How do you normal-approximate a Binomial(200, 0.1) probability with a continuity correction?

## Why the CLT applies to a binomial A `Binomial(n, p)` random variable counts successes in n independent trials, each succeeding with probability p. That count is literally a sum of n i.i.d. Bernoulli(p) variables, each with mean p and variance `p(1-p)`. Since a Bernoulli has finite variance, the Central Limit Theorem applies to the sum: for large n the count is approximately normal with - mean `np` - variance `np(1-p)` - standard deviation `sqrt(np(1-p))` For `n = 200, p = 0.1`: mean 20, variance 18, standard deviation about 4.243. ## The continuity correction The binomial is **discrete** — it puts probability mass on the integers 0, 1, 2, ... The normal is **continuous** — it spreads probability smoothly and assigns zero probability to any single point. Bridging them requires deciding how much of the continuous density belongs to each integer, and the natural convention gives integer k the interval from `k - 0.5` to `k + 0.5`. This produces the correction rules: - `P(X <= k)` becomes `P(Normal <= k + 0.5)` - `P(X < k)` becomes `P(Normal <= k - 0.5)` - `P(X >= k)` becomes `P(Normal >= k - 0.5)` - `P(X > k)` becomes `P(Normal >= k + 0.5)` - `P(X = k)` becomes `P(k - 0.5 <= Normal <= k + 0.5)` The rule that keeps you out of trouble: the corrected boundary always moves so as to **include** the integer you meant to include, and to **exclude** the one you meant to exclude. Note that `P(X <= k)` and `P(X < k)` differ for a discrete variable, which is exactly why they map to different corrected boundaries. ## The worked numbers Approximate `P(X <= 25)` for `Binomial(200, 0.1)`. 1. Mean `np = 200 * 0.1 = 20`. 2. Standard deviation `sqrt(200 * 0.1 * 0.9) = sqrt(18) ~ 4.243`. 3. Corrected boundary: 25.5. 4. `z = (25.5 - 20) / 4.243 ~ 1.296`. 5. Normal lower-tail probability at 1.296 is about **0.9026**. The exact binomial value is **0.8995**, so the corrected approximation is off by about 0.003. Skipping the correction gives `z = (25 - 20) / 4.243 ~ 1.179` and about **0.8807**, off by about 0.019 — more than six times the error. The correction matters most exactly where people are tempted to skip it: modest n, and boundaries near the centre where a half-unit is a meaningful slice of the distribution. ## When the approximation is usable The standard guideline is `np >= 10` **and** `n(1-p) >= 10`. Both are needed because a binomial is skewed whenever p is far from 0.5, and the skew shows up in whichever tail is squeezed against a boundary. The count cannot go below 0 or above n, so when np is small the distribution piles up against zero and cannot look symmetric no matter what the normal curve says. For `Binomial(200, 0.1)`: `np = 20` and `n(1-p) = 180`, both comfortably above 10, so the approximation is sound — as the 0.003 error confirms. Contrast `Binomial(20, 0.05)`, where `np = 1`: the normal approximation would put meaningful probability on negative counts, which is nonsense. For very small np the appropriate limit is a different one entirely — the count converges to a Poisson rather than to a normal. ## Practical notes **Tail accuracy is worse than central accuracy.** The approximation is best near the mean and degrades in the far tails, precisely where extreme probabilities are being estimated. Treat an approximated probability of, say, one in a million with scepticism. **Proportions follow the same algebra.** If you care about the observed proportion `X / n` rather than the count, divide everything by n: the mean becomes p and the standard deviation becomes `sqrt(p(1-p)/n)`. The continuity correction becomes a shift of `0.5 / n` on the proportion scale, which is easy to forget. **Say why you are approximating.** With modern computation the exact binomial probability is directly computable, so the honest framing is that the normal approximation is valuable for reasoning, for closed-form algebra like sample-size formulas, and for intuition about how far a count can drift — not because the exact number is unreachable.