Approximating a binomial with a normal

When n is large and p not too extreme, approximating B(n, p) by N(np, np(1−p)) — and applying the continuity correction (±0.5) when moving from the discrete binomial to the continuous normal.

Statistics · Statistical distributions

Approximating a binomial with a normal

When n is large and p not too extreme, approximating B(n, p) by N(np, np(1−p)) — and applying the continuity correction (±0.5) when moving from the discrete binomial to the continuous normal.

Why it works

For a binomial with a large

n

, working out cumulative probabilities term by term is painful. But as

n

grows, the binomial's bar chart settles into a bell shape — so a normal distribution can stand in for it.

The approximation matches the binomial's mean and variance:

X \sim B(n, p) \;\approx\; Y \sim N\big(np,\; np(1-p)\big).

That is, mean

\mu = np

and variance

\sigma^2 = np(1-p)

(so

\sigma = \sqrt{np(1-p)}

). It works well when $n$ is large and $p$ is not too close to $0$ or $1$ — a common rule of thumb is

np > 5

and

n(1-p) > 5

, so the bell isn't cut off at an end.

The continuity correction. The binomial is discrete (whole numbers), the normal is continuous. To bridge the gap, each whole number

k

is treated as the strip from

k - 0.5

k + 0.5

. So when you rewrite a binomial probability as a normal one, you shift the boundary by

0.5

in the direction that keeps the right values in:

$P(X \le k) \to P(Y < k + 0.5)$ and $P(X < k) \to P(Y < k - 0.5)$ ;
$P(X \ge k) \to P(Y > k - 0.5)$ and $P(X > k) \to P(Y > k + 0.5)$ ;
$P(X = k) \to P(k - 0.5 < Y < k + 0.5)$ .

The reliable check: ask whether

k

itself is included, and nudge the boundary by

0.5

so it stays inside (for

\le, \ge, =

) or outside (for

<, >