Topic 5.5 Sampling Distributions for Sample Proportions: describe the mean, standard deviation, and shape of the sampling distribution of a sample proportion, and check the conditions (10% and large counts) for the normal model.

What this topic is asking

The College Board (Topic 5.5) wants you to describe the mean, standard deviation, and shape of the sampling distribution of a sample proportion $\hat{p}$, and to check the conditions (the $10\%$ condition and the large-counts condition) that allow the normal model.

Center and spread of the sample proportion

The mean result, $\mu_{\hat{p}} = p$, says $\hat{p}$ is unbiased: its sampling distribution centers on the true proportion. The standard deviation formula comes straight from the binomial: the count of successes has standard deviation $\sqrt{np(1-p)}$, and dividing the count by $n$ to get a proportion divides the standard deviation by $n$, giving $\dfrac{\sqrt{np(1-p)}}{n} = \sqrt{\dfrac{p(1-p)}{n}}$. The key feature is the $\sqrt{n}$ in the denominator: larger samples give a smaller spread, so $\hat{p}$ clusters more tightly around $p$.

The two conditions

The two conditions do different jobs. The $10\%$ condition protects the standard deviation formula (independence). The large-counts condition protects the shape (normality). A full answer checks both, because using the normal model without large counts, or the standard deviation formula on a too-large fraction of the population, is unjustified.

Why the shape becomes normal

The large-counts condition is the proportion's version of the central limit theorem. A sample proportion is really a sample mean of $0$s and $1$s (failure/success), so the same averaging logic applies: with enough trials, the distribution of $\hat{p}$ smooths into a bell shape. When $p$ is near $0$ or $1$, the underlying binomial is skewed, so you need a larger $n$ before the normal approximation is good, which is exactly what $np \ge 10$ and $n(1-p) \ge 10$ enforce (they demand enough expected successes and failures). When both counts are at least about $10$, the skew has washed out and the normal model is accurate. This is why the condition is symmetric in successes and failures: a rare event ($p$ small) needs a bigger sample to accumulate $10$ expected successes, and a near-certain event ($p$ large) needs a bigger sample to accumulate $10$ expected failures. Recognizing that the proportion is a mean in disguise links Topic 5.5 to the CLT and explains why the same "large enough sample" theme recurs.

Using the distribution

Once the conditions confirm that $\hat{p}$ is approximately normal with mean $p$ and standard deviation $\sqrt{\dfrac{p(1-p)}{n}}$, every probability question about $\hat{p}$ is a normal-model calculation: standardize with $z = \dfrac{\hat{p} - p}{\sqrt{p(1-p)/n}}$ and read the area from the standard normal. This is the direct foundation of inference for proportions in Unit 6: a confidence interval for $p$ uses this same standard deviation (estimated as a standard error), and a significance test computes exactly this z-score under an assumed $p$. So Topic 5.5 is not an isolated calculation but the template for the proportion procedures that follow. A complete exam answer states the center and spread, checks both conditions, standardizes, finds the area, and interprets the result in context (for example, "there is about a $6.7\%$ chance a sample of this size gives a proportion above $0.46$"), the same structured workflow used throughout the inference units.

Try this

Q1. For $n = 400$ and $p = 0.5$, find the mean and standard deviation of $\hat{p}$. [2 points]

• Cue. $\mu_{\hat{p}} = 0.5$; $\sigma_{\hat{p}} = \sqrt{\dfrac{0.5 \times 0.5}{400}} = \sqrt{0.000625} = 0.025$.

Q2. State the large-counts condition and what it guarantees. [1 point]

• Cue. $np \ge 10$ and $n(1-p) \ge 10$; it guarantees the sampling distribution of $\hat{p}$ is approximately normal.