The CLT for Sample Proportions

Why Sample Proportions Can Be Approximately Normal

A sample proportion is built from repeated success-or-failure outcomes. For example, a survey may classify each response as supporting or not supporting a proposal. Although each outcome has only two possible values, the proportion of successes across many independent observations can take many values. The Central Limit Theorem explains why the distribution of that proportion can be approximated by a normal distribution, provided the population proportion is not too close to zero or one and the sample is large enough.

The approximation describes the sampling distribution: the distribution of the sample proportion over repeated samples under the same model. It does not say that the individual outcomes become normal, nor does it guarantee that every observed sample proportion will look normal. The focus here is the model for a sample proportion and the conditions that make its normal approximation plausible.

From Bernoulli Outcomes to a Sample Proportion

Suppose each of $n$ trials has two outcomes, called success and failure, and the probability of success is $p$ on every trial. Represent trial $i$ by an indicator random variable $X_i$: it equals 1 for success and 0 for failure. The sample proportion of successes is the number of successes divided by $n$, which is also the average of the indicator variables.

Definition: For independent Bernoulli trials with common success probability $p$, the sample proportion is $$ \hat{p}=\frac{X_1+X_2+\cdots+X_n}{n}. $$ Each $X_i$ equals 1 with probability $p$ and 0 with probability $1-p$. The population proportion $p$ is a fixed parameter; $\hat{p}$ is a statistic that varies from sample to sample.

The Bernoulli model has $E(X_i)=p$ and $\operatorname{Var}(X_i)=p(1-p)$. Because the trials are independent, the expected number of successes is $np$, and the variance of that count is $np(1-p)$. Dividing the count by $n$ gives the sample proportion, so its mean is $p$ and its variance is $p(1-p)/n$. Its standard deviation is called its standard error.

Key result: Under the independent, common-$p$ Bernoulli model, $$ E(\hat{p})=p,\qquad \operatorname{Var}(\hat{p})=\frac{p(1-p)}{n},\qquad SE(\hat{p})=\sqrt{\frac{p(1-p)}{n}}. $$

These mean and variance statements hold exactly for any positive integer $n$, not only when a normal approximation is suitable. The CLT adds a separate claim: when the number of trials is sufficiently large relative to the shape of the Bernoulli distribution, the standardized sample proportion is approximately standard normal. In the usual notation, the approximation is

\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}\ \approx\ N(0,1), \qquad \hat{p}\ \approx\ N\!\left(p,\frac{p(1-p)}{n}\right).

In the second expression, the two parameters of $N$ are the mean and variance. Thus the approximate center is $p$, and the approximate spread is $\sqrt{p(1-p)/n}$. Increasing $n$ reduces this spread, but sample size alone does not make every normal approximation equally accurate: the value of $p$ matters too.

Why the CLT Gives This Approximation

The relevant version of the Central Limit Theorem applies to independent, identically distributed random variables with finite variance. Bernoulli indicators meet those conditions when every trial is independent and has the same success probability $p$. If $0<p<1$, each indicator has positive, finite variance $p(1-p)$. The theorem says that the sum, after centering and scaling by its standard deviation, approaches a standard normal distribution as $n$ grows.

Model and conditions: The CLT approximation for $\hat{p}$ is based on $n$ independent trials, each with the same probability $p$ of success. For a nondegenerate Bernoulli model, $0<p<1$. A normal approximation also requires $n$ to be large enough for the distribution not to remain strongly discrete or skewed. When $p=0$ or $p=1$, $\hat{p}$ is constant, so there is no nondegenerate normal approximation.

The derivation is a direct application of the CLT to the indicators. The sum has mean $np$ and standard deviation $\sqrt{np(1-p)}$. Standardizing that sum, and then using $\hat{p}=(X_1+\cdots+X_n)/n$, gives

\frac{\sum_{i=1}^{n}X_i-np}{\sqrt{np(1-p)}} = \frac{n\hat{p}-np}{\sqrt{np(1-p)}} = \frac{\hat{p}-p}{\sqrt{p(1-p)/n}}.

The left side is the standardized sum to which the CLT applies. The final equality shows that standardizing the sum is exactly the same as centering the sample proportion at $p$ and dividing by its standard error. This is why the sample proportion has the stated approximate normal distribution. The argument depends on independence and a common $p$; if those assumptions fail, this application of the CLT does not automatically follow.

Checking Whether the Approximation Is Plausible

The Bernoulli distribution is most symmetric when $p$ is near 0.5. When $p$ is close to zero, most trials are failures and successes are relatively rare; when $p$ is close to one, most trials are successes and failures are relatively rare. In either case, the sample proportion can be concentrated near a boundary and skewed. A normal curve is symmetric and extends in both directions, so it may be a poor description of such a distribution.

A useful introductory diagnostic is to calculate the expected numbers of successes and failures:

np \quad\text{and}\quad n(1-p).

Both should be reasonably large for a normal approximation to be credible. A common rule of thumb uses at least 10 expected successes and 10 expected failures, but this is a heuristic, not a theorem that guarantees a particular approximation error. The distance of $p$ from 0 and 1, sample size, and the part of the distribution being approximated can all affect accuracy. In practice, expected counts help identify obvious problems; they do not prove that an approximation is exact.

Worked Example: A Proportion Near the Center

Suppose a hypothetical process produces independent items, each conforming to a specification with probability $p=0.40$. A sample contains $n=200$ items. Describe the approximate sampling distribution of the sample proportion of conforming items, and check whether the normal approximation is plausible.

