Lesson 15: Sampling Distributions — Proportions

(a) Warm-Up

Imagine the whole country's voters, and exactly p = 0.52 of them favor a ballot measure. You don't get to ask everyone — you take a poll of n = 1000 people and compute the proportion who say "yes." Maybe you get 0.49. Your rival pollster takes a different 1000 people and gets 0.54. A third gets 0.51.

None of those numbers is 0.52, and none of them is "wrong." They're just different samples. Here's the question this whole lesson answers: if every polling firm in the world took its own sample of 1000, what would the pile of all their p̂ values look like? Would they cluster tightly around 0.52, or spray all over the place? Would the pile be bell-shaped?

That pile — the distribution of p̂ across all possible samples — is called a sampling distribution, and understanding its center, spread, and shape is the bridge between probability (Lessons 9–14) and inference (Lessons 19+). Every confidence interval and hypothesis test you'll ever build sits on top of it.

(b) Core Concept

Three distributions, and why people confuse them

Before any formulas, lock in a distinction that the AP exam loves to test:

1. The population distribution. This describes the individuals. If 52% of voters favor a measure, the population distribution is just: 52% "yes," 48% "no." It has one fixed center, the true parameter p. It does not change when you take a sample.

1. The distribution of a single sample. Take one sample of 1000 voters and graph their responses. About 49% (or 54%, or whatever your sample gave) say "yes." This is a snapshot of the individuals you happened to grab. As n grows, this snapshot looks more and more like the population distribution — that's a different idea from what comes next.

1. The sampling distribution of a statistic. Take every possible sample of size n, compute the statistic p̂ for each one, and graph all those p̂ values. This is a distribution of a statistic, not of individuals. It's the thing that tells you how much p̂ bounces around from sample to sample.

Key idea: A population distribution and a single sample's distribution are made of people. A sampling distribution is made of statistics — one dot per possible sample.

A sampling distribution is the distribution of a statistic (here, p̂) over all possible samples of a fixed size n drawn from the population. We never actually build it by taking infinitely many samples — we use probability theory to describe it. For p̂, that description has three parts: center, spread, and shape.

Center: μ_p̂ = p (unbiased)

The mean of the sampling distribution of p̂ equals the true population proportion:

μ_p̂ = p

Across all possible samples, the sample proportions average out to exactly the parameter. Because the center of the sampling distribution lands on the parameter, we say p̂ is an unbiased estimator of p. It doesn't systematically run high or low. Individual samples miss, but they miss in both directions equally.

Spread: σ_p̂ = sqrt(p(1−p)/n) (needs the 10% condition)

The standard deviation of the sampling distribution measures the typical sample-to-sample distance from p:

σ_p̂ = sqrt( p(1−p) / n )

Two things to notice. First, the n in the denominator means bigger samples give smaller spread — more data, less bounce. Second, this exact formula assumes you're sampling with replacement (or from an infinite population). Real samples are without replacement, which actually makes p̂ slightly less variable. The formula is close enough as long as the sample is a small slice of the population:

10% condition: n ≤ 0.10N. The sample size is at most 10% of the population size. When this holds, the with/without-replacement difference is negligible and σ_p̂ = sqrt(p(1−p)/n) is valid.

This is why pollsters can sample 1000 people from a population of millions — 1000 is way under 10% of millions.

Shape: approximately Normal when Large Counts holds

The shape depends on n and p. For small samples the sampling distribution can be skewed, but as n grows it becomes approximately Normal. The check:

Large Counts condition: np ≥ 10 and n(1−p) ≥ 10. You expect at least 10 successes and at least 10 failures.

When Large Counts holds, the sampling distribution of p̂ is approximately:

p̂ ≈ Normal( μ_p̂ = p, σ_p̂ = sqrt(p(1−p)/n) )

[GRAPH: Three overlaid bell-shaped curves all centered at the same value p = 0.60, showing the sampling distribution of p̂ for three sample sizes. X-axis: "Sample proportion p̂" (0.40 to 0.80). Y-axis: "Density." The flattest, widest curve is labeled "n = 50 (σ_p̂ ≈ 0.069)"; a taller, narrower curve is labeled "n = 150 (σ_p̂ = 0.04)"; the tallest, narrowest curve is labeled "n = 600 (σ_p̂ ≈ 0.020)." All three peak at 0.60. A caption reads: "Same center (p = 0.60), shrinking spread as n grows — the sampling distribution narrows around the parameter."]

