Normal approximation to the binomial

The normal approximation replaces a binomial count $X \sim B(n, p)$ with a normal curve carrying the binomial's own parameters, mean $\mu = np$ and standard deviation $\sigma = \sqrt{np(1-p)}$ (mu and sigma). It is licensed by the Large Counts condition, $np \ge 10$ and $n(1-p) \ge 10$: the expected successes and the expected failures each have to reach 10. Both sides are required because a binomial squeezed against 0 or against $n$ is skewed, and no symmetric curve fits a skewed shape.

Watch it fail when the condition fails. With $n = 20$ and $p = 0.1$, $np = 2$, so the mean sits only $2 / \sqrt{1.8} \approx 1.49$ standard deviations above the floor at 0. The exact binomial gives $P(X \le 1) = 0.3917$. The fitted normal, with $\sigma = \sqrt{1.8} \approx 1.342$, puts $1$ at $z = \frac{1 - 2}{1.342} \approx -0.745$ and returns 0.2280. A miss of 16 percentage points is not rounding.

Passing the check buys shape, not precision, and that is the sentence to hold onto. Take $n = 100$ and $p = 0.5$, where $np = n(1-p) = 50$ and the fit is about as good as it gets. The exact probability that the count lands 10 or more away from 50 is 0.0569. The normal model, with $\sigma = 5$, reads that same event as the two tails past $z = \pm 2$ and returns 0.0455. Each number is right about its own model, so a p-value quoted as 0.0455 belongs to the normal model rather than to the binomial, and at $\alpha = 0.05$ the two models reach opposite verdicts on identical data. Cutting the curve at 40.5 and 59.5 instead, the half-unit correction, brings it to 0.0574.

Calculators return exact binomial probabilities in a keystroke, so take numbers from there. The approximation earns its keep as the reason a sample proportion has a roughly normal sampling distribution, which is what topic 3.2 of Unit 3, Sampling Distributions for Sample Proportions, is built on.