Model check. The model states that trials are independent and have the same probability $0.40$ of conformity. These are assumptions of the hypothetical setup. The outcomes are Bernoulli indicators, and $0<0.40<1$, so each has finite positive variance.

Expected counts. The expected number of conforming items is $np=200(0.40)=80$. The expected number not conforming is $n(1-p)=200(0.60)=120$. Both are well above the common threshold of 10, so the rule of thumb supports using a normal approximation.

Center and standard error. The exact mean of $\hat{p}$ is $p=0.40$. Its standard error is $$ SE(\hat{p})=\sqrt{\frac{p(1-p)}{n}} =\sqrt{\frac{(0.40)(0.60)}{200}} =\sqrt{0.0012} \approx 0.03464. $$

Therefore, the sampling distribution is reasonably approximated by a normal distribution with mean $0.40$ and variance $0.0012$, or standard deviation about $0.03464$. The standard deviation is in proportion units. This describes the approximate distribution across repeated samples, not the value that must occur in any one sample.

Worked Example: A Population Proportion Near Zero

Imagine a hypothetical screening process in which an outcome occurs independently on each trial with probability $p=0.03$. For $n=100$ trials, assess whether the CLT normal approximation for the sample proportion is well supported by the expected-count diagnostic.

Check the model assumptions. Treating trials as independent with a common success probability of $0.03$ is part of the hypothetical model. Since $0<0.03<1$, the Bernoulli variance is positive and finite.

Calculate expected counts. The expected number of successes is $np=100(0.03)=3$. The expected number of failures is $n(1-p)=100(0.97)=97$. The expected successes are below 10, so the usual diagnostic warns against relying on a normal approximation.

The exact mean and standard error are still defined: $$ E(\hat{p})=0.03,\qquad SE(\hat{p})=\sqrt{\frac{(0.03)(0.97)}{100}} =\sqrt{0.000291} \approx 0.01706. $$ The issue is not that these formulas fail. Rather, with only three expected successes, the sampling distribution may be strongly affected by discreteness and right skew, while a normal curve is symmetric. A large expected number of failures does not compensate for the small expected number of successes. The diagnostic therefore gives reason to question the approximation; it does not claim that every normal-based calculation would have the same degree of error.

Worked Example: A Proportion Near One

In another hypothetical setting, suppose independent trials have a common probability $p=0.85$ of success, and $n=300$ trials are observed. Assess the normal approximation and state its approximate center and spread.

Check the model. Independence and a common success probability of $0.85$ are assumptions in this setting. Each outcome is Bernoulli, and $p$ is strictly between 0 and 1.

Check both expected counts. The expected number of successes is $np=300(0.85)=255$. The expected number of failures is $n(1-p)=300(0.15)=45$. Both exceed 10, so despite $p$ being closer to one than to one-half, the rule of thumb supports a normal approximation.

The exact mean is $0.85$, and the standard error is $$ SE(\hat{p})=\sqrt{\frac{(0.85)(0.15)}{300}} =\sqrt{0.000425} \approx 0.02062. $$ Thus the approximate distribution is normal with mean $0.85$ and variance $0.000425$, and standard deviation about $0.02062$. The calculation illustrates why checking both success and failure counts matters: a high $p$ alone does not rule out a normal approximation when the sample is large enough to produce a substantial expected number of failures as well.

What the Approximation Does—and Does Not—Say

For a fixed population proportion $p$, the sample proportion is centered at $p$, and its standard error decreases as $n$ increases. Specifically, its standard error is proportional to $1/\sqrt{n}$. Multiplying the sample size by four therefore halves the standard error, assuming the same $p$. This shrinking spread is a statement about sampling variability, not about a change in the underlying population proportion.

The normal approximation is a statement about the shape of the sampling distribution, not a claim that the observed data are normally distributed. Individual observations remain binary. Nor does the CLT guarantee that a particular sample proportion equals $p$; sampling outcomes still vary. A normal model may be useful for describing that variation when the model conditions and approximation checks are adequate.

The formulas here use the population parameter $p$ to describe the sampling distribution. In a real problem, $p$ may be unknown, but estimating it or using the approximation to calculate probabilities are separate tasks. The key point at this stage is to distinguish the exact mean and standard error formulas from the additional, approximate normal-shape claim.

Common Mistake

A frequent error is to check only that the sample size is “large” and then declare the sample proportion normal. For example, $n=100$ sounds substantial, but if $p=0.01$, the expected number of successes is only $100(0.01)=1$. The distribution can then be highly concentrated near zero and skewed, unlike a normal curve.

The correction is to state the Bernoulli assumptions and assess both $np$ and $n(1-p)$. A large $n$ helps, but it must be large enough in relation to how rare either outcome is. The expected-count rule is a practical screen, not an exact boundary between valid and invalid approximations.

Key takeaway: For independent Bernoulli trials with common success probability $p$, $\hat{p}$ has exact mean $p$ and standard error $\sqrt{p(1-p)/n}$. The CLT gives an approximately normal sampling distribution when the sample is sufficiently large and neither successes nor failures are too rare.

Check Your Understanding

Try these questions using the model, formulas, and diagnostics from this tutorial.

For independent Bernoulli trials with common success probability $p$, explain why the sample proportion is an average of indicator variables.
A hypothetical process has $p=0.25$ and $n=80$. Calculate the expected success and failure counts, then say whether the common rule of thumb supports a normal approximation.
State the exact standard error of a sample proportion when $p=0.60$ and $n=150$, leaving your answer in simplified square-root form or calculating it.
Why is a large sample size by itself not enough to guarantee a good normal approximation when $p$ is close to zero?
What assumptions about the trials are needed to apply the CLT argument given here?

What You'll Learn