When is the Normal a good stand-in for a Binomial or Poisson?

A Binomial or Poisson puts all its probability on whole numbers. The bar for $X = 7$ has width 1 and runs from 6.5 to 7.5, so when a smooth Normal curve stands in for the bars, the matching piece of area is the strip from 6.5 to 7.5:

Without the correction, $P(X = x)$ becomes the Normal probability of a single point, which is exactly 0, and $P(X \le x)$ stops at the middle of the last bar and misses half of it. The same half-unit fixes every kind of question:

• $P(X \le a) \approx P_N(X < a + 0.5)$,   $P(X \ge a) \approx P_N(X > a - 0.5)$,   $P(a \le X \le b) \approx P_N(a - 0.5 < X < b + 0.5)$.

The Normal used is the one with the same mean and SD as the count:

• Binomial: mean $np$, SD $\sqrt{np(1-p)}$. The approximation is reasonable when $np \ge 5$ and $n(1-p) \ge 5$.
• Poisson: mean $\lambda$, SD $\sqrt{\lambda}$. The approximation is reasonable when $\lambda > 10$.

Why it works at all: a Binomial count is a sum of $n$ independent 0/1 trials, and a Poisson count with a large mean behaves like a sum of many small independent counts. Sums of many independent pieces are approximately Normal, so this is the central limit theorem from the previous explorer again. The rules of thumb say when there are "enough" pieces for the skew and the boundary at 0 to stop mattering.

1. The lecture's example. The page opens on Binomial(30, 0.25), as on the slide: $np = 7.5$ and $n(1-p) = 22.5$, so the rule is met. Compare the three rows of the table for $P(X \le 5)$: exact 20.3%, Normal with the correction 20.0%, Normal without it 14.6%. The correction turns a 5.7-point miss into a 0.3-point one.
2. Single values need the correction. Switch to $P(X = a)$. Without the correction the Normal gives exactly 0, since a continuous distribution puts no probability on a single point. With it, the strip from $a - 0.5$ to $a + 0.5$ closely matches the bar: for $P(X = 7)$, 16.3% against the exact 16.6%.
3. Break the rule. Set $n = 30$ and $p = 0.05$ ($np = 1.5$). The bars pile up against 0 and skew right, and the Normal spills probability below 0, where no count can be. Even with the correction, $P(X = 0)$ comes out 15.4% against an exact 21.5%, and the worst error in $P(X \le a)$ is 5.4 points, against 1.4 points for the lecture's example. Now raise $n$ to 100 ($np = 5$) and the worst error halves to 2.7 points.
4. p near 1 is the mirror image. Try $p = 0.95$: now $n(1-p)$ is the small one, and the skew runs the other way.
5. Poisson. Switch to Poisson and slide $\lambda$ from 2 to 30. Below about 10 the distribution is visibly skewed and the Normal fits poorly; above it the bars and the curve nearly coincide. The worst corrected error falls from 4.4 points at $\lambda = 2$ to 2.1 at 10 and 1.2 at 30. The slide's R check is hist(rpois(10000, 50)) against hist(rnorm(10000, 50, sqrt(50))).
6. Watch the worst error. The line under the table reports the largest error in $P(X \le a)$ over every $a$, with and without the correction. It is a quick summary of how good the approximation is for the current settings.