Deriving the Normal Approximation to Binomial Counts

From Individual Outcomes to a Count

A count of successes in repeated trials can be viewed in two ways: as one binomial random variable, or as the sum of one indicator for each trial. The second view explains why a normal curve can approximate a binomial distribution. Each individual outcome contributes a small amount to the total, and the Central Limit Theorem describes the standardized sum when the outcomes are independent and share the same probability of success.

The approximation is useful because normal models are often easier to work with than a binomial distribution, especially when the number of trials is large. Its center and spread are not guessed from a graph: they follow directly from the mean and variance of the individual outcomes.

Definition: A Bernoulli random variable records the result of one trial with two outcomes. If success has probability \(p\), the indicator \(I_i\) for trial \(i\) equals 1 on success and 0 otherwise, so \(P(I_i=1)=p\) and \(P(I_i=0)=1-p\). For \(n\) independent trials with the same success probability \(p\), the total number of successes is \(X=\sum_{i=1}^{n} I_i\), and \(X\) has a binomial distribution with parameters \(n\) and \(p\).

A binomial model therefore makes three substantive assumptions: the number of trials \(n\) is fixed; each trial has two possible outcomes, designated success and failure; and the trials are independent with the same success probability \(p\). The success label can stand for any outcome of interest; it does not imply that the outcome is desirable.

Deriving the Mean and Standard Deviation

First find the mean and variance of one indicator. Since \(I_i\) is 1 with probability \(p\) and 0 otherwise, its expected value is \(p\). Also, \(I_i^2=I_i\), because squaring either 0 or 1 leaves it unchanged. Thus its variance is \(p(1-p)\).

E(I_i)=1\cdot p+0\cdot(1-p)=p

\operatorname{Var}(I_i)=E(I_i^2)-[E(I_i)]^2 =p-p^2=p(1-p)

Now add the \(n\) indicators to form \(X\). Expected values add whether or not the variables are independent, so the count has mean \(np\). For independent variables, variances add; this is where independence is essential to the variance calculation. The count’s variance is \(np(1-p)\), and its standard deviation is the square root of that variance.

E(X)=E\left(\sum_{i=1}^{n} I_i\right) =\sum_{i=1}^{n}E(I_i)=np

\operatorname{Var}(X)=\sum_{i=1}^{n}\operatorname{Var}(I_i) =np(1-p), \qquad SD(X)=\sqrt{np(1-p)}

Key result: If \(X\) counts successes in \(n\) independent trials with the same success probability \(p\), its exact mean is \(np\), its exact variance is \(np(1-p)\), and its exact standard deviation is \(\sqrt{np(1-p)}\). These are the center and spread used in its normal approximation.

Why the Normal Approximation Follows

The Central Limit Theorem applies to a sum of independent, identically distributed random variables with finite variance. Each Bernoulli indicator has finite variance \(p(1-p)\). When \(0<p<1\) and \(n\) increases, the theorem says that the count, after subtracting its mean and dividing by its standard deviation, approaches a standard normal distribution.

\frac{X-np}{\sqrt{np(1-p)}}\ \xrightarrow{d}\ N(0,1)

The notation \(\xrightarrow{d}\) means convergence in distribution: as \(n\) grows, probabilities for the standardized count are increasingly well described by the standard normal model. Reversing the standardization gives the familiar approximation for the unstandardized count:

X\ \approx\ N\!\left(np,\ np(1-p)\right)

Here the second parameter of \(N(\mu,\sigma^2)\) is a variance. Thus the approximate normal model has mean \(np\) and standard deviation \(\sqrt{np(1-p)}\), not \(np(1-p)\). The binomial count itself remains discrete: it can only equal an integer from 0 through \(n\). The normal model is continuous and is an approximation to the binomial distribution, not a claim that the count can take noninteger values.

Worked Example: A Count of Defective Parts

Suppose, hypothetically, that 80 parts are inspected, and each part independently has probability \(0.04\) of being defective. Let \(X\) be the number of defective parts. The stated model has a fixed number of trials, two outcomes per part, a common probability, and independent outcomes, so it is binomial with \(n=80\) and \(p=0.04\).

Represent the defect status of part \(i\) by \(I_i\), equal to 1 if defective and 0 otherwise. Then \(X=\sum_{i=1}^{80} I_i\). The exact expected count is

E(X)=np=80(0.04)=3.20\text{ parts}

The exact variance and standard deviation are

\operatorname{Var}(X)=np(1-p) =80(0.04)(0.96)=3.072\text{ parts}^2

SD(X)=\sqrt{3.072}=1.7527\text{ parts}

The corresponding normal approximation is \(X\approx N(3.20,3.072)\), with mean 3.20 parts and standard deviation 1.7527 parts. The mean need not be an integer: it is the long-run average count over repeated groups of 80 parts, not a possible outcome for one group.

