AP Statistics · Unit 3 · Topic 3.2

Sampling Distributions for Sample Proportions Study Lab

Every random sample gives a slightly different sample proportion p̂. The sampling distribution of p̂ describes all those values: it is centered at p, has standard deviation √(p(1 − p)/n), and is approximately normal when the Large Counts condition holds.

1. Build the sampling distribution of p̂

Set the true population proportion and the sample size, then take random samples. The histogram shows the p̂ values; the black curve is the normal model the formulas predict.

μ_p̂ = p σ_p̂ = √( p(1 − p) ÷ n ) np ≥ 10 and n(1 − p) ≥ 10

Preset situations

Population

True proportion p

Sample size n

Population size N Used for the 10% condition

Probability question: P(p̂ ≥ )

Take samples

Simulated p̂ values

Simulated p̂P(p̂ ≥ cutoff) regionTrue p

Mean of p̂ values (theory: p)

SD of p̂ values (theory: σ_p̂)

P(p̂ ≥ cutoff), simulation

P(p̂ ≥ cutoff), normal model

Show your work: normal calculation

2. Key ideas

Center

μ_p̂ = p

On average, p̂ equals the true proportion. That makes p̂ an unbiased estimator of p.

Spread

σ_p̂ = √(p(1 − p)/n)

Spread shrinks as n grows: 4 times the sample size gives half the spread. It is largest when p = 0.5.

Shape

Large Counts

If np ≥ 10 and n(1 − p) ≥ 10, the sampling distribution is approximately normal. Otherwise it is skewed toward the side farther from 0 or 1.

10% condition

When sampling without replacement

Check n ≤ 0.1N. Then the observations are close enough to independent that the σ formula works.

Probability

Use the normal model

Standardize: z = (p̂ − p) ÷ σ_p̂, then find the area, or use normalcdf with mean p and SD σ_p̂.

Three distributions

Population, sample, sampling distribution

The population has one p. A sample has its own p̂. The sampling distribution shows how p̂ varies over all samples of size n.

3. Quick reference

Typical questions and their setups. Always check conditions before using the normal model.

Question	Setup	Work	Answer
p = 0.4, n = 60. Find μ and σ of p̂.	μ = p; σ = √(p(1 − p)/n)	√(0.4 · 0.6 ÷ 60) = √0.004	μ = 0.4, σ ≈ 0.0632
Is the shape approximately normal?	np ≥ 10, n(1 − p) ≥ 10	60(0.4) = 24, 60(0.6) = 36	Yes
Find P(p̂ > 0.5).	z = (0.5 − 0.4) ÷ 0.0632	z ≈ 1.58	≈ 0.057
p = 0.05, n = 80. Normal?	np = 4	4 < 10	No, skewed right
What n makes σ = 0.02 when p = 0.5?	√(0.25/n) = 0.02	n = 0.25 ÷ 0.0004	n = 625

4. On the FRQ

How to earn the points

Describe a sampling distribution with shape, center, and spread, each justified: "approximately normal because 60(0.4) = 24 ≥ 10 and 60(0.6) = 36 ≥ 10."
Show the σ formula with numbers plugged in, not just the result.
For a probability, label the distribution: "Using N(0.4, 0.0632), P(p̂ > 0.5) = 0.057." Calculator syntax alone does not earn credit.
Name both conditions and what each does: Large Counts for shape, 10% for the standard deviation formula.
Interpret in context: "There is about a 5.7% chance that more than half of a random sample of 60 students will say yes."

Common mistakes

Using p̂ instead of p in the Large Counts check when p is known. For sampling distributions, use the true p.
Mixing up the 10% condition (independence, σ formula) with Large Counts (normal shape).
Forgetting the square root in σ = √(p(1 − p)/n).
Saying a larger sample makes the sampling distribution "more normal" without the Large Counts numbers to back it up.
Confusing the distribution of one sample's data (yes/no answers) with the sampling distribution of p̂.

5. Practice questions

AP-style multiple choice. Pick an answer to see the explanation.

0 / 0 correct

6. Key terms

The vocabulary for topic 3.2. Read the list, flip through flashcards, or quiz yourself. For every term in the unit, see the Unit 3 terminology guide.