Shape of the Sampling Distribution When p Is Near 0 or 1

Why a Small Proportion Can Produce a Skewed Sampling Distribution

In Checking Normality of \(\hat{p}\) with \(np\) and \(n(1-p)\), you learned to use the Large Counts condition to check whether a Normal model is supported. This tutorial looks at the shape behind that check. When the population proportion is \(p=0.05\), a small sample often contains zero or only a few successes. The sample proportion \(\hat{p}\) therefore piles up near zero and can have a long tail toward larger values: it is right-skewed.

Let \(X\) be the number of successes in a random sample of size \(n\). Under a suitable binomial model, \(X\) counts the successes and \(\hat{p}=X/n\). The possible values of \(\hat{p}\) are consequently separated by steps of \(1/n\). For example, with \(n=20\), the possible proportions are \(0,\ 0.05,\ 0.10,\) and so on. When \(n\) is small and \(p\) is low, zero and a few small counts can be common, while larger counts stretch the distribution to the right.

The mean of the sampling distribution remains \(p\), and its standard deviation is \(\sqrt{p(1-p)/n}\) under the conditions discussed earlier. Those facts describe its center and spread, not its shape. A distribution can have the correct mean and standard deviation and still be too skewed for a Normal model to be a good approximation.

Small Sample: Many Proportions Near Zero

Consider \(p=0.05\) and \(n=20\). The expected number of successes is \(np=1\), while the expected number of failures is \(n(1-p)=19\). The Large Counts condition is not met because the expected success count is below 10. We can see what that shortage means for shape by considering the binomial count \(X\).

For a binomial random variable, the probability of exactly \(x\) successes is \(\binom{n}{x}p^x(1-p)^{n-x}\). Since \(\hat{p}=X/20\), probabilities for counts translate directly into probabilities for sample proportions.

What Changes When the Sample Gets Larger?

Keep \(p=0.05\), but increase the sample size. A larger \(n\) makes both expected counts larger, and it gives the sample proportion more possible values because its increments are \(1/n\). The probability no successes occur also falls sharply: \(P(X=0)=(1-p)^n=0.95^n\). As samples become more likely to include a range of success counts, the sampling distribution fills out around its center instead of piling up at zero.

The next example compares \(n=20\) with \(n=400\). The point is not that the larger distribution becomes perfectly symmetric; \(\hat{p}\) is still discrete and cannot be negative. Rather, its shape is much more nearly symmetric, with a Normal model better supported by the Large Counts condition.

The Change in Shape Is Gradual, Not a Switch

The Large Counts threshold is a practical rule, not a boundary where skewness suddenly disappears. For \(p=0.05\), the success-count requirement \(np\geq10\) requires \(n\geq200\). The failure-count requirement is much less demanding because \(1-p=0.95\). At \(n=200\), the expected counts are exactly 10 successes and 190 failures, so the Large Counts condition is met. But passing a rule of thumb does not claim perfect symmetry, nor does it guarantee that every Normal approximation will be equally accurate.

Compare \(n=199\) and \(n=200\). At \(n=199\), the expected success count is \(199(0.05)=9.95\), so the condition is not met. At \(n=200\), it is \(200(0.05)=10\), and the failure count is \(200(0.95)=190\), so both parts pass. The change in the actual distribution from 199 to 200 observations is small; only the rule’s pass-or-fail status changes at that point. Larger samples beyond this threshold generally make the approximation more convincing.

Near Zero and Near One: Opposite Directions of Skew

The same reasoning applies when \(p\) is near 1. In that case, successes are common and failures are rare. The sample proportion of successes may pile up near 1, with a thinner tail extending toward smaller proportions; this is left skew. For example, if \(p=0.95\) and \(n=20\), the expected number of failures is \(20(0.05)=1\). Looking at the failure count makes the shape easier to understand: most samples have few failures, while some have more.

When \(p=0.05\), it is the successes that are rare, producing a tail toward larger sample proportions. When \(p=0.95\), it is the failures that are rare, producing a tail toward smaller sample proportions. In either case, increasing \(n\) increases the expected count of the rarer outcome and supports a more nearly symmetric sampling distribution.

Common Mistakes and AP Exam Communication

• Assuming the mean tells the whole story. The mean of \(\hat{p}\) is \(p\), but that does not imply a symmetric shape. Describe shape separately from center and spread.
• Calling a small-\(n\), small-\(p\) distribution symmetric because it is centered at \(p\). With \(p=0.05\) and \(n=20\), the possible proportions are bounded below by zero and many samples have zero successes. Explain the resulting right tail.
• Mixing up the sample data and the sampling distribution. The shape being discussed is the distribution of \(\hat{p}\) across repeated samples of the same size, not the distribution of individual successes and failures in one sample.
• Claiming a larger sample makes the distribution exactly Normal. Say that it becomes more nearly symmetric and that the Large Counts condition supports a Normal approximation. The sample proportion remains discrete and bounded.
• Treating the threshold as an abrupt shape change. The Large Counts condition is a decision rule. The distribution changes gradually as \(n\) grows, even though the rule changes from failing to passing at a particular sample size.
• Checking only expected successes. State both \(np\) and \(n(1-p)\). The rarer outcome usually controls how large \(n\) must be.

Key Takeaway

When the population proportion is near zero and the sample is small, the sampling distribution of \(\hat{p}\) can cluster near zero and stretch to the right. Increasing the sample size raises the expected number of rare outcomes and generally makes the distribution more nearly symmetric. The Large Counts condition helps assess whether a Normal model is reasonable, but it does not guarantee exact symmetry.