The Normal Approximation to a Binomial Distribution

Why a Binomial Count Can Look Normal

A binomial random variable counts successes across repeated independent trials. When there are many trials and neither outcome is too rare, the count is the sum of many small contributions. The central limit theorem explains why its standardized distribution can be close to a standard normal distribution. In practice, this provides a convenient way to approximate binomial probabilities without summing many probability masses.

The approximation is not a new exact distributional identity. It replaces a discrete distribution by a continuous one, so its reliability depends on the binomial parameters and on the probability being calculated. The starting point is to match the binomial mean and variance exactly.

Definition: If $X\sim\operatorname{Binomial}(n,p)$, then $X$ is the number of successes in $n$ independent Bernoulli trials, each with success probability $p$. Write $q=1-p$. For $0<p<1$, $$ P(X=k)=\binom{n}{k}p^kq^{n-k},\qquad k=0,1,\ldots,n, $$ and $$ E[X]=np,\qquad \operatorname{Var}(X)=npq. $$ The moment-matched normal approximation uses $Y\sim N(np,npq)$, or equivalently the standardized count $$ Z_n=\frac{X-np}{\sqrt{npq}}, $$ which is approximated by $N(0,1)$.

The second argument of $N(np,npq)$ is the variance, not the standard deviation. Thus the standard deviation used to standardize a count is $\sqrt{npq}$. The approximation is most useful when the normal curve places most of its mass over the range of possible counts and the binomial distribution is not strongly skewed.

Standardizing Count Intervals

For a count interval, translate each count boundary using the same mean and standard deviation. The direct, unadjusted normal approximation to an integer interval is

P(a\le X\le b) \approx \Phi\!\left(\frac{b-np}{\sqrt{npq}}\right) - \Phi\!\left(\frac{a-np}{\sqrt{npq}}\right),

where $a$ and $b$ are integers and $\Phi$ is the standard normal cumulative distribution function. For example, a direct approximation to $P(X\ge k)$ is

P(X\ge k)\approx 1-\Phi\!\left(\frac{k-np}{\sqrt{npq}}\right).

These formulas approximate the count event by the corresponding event for a continuous normal variable. Since a normal variable assigns probability zero to any single point, the endpoint convention does not change the normal probability. At finite $n$, however, the binomial assigns positive probability to individual integers, so a normal approximation is not exact. Use the same boundary convention throughout a calculation; do not silently shift one endpoint by a count.

Worked Example: A Central Interval for a Fair-Coin Count

Suppose a fair coin is tossed $100$ times and $X$ is the number of heads. Approximate $P(45\le X\le55)$ using the direct standardized boundaries.

Here $n=100$, $p=q=0.5$, so $$ np=100(0.5)=50,\qquad npq=100(0.5)(0.5)=25,\qquad \sqrt{npq}=5. $$ The standardized endpoints are $$ \frac{45-50}{5}=-1,\qquad \frac{55-50}{5}=1. $$ Therefore $$ P(45\le X\le55)\approx\Phi(1)-\Phi(-1) =0.8413-0.1587=0.6826\approx0.6827. $$ The last rounding uses the usual tabulated values $\Phi(1)\approx0.8413$ and $\Phi(-1)\approx0.1587$; the probability is about $68.3\%$.

Worked Example: An Upper-Tail Approximation

Let $X\sim\operatorname{Binomial}(100,0.633067)$. Approximate $P(X\ge70)$ using the direct standardized threshold. This parameter is chosen to make the numerical standardization transparent; it is a valid success probability between zero and one.

The mean and variance are $$ \mu=np=100(0.633067)=63.3067, $$ $$ \sigma^2=np(1-p)=100(0.633067)(0.366933)\approx23.2293, \qquad \sigma\approx4.8197. $$ Thus the standardized threshold is $$ z=\frac{70-63.3067}{4.8197}\approx1.3887. $$ Consequently, $$ P(X\ge70)\approx1-\Phi(1.3887)\approx1-0.9175=0.0825. $$ This is an approximate tail probability of $8.25\%$. The normal CDF value rounds to $0.9175$, so the corresponding tail rounds to $0.0825$.

Why the Approximation Works

Write $X=B_1+\cdots+B_n$, where the $B_i$ are independent Bernoulli random variables with success probability $p$. Then $E[B_i]=p$, $\operatorname{Var}(B_i)=pq$, and $$ \frac{X-np}{\sqrt{npq}} = \frac{\sum_{i=1}^n(B_i-p)}{\sqrt{npq}}. $$ For fixed $p$ strictly between zero and one, this is the standardized sum in the central limit theorem for independent, identically distributed variables with finite positive variance. The theorem gives convergence in distribution to $N(0,1)$.

Theorem: Let $X_n\sim\operatorname{Binomial}(n,p)$ for a fixed $p\in(0,1)$, and set $q=1-p$. Then, as $n\to\infty$, $$ \frac{X_n-np}{\sqrt{npq}}\xrightarrow{d}N(0,1). $$ More generally, if $p=p_n\in(0,1)$ may vary with $n$, the same convergence holds whenever $np_n(1-p_n)\to\infty$.

