Normal Approximation Conditions For A Proportion

Why Check the Normal Approximation?

A sample proportion summarizes the fraction of observations with a specified outcome: for example, the proportion of hypothetical policy applications that are complete or the proportion of devices that pass a quality check. When observations follow a binomial sampling model, the count of successes has an exact distribution. A normal distribution can sometimes provide a convenient approximation, but it is not equally reliable for every sample size and success probability. The key diagnostic is whether the sample is expected to contain enough successes and enough failures.

This check is a condition assessment, not a guarantee that the approximation is accurate in every detail. It is especially important when a proportion is near zero or one, because then one of the two outcome categories may be rare even in a fairly large sample. The discussion here concerns the sampling distribution and the adequacy check; it does not derive or calculate a particular confidence interval.

The Binomial Model and the Sample Proportion

Definition: Suppose $n$ observations are independent, and each observation has the same probability $p$ of being a success. Let $X$ be the number of successes. Then $X\sim\operatorname{Bin}(n,p)$, where $0\leq p\leq 1$, and the sample proportion is $\hat p=X/n$. Under this model, $$ E[X]=np,\qquad \operatorname{Var}(X)=np(1-p), $$ $$ E[\hat p]=p,\qquad \operatorname{Var}(\hat p)=\frac{p(1-p)}{n}. $$

The assumptions matter. The outcomes must be coded as success or failure, the observations must be independent, and the success probability must be the same for every observation. If observations are dependent, or probabilities vary substantially across observations, the simple binomial model may not apply; passing a count rule does not repair a misspecified model.

Write $q=1-p$ when convenient. The expected number of successes is $np$, and the expected number of failures is $nq=n(1-p)$. If both expected counts are sufficiently large, the distribution of $X$, and therefore the distribution of $\hat p=X/n$, is often approximated by a normal distribution. The variance formula above gives the scale of that approximation:

\hat p \ \text{is approximately } N\left(p,\frac{p(1-p)}{n}\right)

Here the second parameter is the variance, consistent with the convention $N(\mu,\sigma^2)$. The approximation is for the sampling distribution over repeated samples under the model, not a claim that the observed proportion itself is normally distributed across its possible values.

Expected-Count Guidelines

A widely used basic guideline is to require at least five expected successes and five expected failures:

np\geq 5 \quad\text{and}\quad n(1-p)\geq 5.

Some courses and applications use the more conservative guideline of at least ten in each category. These are practical rules of thumb, not sharp mathematical boundaries: an approximation does not suddenly become accurate when a count changes from 4.9 to 5.0. When the expected counts are comfortably larger, the normal approximation is generally more credible; when one count is small, it may be poor. State which guideline is being applied rather than treating the threshold as universal.

Key check: Assess both $np$ and $n(1-p)$. A large sample size alone is not enough: if $p$ is very close to zero or one, the expected count in the rare category can remain small.

Why the Two Counts Control the Check

To see the reason, write each observation as an indicator $I_i$, equal to 1 for a success and 0 for a failure. Under the binomial assumptions, these indicators are independent and each has mean $p$ and variance $p(1-p)$. The total count is their sum, $X=I_1+\cdots+I_n$. Adding the means and variances of independent indicators gives $E[X]=np$ and $\operatorname{Var}(X)=np(1-p)$.

The central limit theorem says, in broad terms, that a sum of many independent contributions tends toward a normal shape after centering and scaling. For a binomial count, however, “many observations” by itself can be misleading: when $p$ is tiny, nearly all observations are failures, and the count of successes can still be concentrated near zero and strongly asymmetric. Requiring both $np$ and $n(1-p)$ to grow helps rule out that extreme imbalance. This explains the diagnostic, though it does not turn a rule of thumb into an exact error bound.

The standard deviation of the sample proportion is $ \sqrt{p(1-p)/n} $. It describes the typical sampling variation under the model, but calculating it does not establish that a normal approximation is appropriate. The expected-count check is a separate step.

Worked Applications

Worked Example: A Rare Quality-Control Outcome

A hypothetical inspection program independently checks $n=80$ components. Suppose, for planning purposes, that the probability a component fails a specified test is $p=0.08$. Determine whether the normal approximation for the failure proportion meets the five-count guideline and the more conservative ten-count guideline.

Method. Treat failure as success for this binomial count. The model is $X\sim\operatorname{Bin}(80,0.08)$, so the expected failure count is $np$, and the expected nonfailure count is $n(1-p)$.

Expected failures: $$ np=80(0.08)=6.4. $$ Expected nonfailures: $$ n(1-p)=80(0.92)=73.6. $$

Both counts are at least 5, so the five-count guideline is met. The expected failure count is less than 10, so the ten-count guideline is not met. Thus the approximation passes the less conservative check but should be treated cautiously under the stricter check. The common mistake to avoid is to see $n=80$ and conclude that a large sample automatically settles the question; the rare-outcome count is only 6.4.

Check: The expected counts sum to $6.4+73.6=80$, the sample size.

