Success-Failure Counts and Proportion Approximation

Why Check Both Kinds of Counts?

A binomial count records how many successes occur in a fixed number of trials. When that count is approximated by a normal distribution, the question is not simply whether the number of trials is large. The success probability matters too: a large number of trials can still produce very few expected successes if success is rare. It is therefore useful to check the expected numbers of both successes and failures before using a normal approximation.

The same check applies when the quantity of interest is a sample proportion. A sample proportion is the success count divided by the number of trials, so its distribution inherits the shape of the count distribution. The check is a practical screening rule, not a proof that an approximation will be accurate for every probability question.

The Binomial Model and the Count Check

For a proposed normal approximation to a binomial count, a common practical rule is to check that both expected counts are at least 10:

Some sources use a different cutoff, such as 5, so the threshold is a convention rather than a universal boundary. This tutorial uses 10 as a cautious, commonly used screening rule. If either count is below 10, the normal approximation should be treated as questionable, not automatically impossible. If both are at least 10, the screen is passed, but that alone does not guarantee accuracy for every probability calculation.

The check concerns expected counts, not the observed count in a particular sample. Before data are collected, the observed number of successes is unknown; the model’s \(n\) and \(p\) determine the expected counts. The parameter \(p\) must be the success probability specified by the model or problem, not an observed sample proportion substituted without justification.

Why Both Counts Matter

When the expected success count is small, there may be too little expected mass spread across the success-count scale for a smooth, symmetric normal curve to describe the binomial distribution well. For example, if \(np\) is small, counts tend to be concentrated near zero and can have a long right tail. In contrast, if \(n(1-p)\) is small, counts tend to be near \(n\), with possible asymmetry toward smaller counts. These statements depend on the expected count, which combines \(n\) and \(p\); a small \(p\) by itself does not imply that successes are concentrated near zero.

The two expected counts can also be seen in the variance of the count. Under the binomial model, the mean is \(np\) and the variance is \(np(1-p)\). A normal distribution is symmetric, whereas a binomial distribution may be asymmetric, especially when one of the expected counts is small. Checking both counts screens for situations where one side of the success-failure scale has little expected representation.

A short calculation helps explain why this rule is informative without making it an exact theorem. Let each trial’s success indicator be 1 for success and 0 for failure. Its mean is \(p\), its variance is \(p(1-p)\), and its third central moment is \(p(1-p)(1-2p)\). Because the count is the sum of \(n\) independent indicators, its variance is \(np(1-p)\), and its third central moment is \(np(1-p)(1-2p)\). Standardizing by the variance to the power \(3/2\) gives the binomial skewness:

The denominator is \(\sqrt{(np)(n(1-p))}\). When either expected count is small, this denominator can be small, allowing greater skewness. The calculation helps explain the count check, but it is not a guarantee: skewness alone does not determine how accurate a normal approximation will be for a specific event.

Applying the Check to Sample Proportions

If \(X\) is the number of successes in \(n\) trials, the sample proportion is \(\hat{p}=X/n\). The count and proportion contain the same information: dividing the count by \(n\) rescales its possible values but does not change the underlying distribution’s shape. Under the binomial model, the sample proportion has mean \(p\) and variance \(p(1-p)/n\). Its normal approximation is therefore screened using the same two expected counts, \(np\) and \(n(1-p)\).

A Small Probability Does Not Settle the Question

It is important to distinguish a small probability \(p\) from a small expected success count \(np\). If \(n\) is sufficiently large, \(np\) can be large even when \(p\) is small. The same is true of the expected failure count: when \(p\) is close to 1, failures can still be plentiful if \(n\) is large enough.

Using the Rule Carefully

For a proposed normal approximation, write down the model and evaluate each expected count separately. Do not infer suitability from \(n\) alone, from \(p\) alone, or from the fact that one of the two expected counts is large. The screen requires both counts to meet the chosen threshold.

A common mistake is to reason, “The sample size is large, so the binomial count must be approximately normal.” A large \(n\) does not by itself ensure that both expected counts are large; for instance, a low probability can still give a small \(np\). The correction is to calculate \(np\) and \(n(1-p)\) explicitly.

The opposite mistake is to reason, “The success probability is small, so the count must be near zero and strongly skewed.” That conclusion ignores the sample size. The correction is again to calculate \(np\): if it is large, the count need not be concentrated near zero. Finally, passing both count checks should not be described as proof of accuracy. It is a widely used preliminary screen, and the quality of a normal approximation can depend on the particular probability being approximated.