The picture is the whole lesson in one image: all three curves are centered at p, so the center never moves. But as n grows from 50 to 600, the curve squeezes inward. Larger samples make p̂ a tighter, more reliable estimate of p.

Putting it together: a fully worked probability

Problem. Suppose 60% of a large city's adults support a new bike-lane plan, so p = 0.60. A reporter surveys an SRS of n = 150 adults. What is the probability that more than 65% of the sample support the plan — that is, P(p̂ > 0.65)?

Step 1 — Check the conditions.

• 10% condition: A large city has far more than 10 × 150 = 1500 adults, so n ≤ 0.10N. ✓ The spread formula is valid.
• Large Counts: np = 150(0.60) = 90 ≥ 10 and n(1−p) = 150(0.40) = 60 ≥ 10. ✓ The shape is approximately Normal.

Step 2 — Find center and spread.

μ_p̂ = p = 0.60
σ_p̂ = sqrt( p(1−p)/n ) = sqrt( (0.60)(0.40)/150 )
     = sqrt( 0.24 / 150 ) = sqrt( 0.0016 ) = 0.04

Step 3 — Standardize.

z = (p̂ − μ_p̂) / σ_p̂ = (0.65 − 0.60) / 0.04 = 0.05 / 0.04 = 1.25

Step 4 — Find the probability. We want the area above z = 1.25.

TI-84: 2nd → VARS (DISTR) → normalcdf
normalcdf(lower: 0.65, upper: 1E99, μ: 0.60, σ: 0.04)
Output: 0.1056
(Equivalently, normalcdf(1.25, 1E99, 0, 1) = 0.1056)

Step 5 — Interpret. There's about a 0.1056 probability — roughly a 10.6% chance — that an SRS of 150 adults from this city yields a sample proportion above 0.65, when the true support is 0.60. So a single sample landing above 65% isn't shocking; it happens about 1 time in 10 just from sampling variability.

(c) Worked Examples

Example 1 (easy) — Center, spread, conditions

A factory's chips are 40% Type A, so p = 0.40. An inspector pulls an SRS of n = 200 chips from a day's production of 50,000.

Strategy: Compute μ_p̂ and σ_p̂, and check both conditions.

Solution:

μ_p̂ = p = 0.40
σ_p̂ = sqrt( (0.40)(0.60)/200 ) = sqrt( 0.24/200 ) = sqrt(0.0012) = 0.0346

• 10% condition: 200 ≤ 0.10(50,000) = 5000. ✓
• Large Counts: np = 200(0.40) = 80 ≥ 10, n(1−p) = 200(0.60) = 120 ≥ 10. ✓

Interpretation: The sampling distribution of p̂ is approximately Normal, centered at 0.40, with a standard deviation of about 0.0346. Sample proportions typically land within a few hundredths of 0.40.

Example 2 (medium) — A left-tail probability

About 25% of a college's students are left-handed-dominant for a certain task, p = 0.25. A researcher takes an SRS of n = 120 students. Find P(p̂ < 0.20).

Strategy: Check conditions, find σ_p̂, standardize, use normalcdf.

Solution:

• 10% condition: assume the college has well over 1200 students. ✓
• Large Counts: np = 120(0.25) = 30 ≥ 10, n(1−p) = 120(0.75) = 90 ≥ 10. ✓

σ_p̂ = sqrt( (0.25)(0.75)/120 ) = sqrt( 0.1875/120 ) = sqrt(0.0015625) = 0.0395
z = (0.20 − 0.25) / 0.0395 = −0.05 / 0.0395 = −1.265
normalcdf(−1E99, 0.20, 0.25, 0.0395) = 0.1030

Interpretation: There's about a 0.1030 probability that fewer than 20% of a sample of 120 students are left-dominant, given the true rate is 25%.

Example 3 (AP-style) — A "between" probability

In a national election the true proportion favoring Candidate X is p = 0.50. A polling firm surveys an SRS of n = 400 likely voters. Find the probability that the sample proportion lands between 0.47 and 0.53 — i.e., P(0.47 < p̂ < 0.53).

Strategy: Conditions → spread → two z-scores → normalcdf between them.