Worked Example: A Balanced Screening Outcome

In a hypothetical screening study, $n=240$ independent participants each have the same probability $p=0.35$ of meeting a specified screening criterion. Check the expected counts and calculate the variance and standard deviation of the sample proportion.

Method. Use $X\sim\operatorname{Bin}(240,0.35)$ and $\hat p=X/240$. First check both outcome counts; then substitute into the sampling variance formula.

Expected successes: $$ np=240(0.35)=84. $$ Expected failures: $$ n(1-p)=240(0.65)=156. $$

Both counts exceed 10, so the expected-count condition is comfortably met under either guideline. The sample-proportion variance is $$ \operatorname{Var}(\hat p)=\frac{(0.35)(0.65)}{240} =\frac{0.2275}{240} =0.0009479167\ldots. $$ Its standard deviation is $$ \sqrt{0.0009479167\ldots}=0.030788\ldots. $$ To four decimal places, the standard deviation is $0.0308$. Therefore, under the binomial assumptions, a normal approximation for $\hat p$ is reasonable by the stated count checks. The common mistake avoided here is checking only the success count; both counts are required, even though this example passes either check.

Check: The expected counts sum to $84+156=240$, and the variance is positive and smaller than 1, as expected for a sample proportion.

Worked Example: A Small Observed Count

A hypothetical pilot sample contains $n=40$ independent applications, of which $x=3$ are missing a required document. Use the observed proportion as a practical diagnostic for a normal approximation to the sample proportion, and state the limitation of this check.

Method. The underlying probability $p$ is not given, so estimate it from the pilot data by $\hat p=x/n$. The estimated success and failure counts are $n\hat p=x$ and $n(1-\hat p)=n-x$.

The observed proportion is $$ \hat p=\frac{3}{40}=0.075. $$ The estimated success count is $$ n\hat p=40(0.075)=3, $$ and the estimated failure count is $$ n(1-\hat p)=40(0.925)=37. $$ The success count is below 5, so the data-based check fails both the five-count and ten-count guidelines. A normal approximation is therefore questionable.

This is a diagnostic based on the observed proportion, not a proof about the unknown population probability. In a different sample, the observed count could differ; when the true $p$ is known for a theoretical assessment, use $np$ and $n(1-p)$ based on that $p$. The common mistake is to use $n=40$ as if it were the expected number in each outcome category. The correct counts are 3 and 37, and they add to 40.

Practical Interpretation and Common Traps

When $p$ is known as part of a model, assess suitability with the model-based expected counts. In an applied data analysis, $p$ is usually unknown. A common practical check substitutes the observed proportion $\hat p$, giving $n\hat p=x$ and $n(1-\hat p)=n-x$. This substitution is a diagnostic convention, not knowledge of the true expected counts. In particular, a small observed count warns that a normal approximation may be unreliable; it does not establish that the population probability is exactly zero or equal to the observed proportion.

For example, suppose $n=80$ and a proposed model gives $p=0.08$. The correct expected counts are $80(0.08)=6.4$ and $80(0.92)=73.6$. An incorrect calculation might compare $n=80$ with the threshold and declare the approximation suitable without considering $p$. That reasoning ignores the rarity of one outcome. The correction is to calculate both expected counts and apply the chosen guideline to each.

Do not check only one category. A large expected success count cannot compensate for a small expected failure count, or vice versa.
Do not confuse observed and expected counts. With a known model probability, use $np$ and $n(1-p)$. With only sample data, the plug-in diagnostic uses $x$ and $n-x$.
Do not treat a threshold as an absolute boundary. Five and ten are common conventions; conclusions near a cutoff should be described as borderline rather than certain.
Do not overlook model assumptions. Dependence or varying success probabilities can invalidate the binomial model even when both counts are large.

Key takeaway: A normal approximation for a binomial sample proportion is most plausible when the sampling model is appropriate and both expected outcome counts are sufficiently large. Always report both counts and identify the guideline used.

Check Your Understanding

For each problem, assess the expected-count condition using the stated guideline. Where requested, distinguish a model-based check from a sample-based diagnostic.

A hypothetical process has $n=90$ independent trials and success probability $p=0.20$. Find both expected counts and decide whether the five-count and ten-count guidelines are met. (Answer: 18 successes and 72 failures; both guidelines are met.)
A study models $n=300$ independent observations with $p=0.015$. Find both expected counts and assess the five-count guideline. (Answer: 4.5 expected successes and 295.5 expected failures; the five-count guideline is not met.)
A pilot sample has $n=64$ observations, with 58 successes. Calculate the plug-in proportion and both estimated counts; assess the two guidelines. (Answer: $\hat p=0.90625$; estimated counts are 58 and 6, so the five-count guideline is met but the ten-count guideline is not.)
A hypothetical model specifies $n=200$ and $p=0.07$. Calculate both expected counts and say whether the normal approximation is comfortably supported by the ten-count guideline. (Answer: 14 successes and 186 failures; yes, both counts are at least 10.)
For $p=0.50$, what is the smallest integer sample size that meets the ten-count guideline for both outcomes? (Answer: $n=20$, giving 10 expected successes and 10 expected failures.)

What You'll Learn