When \(n\) is large, adding dozens of binomial terms by hand is hopeless, but the bars of the distribution trace out a normal curve. Match the mean and variance, apply a continuity correction, and one standardisation does the job.
If \(X \sim \text{B}(n, p)\) with \(np > 5\) and \(nq > 5\) (where \(q = 1 - p\)), then approximately \(X \sim \text{N}(np, npq)\). Each whole number \(r\) is represented by the interval from \(r - 0.5\) to \(r + 0.5\): this is the continuity correction.
Fig. 1: B(60, 0.4) with the normal curve N(24, 14.4) drawn to scale. The bar for 20 runs from 19.5 to 20.5, so 'X ≤ 20' means all the area to the left of 20.5
Section 01
When the bars look normal.
A binomial distribution is roughly bell-shaped when \(n\) is large and \(p\) is not too close to 0 or 1. The syllabus condition is \(np > 5\) and \(nq > 5\). The approximating normal has the SAME mean and variance as the binomial.
Worked example Setting up the approximation
\(X \sim \text{B}(60, 0.4)\). Show that a normal approximation is suitable and state its distribution.
Answer: \(X \approx Y \sim \text{N}(24, 14.4)\), with \(\sigma = \sqrt{14.4} = 3.795\)
Worked example Not suitable
Can \(\text{B}(200, 0.98)\) be approximated by a normal distribution?
\(np = 196 > 5\), but \(nq = 200 \times 0.02 = 4\), which is not greater than 5.
The distribution is squashed against \(X = 200\) and is strongly skewed.
Answer: No, since \(nq = 4 < 5\)
Examiner trap
Using \(\sqrt{np}\) or \(npq\) as the standard deviation. The variance is \(npq = 14.4\), so \(\sigma = \sqrt{14.4} = 3.795\).
Section 02
The continuity correction.
A binomial variable takes whole numbers only; a normal variable is continuous. In Fig. 1 the bar for each whole number \(r\) spans \(r - 0.5\) to \(r + 0.5\), so translate every inequality to cover exactly the right bars.
Binomial \(X\)
Normal \(Y\)
Bars included
\(X \le 20\)
\(Y < 20.5\)
up to and including 20
\(X < 20\)
\(Y < 19.5\)
up to 19
\(X \ge 30\)
\(Y > 29.5\)
30 and above
\(X > 30\)
\(Y > 30.5\)
31 and above
\(X = 25\)
\(24.5 < Y < 25.5\)
just 25
Worked example At most
\(X \sim \text{B}(60, 0.4)\). Use a suitable approximation to find \(\text{P}(X \le 20)\).
Answer: 0.178 (3 s.f.); the exact binomial value is 0.1786
Worked example At least
For the same \(X\), find \(\text{P}(X \ge 30)\).
\(X \ge 30\) includes the bar for 30, so use \(\text{P}(Y > 29.5)\).
\(z = \dfrac{29.5 - 24}{3.795} = 1.449\).
\(1 - \Phi(1.449) = 1 - 0.9263 = 0.0737\).
Answer: 0.0737 (3 s.f.); exact value 0.0746
Examiner trap
Correcting in the wrong direction. Ask 'is the boundary value included?' For \(X \ge 30\) it is, so you go DOWN to 29.5 to take in its whole bar.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What are the conditions for approximating \(\text{B}(n, p)\) by a normal distribution?
Quick check+10 XP
\(X \sim \text{B}(50, 0.3)\). Which normal distribution approximates \(X\)?
Quick check+10 XP
With a continuity correction, \(\text{P}(X < 12)\) becomes:
Quick check+10 XP
With a continuity correction, \(\text{P}(X \ge 12)\) becomes:
Section 03
Between, and exactly.
Strict and non-strict inequalities at each end are corrected separately. A single value becomes a thin interval of width 1, because a continuous variable has zero probability at any single point.
Worked example Between
\(X \sim \text{B}(60, 0.4)\). Use a suitable approximation to find \(\text{P}(20 < X < 28)\).
Whole numbers 21 to 27, so \(\text{P}(20.5 < Y < 27.5)\).
Answering 0 for \(\text{P}(X = 25)\). That is true for a continuous variable, but \(X\) is binomial; the continuity correction turns 25 into the interval 24.5 to 25.5.
Section 04
In context.
In a word problem, define \(X\) and its binomial distribution first, check \(np > 5\) and \(nq > 5\), then state the normal approximation and its continuity-corrected inequality before standardising.
Worked example Late flights
12% of flights from an airport are late. Of 150 flights chosen at random, find the probability that more than 25 are late.
\(X \sim \text{B}(150, 0.12)\): \(np = 18\), \(nq = 132\), both greater than 5.