What is the normal approximation to the binomial?

What the approximation says

A binomial random variable counts successes in $n$ independent trials with a fixed success probability $p$. Its probability histogram is a set of bars, one per whole-number count from 0 to $n$. As $n$ grows, that staircase smooths out into a bell shape, and you can compute probabilities by treating the count as normal instead of adding bars.

The approximating curve uses the binomial's own mean and standard deviation:

So for $X \sim B(100, 0.4)$, read "binomial with $n = 100$ and $p = 0.4$", the mean is $\mu = 100(0.4) = 40$ successes and the standard deviation is $\sigma = \sqrt{100(0.4)(0.6)} = \sqrt{24} \approx 4.899$ successes. To find $P(X \le 45)$ you would standardize 45 against that curve and read off a normal probability.

The reason this works is the same reason the central limit theorem works. A binomial count is a sum of $n$ independent Bernoulli trials, each worth 0 or 1, and sums of many independent pieces tend toward a normal shape.

The large-counts condition, and why 10

The approximation is safe only when both tails of the binomial have room. The standard check is

Read $np$ as the expected number of successes and $n(1-p)$ as the expected number of failures. Both must clear 10.

Both sides matter because a binomial is skewed when $p$ sits near 0 or near 1. The count cannot go below 0 or above $n$, so if the mean is close to either wall the distribution has to pile up against it and stretch out on the other side. With $n = 20$ and $p = 0.1$, the mean is only 2 successes, and the standard deviation is $\sqrt{1.8} \approx 1.342$, so the mean sits about 1.49 standard deviations above the hard floor at 0. No symmetric curve fits that.

Ten is a convention, not a theorem. It puts the mean roughly three standard deviations clear of both walls, which is enough room for a bell shape to fit without spilling past the ends. Why 10 successes and 10 failures works through that reasoning in detail. Note that the AP Statistics version of the check uses 10; some textbooks use 5, which is looser and less safe.

The same question worked both ways

Take $X \sim B(100, 0.4)$ and ask for $P(X \le 45)$. Both counts clear the condition: $np = 40$ and $n(1-p) = 60$.

Exact binomial. Adding the bars from 0 to 45, which is what binomcdf(100, 0.4, 45) does, gives

Normal approximation. Standardize with $\mu = 40$ and $\sigma \approx 4.899$:

and the standard normal probability below $z = 1.0206$ is 0.84628.

So the exact answer is 86.89% and the approximation is 84.63%. The approximation is off by 2.26 percentage points, and it is too small. That is not a rounding difference; it is the method itself. If a question asked whether the probability exceeds 0.85, the two approaches would give opposite answers.

This is why the approximation is a way of thinking about shape, not a shortcut for getting a number. Your calculator computes the exact binomial instantly, so use it. Binompdf vs binomcdf covers which command to press.

Why it missed, and the half-unit fix

The gap has a specific cause. The binomial lives on whole numbers, and $P(X \le 45)$ includes the entire bar at 45. The normal curve is continuous, and cutting it at exactly 45 slices that bar down the middle, throwing away its right half.

The repair is to cut at 45.5 instead, the outer edge of the bar. That half-unit shift is called the continuity correction:

which gives 0.86921, against the exact 0.86891. The error drops from 2.26 percentage points to 0.03 of a percentage point.

The direction of the shift follows from which bars you want. For $P(X \le 45)$ add 0.5 to reach 45.5. For $P(X \ge 45)$ subtract 0.5 to reach 44.5, since you want the bar at 45 included on the other side. Getting the direction backwards makes the answer worse than no correction at all.

The correction also fixes something the plain approximation cannot do. A continuous curve assigns probability 0 to any single value, so it says $P(X = 40) = 0$, while the exact binomial gives 0.08122. Treating the value 40 as the interval from 39.5 to 40.5 under the normal curve returns 0.08129, which is right to three decimal places.

How big the error gets

Passing the large-counts check does not mean the error is negligible. It means the shape is roughly right. Here is the same style of question at three sample sizes, each with the exact binomial probability alongside the plain normal approximation.

The first row barely satisfies the condition, with $np = 10$ exactly and $n(1-p) = 15$, and it is off by more than five percentage points. The error shrinks roughly in proportion to $\frac{1}{\sqrt{n}}$, so buying one more decimal place of accuracy costs about a hundredfold increase in sample size.

Tails are worse in relative terms. For $n = 100$ and $p = 0.4$, the exact $P(X \ge 50)$ is 0.02710, while the plain normal approximation gives 0.02061. In absolute terms that is only 0.65 of a percentage point, but it is 24% of the true probability, which matters a great deal when the number in question is a p-value.

Where this actually matters

You will rarely need the normal approximation to answer a question, because technology returns exact binomial probabilities. It matters for three other reasons.

It is why inference for proportions works. The sample proportion $\hat{p}$, read "p-hat", is a binomial count divided by $n$. Dividing by a constant does not change the shape, so once the binomial count is approximately normal, so is $\hat{p}$, with mean $p$ and standard deviation $\sqrt{\frac{p(1-p)}{n}}$. Every proportion $z$-interval and $z$-test on the AP exam stands on that fact, which is why the Large Counts condition reappears there.

It explains the conditions you are asked to check. When you write "$np \ge 10$ and $n(1-p) \ge 10$" on a free-response question, this is the reason. The condition is not paperwork; it is the sentence that licenses using a normal curve.

It tells you what a binomial looks like. Knowing that $B(100, 0.4)$ is centered at 40 with a standard deviation near 4.9 lets you see at a glance that a result of 55 successes is about three standard deviations high and therefore surprising, without computing anything.

For the sampling-distribution version of the same idea, see sampling distributions explained, and the normal distribution calculator will do the $z$ arithmetic.

Mistakes to avoid

• Checking only $np$. Both $np$ and $n(1-p)$ must be at least 10, and it is the smaller one that decides.
• Using $\sqrt{np}$ as the standard deviation. It is $\sqrt{np(1-p)}$.
• Reporting the approximation when the exact binomial is one calculator command away. Give the exact value; mention the approximation only if the question asks for it.
• Applying the half-unit correction in the wrong direction. Add 0.5 for a "less than or equal to" bound, subtract 0.5 for a "greater than or equal to" bound.
• Treating a passed condition as a promise of accuracy. At $n = 25$ with $p = 0.4$ the condition passes and the plain approximation is still off by more than five percentage points.
• Approximating a count that is not binomial in the first place. Check the four binomial conditions first, using how to tell if a situation is binomial.