The Sampling Distribution of a Sample Proportion — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: Center, spread, and shape of the sampling distribution of $\hat{p}$, the 10% condition, the Large Counts Normal approximation condition, probability calculation for $\hat{p}$, bias of the sample proportion, and notation conventions for parameters and statistics.

You should already know: The difference between a population parameter and a sample statistic. Properties of the Normal distribution and z-score probability calculations. How to draw a random sample from a population.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Statistics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is The Sampling Distribution of a Sample Proportion?

When we collect a random sample from a population with a categorical trait (e.g., voter support, defective products, flower color), we calculate a sample proportion $\hat{p}$ to estimate the true population proportion $p$. The sampling distribution of a sample proportion is the probability distribution of every possible value of $\hat{p}$ calculated from all random samples of the same size $n$ drawn from the same population.

In AP Statistics, this topic is part of Unit 5: Sampling Distributions, which accounts for 10-15% of the total AP exam score; this subtopic makes up roughly half of the Unit 5 weight, or 5-8% of your total exam score. It appears on both multiple choice (MCQ) and free response (FRQ) sections, often as a standalone question or as a prerequisite step for inference questions on proportions.

Notation conventions are standardized: $p$ is the fixed, unknown population proportion (parameter), $\hat{p}$ is the sample proportion (statistic, varies per sample), $n$ is sample size, ${\mu }_{\hat{p}}$ is the mean of the sampling distribution, and ${\sigma }_{\hat{p}}$ is the standard deviation (often called the standard error) of the sampling distribution. This topic is sometimes shortened to "sampling distribution of $\hat{p}$" in course materials.

2. Center and Spread of the Sampling Distribution of $\hat{p}$

The two most important basic properties of any sampling distribution are its center (average value across all samples) and its spread (how much $\hat{p}$ varies from sample to sample). For the sampling distribution of $\hat{p}$, the center is always equal to the true population proportion: $${\mu }_{\hat{p}}=p$$ This property means $\hat{p}$ is an unbiased estimator of $p$: on average, across all possible random samples of the same size, the sample proportion hits the true population proportion exactly. There is no systematic overestimation or underestimation of $p$ when random sampling is used.

For spread, the standard deviation (standard error) of $\hat{p}$ follows the formula: $${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$$ Intuition for this formula: $p(1-p)$ reaches its maximum when $p=0.5$, so proportions near 0.5 have more variability than proportions close to 0 or 1. For example, if 99% of a population has a trait, almost all samples will get proportions between 97% and 100%, so spread is very small. The sample size $n$ is in the denominator: larger samples produce less variable estimates, so to cut the standard error in half, you need a sample 4 times as large.

Worked Example

A small city has 62% of its registered voters who are registered as Democrats. A researcher takes a random sample of 75 registered voters to estimate the proportion of Democratic voters. Calculate the mean and standard deviation of the sampling distribution of $\hat{p}$.

1. Identify parameters: $p=0.62$ (true population proportion of Democrats), $n=75$ (sample size).
2. The mean of the sampling distribution is ${\mu }_{\hat{p}}=p=0.62$, by definition of an unbiased estimator.
3. Check the 10% condition to confirm the standard deviation formula is valid: the city has far more than $10\ast 75=750$ registered voters, so the condition holds.
4. Calculate the standard deviation: $${\sigma }_{\hat{p}}=\sqrt{\frac{(0.62)(0.38)}{75}}=\sqrt{\frac{0.2356}{75}}\approx 0.056$$

Exam tip: Always distinguish between $\hat{p}$ (a single value from your sample) and ${\mu }_{\hat{p}}$ (the mean of all possible $\hat{p}$ values) — mixing up these two notations is a common point deduction on FRQs.

3. Conditions for a Normal Sampling Distribution

To use the Normal distribution to calculate probabilities for $\hat{p}$ (required for most AP exam problems on this topic), two conditions must be checked and satisfied. Each condition serves a separate purpose:

1. 10% Condition: When sampling without replacement (the standard case in almost all real-world problems), the sample size $n$ must be no more than 10% of the total population size $N$, or $n\le 0.1N$. The formula for ${\sigma }_{\hat{p}}$ assumes independent observations. Sampling without replacement introduces small dependence between observations, but this dependence is negligible if the sample is less than 10% of the population, so the formula remains valid.
2. Large Counts Condition (Normal Condition): The expected number of successes and failures in the sample must both be at least 10, or $np\ge 10$ and $n(1-p)\ge 10$. This rule ensures that the sampling distribution of $\hat{p}$ is close enough to Normal that we can use Normal probability calculations instead of the more complex exact binomial distribution. Note that the AP Statistics CED requires 10, not the 5 that appears in some older textbooks — always use 10 on the exam.

