Why the Shape Changes Near a Boundary
A sample proportion is always between 0 and 1, but its sampling distribution need not look like a smooth, symmetric mound. When the population proportion is near zero, most samples may contain few successes, with a smaller number containing more. When the population proportion is near one, the pattern reverses: most samples contain few failures, with a tail toward samples containing more failures. Near one-half, the distribution is often more symmetric. These differences matter because the shape tells us what sample-to-sample outcomes are typical and which are unusual.
The key is to distinguish the population proportion, \(p\), from the random sample proportion, \(\hat{p}\). The value of \(p\) describes the population or sampling model. The value of \(\hat{p}\) varies from sample to sample. The question here is how that variation is distributed when \(p\) is near the endpoints 0 or 1, or near the midpoint \(1/2\).
The distribution is discrete: it places probability at a finite set of possible proportions, not at every point on a continuous line. Its shape depends on \(n\) and \(p\). In this tutorial, “skewed” describes an uneven distribution with a longer tail toward one side; “symmetric” means corresponding values on either side of the center have equal probabilities. Neither description changes the possible range \(0\leq\hat{p}\leq1\).
Exact Shape Near Zero
When \(p\) is small, success is uncommon on each trial. Samples with zero or a few successes can be much more probable than samples with many successes. Since \(\hat{p}=X/n\), probability concentrated among small counts produces values near zero and leaves a tail extending toward larger proportions. This is right-skewness.
Worked Example: A Population Proportion Near Zero
Consider a hypothetical model with \(n=10\) independent trials and success probability \(p=0.04\). The sample proportion can be \(0,0.1,0.2,\ldots,1\), but the lower values are most likely. Calculate the probabilities of zero, one, and two successes, and the probability of three or more.
For zero successes, $$ P(X=0)=\binom{10}{0}(0.04)^0(0.96)^{10}=0.96^{10}\approx0.6648. $$ Thus about 66.48% of samples have no successes, corresponding to \(\hat{p}=0\).
For one success, $$ P(X=1)=\binom{10}{1}(0.04)(0.96)^9 =10(0.04)(0.96)^9\approx0.2770. $$ For two successes, $$ P(X=2)=\binom{10}{2}(0.04)^2(0.96)^8 =45(0.04)^2(0.96)^8\approx0.05194. $$ The probability of three or more successes is the complement of the first three exact probabilities: $$ P(X\geq3)=1-\left[0.96^{10}+10(0.04)(0.96)^9+45(0.04)^2(0.96)^8\right]\approx0.006214. $$ The exact terms inside the brackets are used before rounding. The results show a large concentration at zero and progressively smaller probabilities for these low counts, with only a small probability remaining for three or more successes. The distribution of \(\hat{p}\) therefore has a pronounced tail to the right.
A high probability at \(\hat{p}=0\) is not an error or a failure of the sampling model. It is a natural consequence of a low success probability and a finite sample size. Even though the population proportion is positive, some samples may contain no successes at all.
Near One: Count Failures to See the Shape
When \(p\) is near one, successes are common and failures are uncommon. Looking directly at the success count can make the pattern less obvious. Instead, let \(Y=n-X\) be the number of failures. If each trial has success probability \(p\), then each has failure probability \(1-p\), so \(Y\sim\mathrm{Binomial}(n,1-p)\). A small number of failures corresponds to a sample proportion of successes near one; more failures correspond to smaller sample proportions. Thus the distribution of \(\hat{p}\) typically has a tail extending left from values near one.
Worked Example: A Population Proportion Near One
Suppose a hypothetical model has \(n=8\) independent trials and \(p=0.90\). Calculate probabilities for six, seven, and eight successes, then determine the probability of five or fewer successes.
For eight successes, $$ P(X=8)=\binom{8}{8}(0.90)^8(0.10)^0=0.90^8\approx0.4305. $$ For seven successes, $$ P(X=7)=\binom{8}{7}(0.90)^7(0.10) =8(0.90)^7(0.10)\approx0.3826. $$ For six successes, $$ P(X=6)=\binom{8}{6}(0.90)^6(0.10)^2 =28(0.90)^6(0.10)^2\approx0.1488. $$ Adding these exact probabilities gives approximately \(0.9619\). Therefore, $$ P(X\leq5)=1-\left[0.90^8+8(0.90)^7(0.10)+28(0.90)^6(0.10)^2\right]\approx0.03809. $$ The most likely values among those calculated are six, seven, and eight successes, which correspond to \(\hat{p}=0.75,0.875,\) and \(1\). The small remaining probability is spread over lower success counts, creating a tail toward the left.
There is a useful mirror relationship here. If the success probability is \(p\), the failure probability is \(1-p\). Counting failures turns a model near \(p=1\) into one near zero. The shape is therefore easier to understand by asking how many failures a sample contains, while remembering that more failures mean a smaller sample proportion of successes.
Near One-Half: Symmetry
At \(p=1/2\), success and failure are equally likely. The exact binomial probabilities for \(k\) successes and \(n-k\) successes are equal. In fact, for any \(n\), $$ P(X=k)=\binom{n}{k}\left(\frac12\right)^k\left(\frac12\right)^{n-k} =\binom{n}{k}\left(\frac12\right)^n =P(X=n-k). $$ Since \(k/n\) and \((n-k)/n\) are equally far from \(1/2\), the distribution of \(\hat{p}\) is symmetric around \(1/2\). It remains discrete, and symmetry does not mean that all possible values are equally likely.