Solution:

• 10% condition: the electorate vastly exceeds 4000. ✓
• Large Counts: np = 400(0.50) = 200 ≥ 10, n(1−p) = 200 ≥ 10. ✓

σ_p̂ = sqrt( (0.50)(0.50)/400 ) = sqrt( 0.25/400 ) = sqrt(0.000625) = 0.025
z₁ = (0.47 − 0.50)/0.025 = −1.20
z₂ = (0.53 − 0.50)/0.025 = +1.20
normalcdf(0.47, 0.53, 0.50, 0.025) = 0.7699

Interpretation: About 77% of all samples of 400 voters give a p̂ within 3 percentage points of the true 50%. This is exactly why a margin of error of "±3 points" is so common — it captures the bulk of sampling variability for n ≈ 400.

Example 4 (concept) — The effect of larger n on spread

Suppose p = 0.30. Compare the spread of the sampling distribution of p̂ for n = 100 versus n = 400.

Strategy: Compute σ_p̂ for each and compare.

Solution:

n = 100:  σ_p̂ = sqrt( (0.30)(0.70)/100 ) = sqrt(0.0021) = 0.0458
n = 400:  σ_p̂ = sqrt( (0.30)(0.70)/400 ) = sqrt(0.000525) = 0.0229

Interpretation: Quadrupling n (100 → 400) halved the standard deviation (0.0458 → 0.0229). Because n is under the square root, to cut the spread in half you must multiply the sample size by four. Precision improves with sample size — but only at the square-root rate, so each extra unit of precision costs more and more data.

(d) Common Mistakes

1. Skipping the condition checks. Students jump straight to normalcdf without verifying Large Counts and the 10% condition. On the FRQ, the conditions are worth points — and if Large Counts fails, the Normal calculation isn't even valid. Always check both, with numbers, before computing.

2. Using σ instead of σ_p̂. A problem might mention a population standard deviation or treat the situation like a single observation. For a proportion the spread of the sampling distribution is sqrt(p(1−p)/n) — not some other σ, and not sqrt(p(1−p)) (that's the SD of a single 0/1 outcome, the case n = 1). Forgetting to divide by n makes your spread far too big.

3. Forgetting the 10% condition. The shape check (Large Counts) gets all the attention, so students forget the spread check. The 10% condition (n ≤ 0.10N) is what justifies the σ_p̂ formula for sampling without replacement. Mention it explicitly — graders look for it.

4. Confusing the population SD with the sampling-distribution SD. "The standard deviation is 0.49" (from sqrt(0.6·0.4)) describes the spread of individuals, not the spread of p̂. The sampling distribution is always tighter — divided by n inside the root. Mixing these up gives a z-score that's off by a factor of sqrt(n).

5. Forgetting p̂ is approximately Normal, not exactly Normal. The Normal model is an approximation justified by Large Counts. Write "approximately Normal." It's a small thing, but it signals you understand why the condition matters.

(e) Practice Problems

A population proportion is p = 0.45. For an SRS of n = 100, the standard deviation of the sampling distribution of p̂ is closest to:

- (A) 0.0025

- (B) 0.0497

- (C) 0.2475

- (D) 0.4975

Which of the following is the center of the sampling distribution of p̂?

- (A) p̂

- (B) p

- (C) sqrt(p(1−p)/n)

- (D) np

A sampling distribution of p̂ is approximately Normal provided that:

- (A) the population is Normal

- (B) n ≥ 30

- (C) np ≥ 10 and n(1−p) ≥ 10

- (D) the sample is larger than 10% of the population

Holding p fixed, which change makes the sampling distribution of p̂ narrower?

- (A) Decreasing n

- (B) Increasing n

- (C) Changing the population size only

- (D) Nothing changes the spread once p is fixed

For p = 0.50, the standard deviation of p̂ for n = 100 is 0.05. To cut that standard deviation to 0.025, the sample size should be:

- (A) 200

- (B) 300

- (C) 400

- (D) 1000

The 10% condition (n ≤ 0.10N) is what justifies which part of the sampling-distribution model for p̂?

- (A) its center

- (B) its spread formula

- (C) its Normal shape

- (D) the value of p

A candy company claims 80% of its bags pass inspection (p = 0.80). An SRS of n = 50 bags is taken from a large warehouse. Find P(p̂ < 0.75). Show your conditions, σ_p̂, the z-score, and the normalcdf call.