Worked Example

A bakery knows that 12% of its chocolate chip cookies contain at least 10 chocolate chips. The bakery manager takes a random sample of 75 cookies from the 2000 cookies baked that day to check quality. Do the conditions for a Normal sampling distribution of $\hat{p}$ hold? Justify your answer.

1. Check the 10% condition: Total population $N=2000$, 10% of $N$ is $0.1\ast 2000=200$. The sample size $n=75<200$, so the 10% condition is satisfied.
2. Check expected successes: $np=75\ast 0.12=9$, which is less than 10.
3. Check expected failures: $n(1-p)=75\ast 0.88=66$, which is greater than 10.
4. Conclusion: Because $np=9<10$, the Large Counts Condition fails. The sampling distribution of $\hat{p}$ is not approximately Normal, so the Normal approximation cannot be used.

Exam tip: On FRQs, you must explicitly name each condition and show your calculation for the check to earn full credit — just saying "conditions are met" earns zero points for the condition step.

4. Calculating Probabilities for a Sample Proportion

Once both conditions are satisfied, the sampling distribution of $\hat{p}$ is approximately Normal with mean ${\mu }_{\hat{p}}=p$ and standard deviation ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$. To find the probability that $\hat{p}$ falls in any range, we convert the value of $\hat{p}$ to a z-score, then use the standard Normal distribution to find the probability, just like we do for any other Normal distribution.

The z-score formula for $\hat{p}$ is: $$z=\frac{\hat{p}-{\mu }_{\hat{p}}}{{\sigma }_{\hat{p}}}=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$$ This is used to answer common exam questions like "what is the probability the sample proportion is within 0.04 of the true proportion?" or "what is the probability the sample proportion is greater than 0.5?"

Worked Example

A cell phone company knows that 18% of its customers will upgrade their phone this year. The company takes a random sample of 200 of its 15000 total customers. What is the probability that more than 20% of the sampled customers will upgrade their phone this year?

1. Check conditions: 10% condition: $200<0.1\ast 15000=1500$, so satisfied. Large Counts: $np=200\ast 0.18=36\ge 10$, $n(1-p)=200\ast 0.82=164\ge 10$, so satisfied.
2. Calculate parameters: ${\mu }_{\hat{p}}=0.18$, ${\sigma }_{\hat{p}}=\sqrt{\frac{(0.18)(0.82)}{200}}\approx 0.0272$.
3. Calculate the z-score for $\hat{p}=0.20$: $z=\frac{0.20-0.18}{0.0272}\approx 0.74$.
4. Find the probability: $P(\hat{p}>0.20)=P(z>0.74)=1-P(z<0.74)=1-0.7704=0.2296$.
5. Conclusion: There is approximately a 23% chance that more than 20% of the sample will upgrade their phone this year.

Exam tip: When $p$ is known (as it always is for sampling distribution problems before inference), always use $p$ to calculate ${\sigma }_{\hat{p}}$ — never use $\hat{p}$ here, that is only for confidence intervals when $p$ is unknown.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $\hat{p}$ instead of $p$ to calculate ${\sigma }_{\hat{p}}$ when $p$ is known for probability calculations. Why: Students confuse this with confidence interval inference, where we do not know $p$ so we use $\hat{p}$ to estimate standard error. Correct move: When the true population $p$ is given, always use $p$ to calculate ${\sigma }_{\hat{p}}$.
• Wrong move: Forgetting the square root around $\frac{p(1-p)}{n}$ when calculating ${\sigma }_{\hat{p}}$. Why: Students memorize the variance $\frac{p(1-p)}{n}$ but forget standard deviation is the square root of variance. Correct move: Every time you calculate ${\sigma }_{\hat{p}}$, double-check that you have a square root around the entire fraction.
• Wrong move: Mixing up the purpose of the 10% and Large Counts conditions, or only checking one. Why: Students memorize the two conditions but do not learn what each checks. Correct move: Remember: 10% = valid standard error for sampling without replacement; Large Counts = Normality of the sampling distribution. Check both explicitly.
• Wrong move: Saying $\hat{p}$ is unbiased just because "the sample is random" without linking to the sampling distribution mean. Why: Students memorize that random sampling gives unbiased estimators, but do not state the definition correctly. Correct move: To confirm $\hat{p}$ is unbiased, state "the mean of the sampling distribution of $\hat{p}$ equals the true population proportion $p$".
• Wrong move: Using 5 instead of 10 for the Large Counts condition. Why: Some older textbooks use 5, but the AP Statistics CED requires 10. Correct move: Always use the cutoff of 10 for expected successes and failures on the AP exam.

6. Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

A toy factory produces rubber balls, and 7% of all balls produced have a manufacturing defect that makes them unsafe. A quality control inspector takes a random sample of 175 balls from the thousands of balls produced in a day. What is the approximate probability that less than 5% of the sampled balls are defective? A) 0.16 B) 0.22 C) 0.78 D) 0.84

Worked Solution: First confirm conditions: 10% condition is satisfied because the population of balls is thousands, so 175 < 10% of the population. Large Counts: $np=175\ast 0.07=12.25\ge 10$, $n(1-p)=175\ast 0.93=162.75\ge 10$, so conditions hold. The sampling distribution is approximately Normal with ${\mu }_{\hat{p}}=0.07$, ${\sigma }_{\hat{p}}=\sqrt{\frac{(0.07)(0.93)}{175}}\approx 0.0193$. The z-score for $\hat{p}=0.05$ is $z=\frac{0.05-0.07}{0.0193}\approx -1.04$. The probability $P(z<-1.04)\approx 0.1492$, which is closest to 0.16. Correct answer: A.

Question 2 (Free Response)

A large university reports that 38% of its undergraduate students live on campus. The student government takes a random sample of 60 undergraduates to ask about their housing. (a) Describe the sampling distribution of the sample proportion of undergraduate students who live on campus, including shape, center, spread, and justification of all conditions. (b) What is the probability that the sample proportion of students living on campus is between 35% and 41%? (c) If the student government increases their sample size to 240 students, how does this change the center, spread, and shape of the sampling distribution, assuming all conditions still hold?

Worked Solution: (a) Conditions Check: 10% condition: The university has far more than $10\ast 60=600$ undergraduates, so the 10% condition is satisfied. Large Counts: $np=60\ast 0.38=22.8\ge 10$, $n(1-p)=60\ast 0.62=37.2\ge 10$, so the Large Counts condition is satisfied. Sampling Distribution Description: Shape: Approximately Normal. Center: ${\mu }_{\hat{p}}=p=0.38$. Spread: ${\sigma }_{\hat{p}}=\sqrt{\frac{(0.38)(0.62)}{60}}\approx 0.0627$.

(b) We want $P(0.35<\hat{p}<0.41)$. Calculate z-scores: $z_1=\frac{0.35-0.38}{0.0627}\approx -0.48$, $z_2=\frac{0.41-0.38}{0.0627}\approx 0.48$. $P(-0.48<z<0.48)=P(z<0.48)-P(z<-0.48)=0.6844-0.3156=0.3688$. The probability is approximately 0.37.

(c) Center: The mean of the sampling distribution ${\mu }_{\hat{p}}$ is still equal to $p=0.38$, so the center does not change. Spread: The new standard deviation is ${\sigma }_{\hat{p}}=\sqrt{\frac{(0.38)(0.62)}{240}}\approx 0.0313$, so spread is cut in half. Shape: The sampling distribution is still approximately Normal, and it is closer to Normal than the distribution for the smaller sample size.

Question 3 (Application / Real-World Style)

A conservation biologist estimates that 25% of oak trees in a large national park are infected with a particular fungus. The biologist takes a random sample of 160 oak trees to test for the fungus. What is the probability that between 20% and 30% of the sampled trees are infected with the fungus? Interpret your result in context.

Worked Solution: Check conditions: The national park has far more than $10\ast 160=1600$ oak trees, so the 10% condition is satisfied. Large Counts: $np=160\ast 0.25=40\ge 10$, $n(1-p)=160\ast 0.75=120\ge 10$, so conditions are satisfied. The sampling distribution is approximately Normal with ${\mu }_{\hat{p}}=0.25$, ${\sigma }_{\hat{p}}=\sqrt{\frac{(0.25)(0.75)}{160}}\approx 0.0342$. Calculate z-scores: $z_{0.20}=\frac{0.20-0.25}{0.0342}\approx -1.46$, $z_{0.30}=\frac{0.30-0.25}{0.0342}\approx 1.46$. $P(-1.46<z<1.46)=0.9279-0.0721=0.8558$. Interpretation: If we repeatedly take random samples of 160 oak trees from this park, about 85.6% of the samples will have a proportion of infected trees between 20% and 30%.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for all inference on categorical proportions, which makes up a large share of AP exam points after Unit 5. Immediately after this, you will study the sampling distribution of a sample mean, which follows the same core logic but applies to quantitative data instead of categorical data. Without mastering the conditions, center, spread, and probability calculation rules for $\hat{p}$, you will struggle to distinguish between proportion and mean inference on the exam, and will lose easy points for missing required condition checks on FRQs. In the bigger picture, this topic is the backbone of confidence intervals for a population proportion and hypothesis tests for a population proportion, which are core topics tested almost every year on the AP Statistics FRQ section.

Confidence Intervals for a Population Proportion Hypothesis Tests for a Population Proportion The Sampling Distribution of a Sample Mean