Worked Example: A Symmetric Proportion Distribution
For a hypothetical model with \(n=6\) and \(p=0.50\), the probabilities for \(X=0,1,\ldots,6\) are the binomial coefficients divided by \(2^6=64\): $$ \begin{array}{c|rrrrrrr} x&0&1&2&3&4&5&6\\ \hline P(X=x)&1/64&6/64&15/64&20/64&15/64&6/64&1/64 \end{array} $$ The corresponding sample proportions are \(0,1/6,2/6,3/6,4/6,5/6,1\). For example, \(P(\hat{p}=1/6)=6/64\) and \(P(\hat{p}=5/6)=6/64\). The probabilities mirror across \(1/2\); the greatest probability is at \(\hat{p}=3/6=0.5\). The distribution is symmetric, but it consists of seven possible values rather than a smooth curve.
A Probability-Ratio Justification
The change in probabilities from one success count to the next provides a compact way to see why the shape depends on \(p\). For \(0\leq k<n\), divide the probability of \(k+1\) successes by the probability of \(k\) successes: $$ \frac{P(X=k+1)}{P(X=k)} =\frac{\binom{n}{k+1}p^{k+1}(1-p)^{n-k-1}} {\binom{n}{k}p^k(1-p)^{n-k}} =\frac{n-k}{k+1}\frac{p}{1-p}. $$ The binomial-coefficient ratio is \(\binom{n}{k+1}/\binom{n}{k}=(n-k)/(k+1)\); canceling the remaining powers gives the final expression. If this ratio exceeds one, the probability increases from \(k\) to \(k+1\); if it is below one, the probability decreases.
For \(p=1/2\), the formula and the exact symmetry show that probabilities rise toward the middle and fall away from it. When \(p\) is small, the factor \(p/(1-p)\) is small, so probabilities tend to decrease quickly as the count moves upward from the low end. When \(p\) is near one, the same factor is large, shifting probability toward high success counts. The other factor, \((n-k)/(k+1)\), also changes with \(k\), so the full distribution depends on both \(n\) and \(p\), not on \(p\) alone.
At the endpoints, the model is degenerate rather than skewed: if \(p=0\), every trial is a failure and \(\hat{p}=0\) with probability one; if \(p=1\), every trial is a success and \(\hat{p}=1\) with probability one. Near, but not exactly at, an endpoint, there is still variation, although a substantial part of the probability can lie at or close to that endpoint.
What Repeated-Sample Simulations Show
A simulation approximates a sampling distribution by repeatedly generating samples under a specified model and recording the resulting \(\hat{p}\) values. To match the binomial model, each simulated sample must have \(n\) independent trials with success probability \(p\), and the same \(n\) and \(p\) must be used for every repetition. A simulation’s relative frequencies will not usually equal the exact binomial probabilities: they vary randomly, especially for outcomes that are uncommon.
Worked Example: Reading a Simulated Distribution Near Zero
Imagine a hypothetical computer simulation of 1,000 repeated samples, each with \(n=20\) independent trials and success probability \(p=0.05\). Suppose the recorded counts of successes are grouped as follows:
| Success count \(X\) | Number of simulated samples | Relative frequency |
|---|---|---|
| 0 | 351 | 0.351 |
| 1 | 368 | 0.368 |
| 2 | 193 | 0.193 |
| 3 | 67 | 0.067 |
| 4 or more | 21 | 0.021 |
The counts total \(351+368+193+67+21=1000\), and the relative frequencies total \(1\). Since \(\hat{p}=X/20\), the first four groups correspond to sample proportions \(0,0.05,0.10,\) and \(0.15\); the final group contains proportions of at least \(0.20\). The simulated distribution places most observations at the low end, with fewer observations farther to the right. This is the same broad right-skewed pattern predicted by the exact binomial model.
The displayed frequencies are simulated results, not exact probabilities and not observations from a real study. A different run would produce different counts. The simulation helps make the sampling pattern visible; the binomial formula gives the exact probabilities under the stated model.
Common Pitfall and Why Shape Matters
Wrong reasoning: “The sample size is fairly large, so the distribution of \(\hat{p}\) must be symmetric and bell-shaped.” A large sample size by itself does not establish that conclusion. If \(p\) is close to zero, samples can still pile up near zero; if \(p\) is close to one, they can pile up near one. In those situations, a symmetric description may miss an important boundary concentration and tail.
Correction: Consider both the sample size and the location of \(p\). Use the exact binomial model when exact probabilities are needed, or inspect a correctly specified repeated-sample simulation to see the pattern. In particular, ask whether samples commonly have very few successes or very few failures. This does not replace checking the assumptions of a model; it explains why the shape should not be inferred from sample size alone.
Shape matters because a statement about “typical” sample-to-sample variation should reflect the distribution that can actually occur. The bounded range and discreteness of \(\hat{p}\) are especially visible near zero and one. Near one-half, symmetry is supported by the model’s equal probabilities for complementary counts. Recognizing these patterns prevents a boundary-heavy distribution from being described as if it were centered and balanced.
Check Your Understanding
Use the binomial model and the shape patterns above to answer these questions.
- If \(n=12\) and \(p=0.03\), which end of the sample-proportion scale is likely to contain most of the probability? In which direction would you expect the tail?
- For a model with \(n=10\) and \(p=0.95\), explain why counting failures can make the distribution’s shape easier to describe.
- When \(p=1/2\), what relationship should hold between \(P(X=2)\) and \(P(X=n-2)\)? Explain how that relationship corresponds to symmetry in \(\hat{p}\).
- Why can a simulation with 1,000 repetitions have relative frequencies that differ from the exact binomial probabilities?
- A colleague says a proportion distribution must be bell-shaped because the sample size is large. What additional feature of the model should be considered, and why?