(In context.) Nationally, about 65% of teens report using a particular app, p = 0.65. A school surveys an SRS of n = 200 of its many students. Find the probability that more than 70% of the sample use the app, P(p̂ > 0.70), and interpret the result.

(In context.) A city says 30% of its residents commute by bike, p = 0.30. A transportation study takes an SRS of n = 150 residents from the large city population. Find P(0.25 < p̂ < 0.35) and interpret it.

A recent poll uses p = 0.52 and n = 600 (from a huge electorate). Find P(p̂ < 0.50). State your conditions and show the z-score.

A quality engineer wants the standard deviation of p̂ to be no more than half its current value. Her current sample size is 100. What is the smallest sample size that achieves this? (Assume p is unchanged.)

A blood-type proportion is p = 0.10 in the population. A clinic takes an SRS of n = 80 patients. Is the Normal model appropriate for the sampling distribution of p̂? Justify with the Large Counts check.

(f) FRQ Practice (10 points · Practice 3 — Analyze Data)

> The problem. A national health survey reports that 12% of teenagers regularly skip breakfast on school days, so the population proportion is p = 0.12. A researcher at a large urban high school district (enrollment over 40,000 students) takes a simple random sample of n = 250 students and records the proportion p̂ who regularly skip breakfast.

>

> (a) Describe the shape, center, and spread of the sampling distribution of p̂. Justify the shape and the spread by checking the appropriate conditions. (5 points)

>

> (b) The researcher worries that her sample might overstate the problem. Find the probability that the sample proportion exceeds 0.15 — that is, P(p̂ > 0.15). Show your work. (3 points)

>

> (c) Interpret the probability from part (b) in the context of this study. (2 points)

Model Response

Part (a).

Center: The mean of the sampling distribution of p̂ is μ_p̂ = p = 0.12. (p̂ is an unbiased estimator of p.)

Spread: The 10% condition is satisfied because the sample of 250 is at most 10% of the district population: 250 ≤ 0.10(40,000) = 4000. ✓ So

σ_p̂ = sqrt( p(1−p)/n ) = sqrt( (0.12)(0.88)/250 )
     = sqrt( 0.1056/250 ) = sqrt(0.0004224) = 0.0206

Shape: The Large Counts condition is satisfied: np = 250(0.12) = 30 ≥ 10 and n(1−p) = 250(0.88) = 220 ≥ 10. ✓ Therefore the sampling distribution of p̂ is approximately Normal.

Summary: approximately Normal, center μ_p̂ = 0.12, spread σ_p̂ ≈ 0.0206.

Part (b).

z = (p̂ − μ_p̂)/σ_p̂ = (0.15 − 0.12)/0.0206 = 0.03/0.0206 = 1.46

P(p̂ > 0.15) = P(z > 1.46)
TI-84: normalcdf(0.15, 1E99, 0.12, 0.0206) = 0.0722

So P(p̂ > 0.15) ≈ 0.0722.

Part (c). If the true proportion of teens who skip breakfast in this district is 0.12, then about 7.2% of all simple random samples of 250 students would produce a sample proportion greater than 0.15. A sample result above 0.15 is somewhat unusual but not extreme — it would happen roughly 1 time in 14 by chance alone, so a single such sample does not strongly suggest the true rate exceeds 0.12.

Scoring Rubric (10 points total)

Part (a) — 5 points

Part (b) — 3 points

Part (c) — 2 points

Where students lose points:

• Part (a): Stating "Normal" without the Large Counts check, or checking only np and forgetting n(1−p). Skipping the 10% condition entirely (very common — it's a full point). Reporting σ as sqrt(0.12·0.88) = 0.325 (the individual SD) instead of dividing by n.
• Part (b): Using σ = sqrt(p(1−p)) = 0.325 in the z-score, which gives z ≈ 0.09 and a wrong probability near 0.46. Finding the left tail (0.928) instead of the right tail. Forgetting 1E99 and using a tiny upper bound.
• Part (c): A "naked" interpretation ("the probability is 0.0722") with no context and no mention of the assumed p = 0.12. Saying the result "proves" the rate is higher — one sample can't prove that.