For the varying-probability statement, represent $X_n$ as a sum of $n$ independent Bernoulli variables with common success probability $p_n$. Put $q_n=1-p_n$ and $s_n^2=np_nq_n$. The third absolute centered moment of one summand is $$ E|B_i-p_n|^3=p_nq_n^3+q_np_n^3 =p_nq_n(q_n^2+p_n^2)\le p_nq_n, $$ because $p_n^2+q_n^2\le(p_n+q_n)^2=1$. Hence the Lyapunov ratio of order three is bounded by $$ \frac{\sum_{i=1}^n E|B_i-p_n|^3}{s_n^3} \le \frac{np_nq_n}{(np_nq_n)^{3/2}} =\frac{1}{\sqrt{np_nq_n}} \longrightarrow0. $$ The Lyapunov central limit theorem therefore gives the stated convergence. This proof also shows why a growing variance, rather than merely a large $n$, is the relevant condition when $p_n$ changes. $\square$

Convergence in distribution justifies interval approximations when the standardized boundaries settle at finite values. Specifically, suppose $a_n$ and $b_n$ are integer bounds and $$ \frac{a_n-np_n}{s_n}\to\alpha,\qquad \frac{b_n-np_n}{s_n}\to\beta,\qquad \alpha<\beta, $$ where $s_n=\sqrt{np_n(1-p_n)}$ and $s_n\to\infty$. The limiting standard normal distribution is continuous, so the probabilities of the standardized count lying below either moving boundary converge to the corresponding values of $\Phi$. Subtracting the lower-tail probability from the upper one gives

P(a_n\le X_n\le b_n)\longrightarrow\Phi(\beta)-\Phi(\alpha).

The one-count distinction between an inclusive lower endpoint and a strict one vanishes after standardization because $1/s_n\to0$. This is an asymptotic statement: it explains the approximation for growing variance, but it does not guarantee small error for every finite $n$.

Worked Example: A Rare-Success Case to Treat Cautiously

Let $X\sim\operatorname{Binomial}(20,0.05)$. A normal approximation can be calculated, but its limitations are apparent when approximating $P(X\le0)$.

The binomial mean, variance, and standard deviation are $$ \mu=20(0.05)=1,\qquad \sigma^2=20(0.05)(0.95)=0.95,\qquad \sigma=\sqrt{0.95}\approx0.9747. $$ Using the direct standardized boundary gives the normal approximation $$ P(X\le0)\approx\Phi\!\left(\frac{0-1}{0.9747}\right) =\Phi(-1.026)\approx0.152. $$ But the exact probability is $$ P(X=0)=(0.95)^{20}\approx0.3585. $$ The discrepancy is large. The binomial count is concentrated on small nonnegative integers and is strongly right-skewed; a symmetric normal curve does not represent that shape well. Here $np=1$, signaling that the expected number of successes is too small for this approximation to be trusted.

Assessing Whether the Approximation Is Plausible

The theorem gives a precise asymptotic condition, but applied work requires a finite-sample check. A common diagnostic is to inspect both expected counts:

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

Both should be reasonably large for the distribution to be neither severely skewed nor dominated by a boundary. Rules of thumb often ask that each expected count be at least $5$ or $10$, but these are diagnostics, not universal error guarantees. A stricter assessment may compare the desired approximation with an exact binomial calculation when that is computationally feasible.

The variance $npq$ is central to the asymptotic condition, but it does not tell the whole finite-sample story. For instance, a moderate variance can coexist with a skewed distribution when one outcome is relatively rare. Also consider where the event lies: a central interval may be approximated well even when a far-tail probability is less accurate. Tail errors matter especially when the approximate probability is very small.

Common Mistakes

Using the variance as the standard deviation. The matched normal distribution is $N(np,npq)$, so the denominator in a standardized count is $\sqrt{npq}$, not $npq$.
Assuming a large $n$ is sufficient. If $p$ is very close to zero or one, one expected count may remain small. Check $np$ and $nq$, or in an asymptotic sequence check that $np_nq_n\to\infty$.
Treating the approximation as exact. A binomial count has only integer values between zero and $n$; a normal variable is continuous and can take values outside that range.
Mixing endpoint conventions. Standardize every boundary using the same stated approximation convention. A change by one count may have a visible effect when the standard deviation is small.
Overinterpreting a rounded CDF value. Calculate the normal CDF and the tail from the same rounding. For example, $1-0.9175=0.0825$, not $0.0824$.

Key takeaway: Approximate a binomial count by $N(np,npq)$, standardize using $\sqrt{npq}$, and evaluate the resulting standard normal probability. Trust the result most when both expected outcome counts are substantial and the event is not an extreme tail; otherwise, consider an exact binomial calculation.

Check Your Understanding

Use the definitions and results above to answer the following.

For $X\sim\operatorname{Binomial}(80,0.25)$, find the matched normal mean, variance, and standard deviation.
Write the direct normal approximation for $P(30\le X\le45)$ when $X\sim\operatorname{Binomial}(100,0.4)$.
What condition on $p_n$ and $n$ ensures the stated normal limit when the success probability varies with $n$?
Why can a binomial distribution with a large number of trials still be poorly approximated by a normal distribution?
For $X\sim\operatorname{Binomial}(30,0.02)$, calculate $np$ and $n(1-p)$. What do these expected counts suggest about using a normal approximation?