Mathematics/Lessons/The Normal Distribution/Approximating a Binomial Distribution
Statistics · The Normal Distribution · Year 2

Approximating a Binomial Distribution

When n is large and p isn't too extreme, a normal curve tracks a binomial distribution closely.

9 min read 2 worked examples

When a normal curve tracks a binomial distribution

Consider the binomial random variable $X\sim \mathrm{B}(n,p)$. It can be difficult to calculate probabilities for $X$ when $n$ is large. In certain circumstances you can use a normal distribution to approximate a binomial distribution.

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A binomial distribution $\mathrm{B}(30,0.5)$ (bars) with its normal approximation $\mathrm{N}(15, 7.5)$ overlaid (curve).

You need to understand the conditions under which this approximation is valid, and learn the relationship between the values of $n$ and $p$ in $\mathrm{B}(n,p)$ and the values of $\mu$ and $\sigma$ in the normal approximation $\mathrm{N}(\mu,\sigma^2)$.

Key fact

If $n$ is large and $p$ is close to $0.5$, then the binomial distribution $X\sim \mathrm{B}(n,p)$ can be approximated by the normal distribution $\mathrm{N}(\mu,\sigma^2)$ where $\mu=np$ and $\sigma=\sqrt{np(1-p)}$.

Hint

The approximation is only valid when $p$ is reasonably close to $0.5$, since the normal distribution is symmetrical — a binomial distribution with $p$ very close to $0$ or $1$ is heavily skewed and won't be well approximated by a symmetric curve.

Worked example 1

Finding μ and σ for the approximation
A call centre resolves 62% of incoming calls within 5 minutes. 150 calls are sampled, and the number resolved within 5 minutes, $X$, is recorded. a Write down a binomial model for $X$. b Explain why $X$ can be approximated with a normal distribution, $Y\sim \mathrm{N}(\mu,\sigma^2)$. c Find the values of $\mu$ and $\sigma$ in this approximation.
a
$X\sim \mathrm{B}(150, 0.62)$.
b
$X$ can be approximated with a normal distribution since $n$ is large and $p$ is reasonably close to $0.5$.
c
$\mu = 150\times0.62 = 93$. $\sigma = \sqrt{150\times0.62\times(1-0.62)} = 5.94$ (3 s.f.).
Use $\mu=np$ and $\sigma=\sqrt{np(1-p)}$.
▪ solved

Continuity correction

The binomial distribution is discrete, but the normal distribution is continuous.

Key fact

If you are using a normal approximation to a binomial distribution, you need to apply a continuity correction when calculating probabilities.

The diagrams below show a binomial variable $X\sim \mathrm{B}(16,0.5)$ being approximated by $Y\sim \mathrm{N}(8, 2^2)$:

6.5 8.0 Y ~ N(8, 2.00²)
$\mathrm{P}(X>6) \approx \mathrm{P}(Y>6.5)$
8.0 9.5 10.5 Y ~ N(8, 2.00²)
$\mathrm{P}(X=10) \approx \mathrm{P}(9.5<Y<10.5)$
3.5 8.0 Y ~ N(8, 2.00²)
$\mathrm{P}(X\le 4) \approx \mathrm{P}(Y<4.5)$

Worked example 2

Applying the continuity correction
At a large university, 40% of students hold a scholarship. A random sample of 200 students is taken, and $X$, the number holding a scholarship, is recorded. Use a suitable approximation to find $\mathrm{P}(X>90)$.
MethodApproximate, then apply the continuity correction
1
$X\sim \mathrm{B}(200,0.4)$ can be approximated by $Y\sim \mathrm{N}(80, 6.93^2)$, since $n$ is large and $p$ is reasonably close to $0.5$.
$\mu=200\times0.4=80$, $\sigma=\sqrt{200\times0.4\times0.6}=6.93$ (3 s.f.).
2
$\mathrm{P}(X>90) \approx \mathrm{P}(Y>90.5)$.
$X$ is discrete and $Y$ is continuous, so shift the boundary by $0.5$ towards the tail you're excluding.
60 65 70 75 80 90.5 95 100 Y ~ N(80, 6.93²)
$\mathrm{P}(Y>90.5)=0.0648$ (3 s.f.) — very close to the exact binomial value of $0.0655$.
3
$\mathrm{P}(X>90) \approx 0.0648$ (3 s.f.).
Compare with the exact binomial probability $\mathrm{P}(X>90)=0.0655$ — the normal approximation is very close.
▪ solved

Check yourself

In a large city, 45% of households recycle regularly. A random sample of 120 households is taken, and $Y$, the number that recycle regularly, is recorded. a Explain why $Y$ can be approximated by a normal distribution. b Find the values of $\mu$ and $\sigma$ for this approximation.

a
$n=120$ is large and $p=0.45$ is reasonably close to $0.5$.
b
$\mu = 120\times0.45 = 54$. $\sigma = \sqrt{120\times0.45\times0.55} = 5.45$ (3 s.f.).