Worked Example: Positive Results in a Hypothetical Assay

Imagine a hypothetical assay applied to 250 independently selected samples. Suppose each sample has the same probability, \(p=0.12\), of producing a positive result under the conditions being modeled. Let \(X\) count positive results. With a fixed sample count and independent outcomes sharing that probability, the binomial model applies.

Let \(I_i=1\) for a positive result on sample \(i\), and \(I_i=0\) otherwise. The total is \(X=\sum_{i=1}^{250} I_i\), so

E(X)=np=250(0.12)=30.0\text{ positive results}

For each sample, the indicator variance is \(p(1-p)=0.12(0.88)=0.1056\). Independence lets these variances add:

\operatorname{Var}(X)=250(0.1056)=26.40\text{ results}^2, \qquad SD(X)=\sqrt{26.40}=5.138\text{ results}

Therefore the count is approximated by \(N(30.0,26.40)\), or equivalently by a normal model centered at 30.0 positive results with standard deviation 5.138 results. This describes the distribution of counts across hypothetical repetitions of the same sampling process; it does not predict that any particular set of 250 samples will contain exactly 30 positives.

Worked Example: Survey Responses and the Count Scale

Suppose, hypothetically, that 400 independently selected people each have probability \(0.53\) of giving a particular response, with the same probability for each person. Let \(X\) be the number giving that response. The fixed number of people, binary response, common probability, and independence specify a binomial model.

Using one indicator per person, \(X=\sum_{i=1}^{400} I_i\). Its mean is

E(X)=400(0.53)=212\text{ responses}

The variance and standard deviation are

\operatorname{Var}(X)=400(0.53)(0.47)=99.64\text{ responses}^2

SD(X)=\sqrt{99.64}=9.982\text{ responses}

The count is approximated by \(N(212,99.64)\). If the result is instead expressed as a fraction of the 400 people, that fraction is \(X/400\); dividing the count’s mean and standard deviation by 400 gives a center of \(212/400=0.530\) and a standard deviation of \(9.982/400=0.02496\). This rescaling does not change how the count approximation was derived: it comes from adding the individual binary outcomes.

What the Approximation Does—and Does Not—Say

The derivation gives an asymptotic reason for the approximation, not a promise that every binomial distribution is already close to normal at a particular sample size. If \(p=0\) or \(p=1\), every outcome is fixed and the count has no variability, so a nondegenerate normal approximation does not follow. When \(p\) is strictly between 0 and 1, the standardized sum approaches normality as \(n\) grows. How accurate it is for a particular finite \(n\) depends on the model’s parameters.

The distinction between exact and approximate quantities is important. The formulas \(np\) and \(np(1-p)\) give the exact mean and variance under the binomial model. The normal shape is an approximation to the distribution around those exact quantities. It does not replace the binomial assumptions, and it does not make dependent trials independent.

Common pitfall: It is incorrect to say, “The count is normally distributed because its mean and standard deviation are known.” Those two values determine the normal model’s location and scale, but the Central Limit Theorem is the reason a normal shape can approximate the binomial count. The justification also requires independent, identically distributed Bernoulli trials and a sufficiently large number of trials for the approximation to be useful.

A second useful check is to keep variance and standard deviation distinct. The variance \(np(1-p)\) is measured in squared count units, while the standard deviation \(\sqrt{np(1-p)}\) is measured in count units. The normal notation \(N(\mu,\sigma^2)\) takes the variance as its second parameter; when describing the spread in words, report \(\sigma\), the standard deviation.

This sum-of-indicators derivation links the binomial count to the Central Limit Theorem directly. It also shows why changing the number of trials or the success probability changes the model’s center and spread in specific, predictable ways: each trial contributes mean \(p\) and variance \(p(1-p)\), and the contributions add across independent trials.

Check Your Understanding

Use the indicator-sum argument and the normal approximation developed above.

For one Bernoulli indicator with success probability \(p\), derive its expected value and variance.
A count is based on 120 independent trials, each with success probability \(0.25\). Find its exact mean, variance, and standard deviation, then state its normal approximation.
Why is independence relevant when deriving the variance of a sum of Bernoulli indicators?
Explain why a binomial count can have a noninteger mean even though its possible observed values are integers.
What does convergence in distribution say about the standardized binomial count as the number of trials increases?

Deriving the Normal Approximation to Binomial Counts

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