A binomial count is a sum of independent 0–1 trials, so by the central limit theorem it is close to normal when $n$ is large. But the binomial lives on whole numbers and the normal curve on all real numbers. The continuity correction bridges the two.
Idea clave
For $X$ binomial with $np(1 - p)$ reasonably large (say at least 10), use $Y$ normal with mean $np$ and standard deviation $\sqrt{np(1 - p)}$, and widen each whole number $k$ to $[k - 0.5,\ k + 0.5]$: $$P(X \le k) \approx P(Y < k + 0.5), \qquad P(X \ge k) \approx P(Y > k - 0.5), \qquad P(X = k) \approx P(k - 0.5 < Y < k + 0.5).$$
Por qué funciona
Draw the binomial pmf as bars of width 1, centered on the whole numbers. The bar for $k$ covers $[k - 0.5, k + 0.5]$ and its area is $P(X = k)$. The normal curve approximates the tops of the bars, so the area under the curve over the same stretch approximates the same probability. For $X \le k$ the bars run through the one at $k$, which ends at $k + 0.5$. Without the correction, a single value would get probability 0, and each "$\le$" question would lose half a bar.
| Event | Corrected normal event |
|---|---|
| $X \le 45$ | $Y < 45.5$ |
| $X < 45$, which is $X \le 44$ | $Y < 44.5$ |
| $X \ge 45$ | $Y > 44.5$ |
| $X > 45$, which is $X \ge 46$ | $Y > 45.5$ |
| $X = 45$ | $44.5 < Y < 45.5$ |
Worked examples
Ejemplo resuelto: At most 45 heads
Flip a fair coin 100 times. Approximate $P(X \le 45)$. Use $\Phi(-0.9) = 0.1841$.
- $np = 50$ and $\sqrt{np(1 - p)} = 5$.
- $P(X \le 45) \approx P(Y < 45.5)$; $z = \frac{45.5 - 50}{5} = -0.9$.
- $\approx 0.1841$. The exact binomial value is $0.1841$ too.
Respuesta: about $0.184$
Ejemplo resuelto: Fewer than 45
Same coin. Approximate $P(X < 45)$. Use $\Phi(-1.1) = 0.1357$.
- $X < 45$ means $X \le 44$, so use $Y < 44.5$.
- $z = \frac{44.5 - 50}{5} = -1.1$, and $P \approx 0.1357$ (exact: $0.1356$).
Respuesta: about $0.136$
Ejemplo resuelto: Exactly 50
Approximate $P(X = 50)$. Use $\Phi(0.1) = 0.5398$.
- $P(X = 50) \approx P(49.5 < Y < 50.5)$, with $z = \pm 0.1$.
- $\approx 0.5398 - 0.4602 = 0.0796$ (exact: $0.0796$).
Respuesta: about $0.080$
Error común
Skipping the correction, or moving the boundary the wrong way. $P(X < 12)$ is $P(X \le 11)$, so the boundary is $11.5$, not $12.5$. Check: list which whole numbers the event includes, and make the normal interval run half a unit beyond the extreme ones.