Sample Mean Unusual?

σ_x̄ = 6/√36 = 1. z = (338 − 340)/1 = −2.0. P(x̄ ≤ 338) = 0.0228. A probability of about 2.3% is small, so a sample mean this low would be unusual if the mean really were 340, which is moderate evidence that the true mean is lower.

≈ 0.0228; moderate evidence against the claim

Problem 20 Comparing Two Clinics

Mean of p̂_A − p̂_B = 0.08. σ = √(0.70 × 0.30/140 + 0.62 × 0.38/120) = √(0.001500 + 0.001963) = 0.0589. z = (0.15 − 0.08)/0.0589 = 1.19. Probability = 0.1171. A gap this large happens about 12% of the time by chance alone, even though the true gap is 0.08.

≈ 0.1171