Actuarial Statistics · Central limit theorem

Normal Approximations to Standard Distributions

Updated 11 October 2026 · Fact-checked

A normal approximation replaces a distribution that is a sum of many independent pieces, such as binomial, Poisson, gamma or chi-squared, with a normal distribution that has the same mean and variance. Match mean and variance, apply continuity correction for discrete cases, standardise, then read the normal tables.

Understand Normal Approximations to Standard Distributions

The central limit theorem (CLT) says that the sum of many independent, identically distributed random variables with finite variance is approximately normal. The more terms you add, the better the fit.

Many standard distributions are sums of simpler ones. A binomial(n, p) is the sum of n Bernoulli variables. A Poisson(λ) with large λ behaves like the sum of many small independent Poisson variables. A gamma(α, λ) with integer α is the sum of α independent exponentials. A chi-squared with k degrees of freedom is the sum of k independent squared standard normals. So the CLT applies to each.

The approximation needs only two numbers: the exact mean and the exact variance of the original distribution. You then say the variable is approximately N(mean, variance) and use normal tables.

Binomial and Poisson are discrete, but the normal is continuous. To fix this, you use a continuity correction: treat the integer r as the interval from r − 0.5 to r + 0.5. Gamma and chi-squared are already continuous, so no correction is needed.

The approximation is good when n is large and p is not too close to 0 or 1 (binomial), or when λ is large (Poisson). Gamma and chi-squared are skewed to the right, so the fit improves as the shape parameter or degrees of freedom grows.

Key rules to remember

Binomial approximation: X ~ Bin(n, p) ≈ N(np, np(1 − p)); Works well for large n with p not near 0 or 1. A common rule of thumb is np and n(1 − p) both above about 5 to 10; it is a guide, not a law.
Poisson approximation: X ~ Poisson(λ) ≈ N(λ, λ); Mean and variance are both λ. Better for larger λ, for example above about 10 to 20.
Gamma approximation: X ~ Gamma(α, λ) ≈ N(α ÷ λ, α ÷ λ²); Uses the rate parameter λ. If the question gives a scale parameter, convert first. Improves as α grows.
Chi-squared approximation: X ~ χ²(k) ≈ N(k, 2k); Chi-squared is gamma with α = k/2 and λ = 1/2. Mean k, variance 2k.
Continuity correction: P(X ≤ r) ≈ P(Z ≤ (r + 0.5 − μ) ÷ σ); P(X ≥ r) ≈ P(Z ≥ (r − 0.5 − μ) ÷ σ); For integer-valued X. P(X = r) ≈ P(r − 0.5 < Y < r + 0.5). Not used for gamma or chi-squared.
Standardisation: Z = (Y − μ) ÷ σ, where σ = √variance; Take the square root of the variance before dividing.

How to solve Normal Approximations to Standard Distributions questions

Use this method for any question asking you to approximate a probability or a percentile using the normal distribution.

1Identify the exact distribution and its parameters, and check whether it is discrete or continuous.
2Write down the exact mean and variance of the distribution.
3State the approximation: X is approximately N(μ, σ²). Say so in words in a written answer.
4If X is discrete, rewrite the event with a continuity correction, for example P(X ≤ 12) becomes P(Y < 12.5).
5Standardise using Z = (Y − μ) ÷ σ, with σ the square root of the variance.
6Read the probability from normal tables, using symmetry for negative z values.
7Check the answer is sensible and comment briefly on the quality of the approximation if the question asks.

Quickest way: Mean, variance, correct, standardise

When to use it: Use this for multiple-choice questions and short written parts where you need a probability quickly.

Write μ and σ² straight from the parameters (np and np(1 − p); λ and λ; α/λ and α/λ²; k and 2k).
Convert the event to a continuous interval. Add or subtract 0.5 only for discrete variables.
Compute z once, round to two decimals, and look up Φ(z).
For an 'at least' event, use 1 − Φ(z). For a 'between' event, subtract two Φ values.

Common mistakes in Normal Approximations to Standard Distributions

Forgetting the continuity correction for binomial or Poisson.
Students focus on the normal formula and forget that the original variable takes only integer values.
Fix: Whenever the original variable is discrete, write the integer as an interval r ± 0.5 before standardising.
Applying the correction in the wrong direction, for example using P(X ≥ 30) ≈ P(Y > 30.5).
Students memorise 'add 0.5' without thinking about which integers are included.
Fix: For X ≥ r, the value r must be included, so use Y > r − 0.5. For X > r, use Y > r + 0.5. Sketch the integers if unsure.
Dividing by the variance instead of the standard deviation.
The variance is the number written in N(μ, σ²), so it is easy to plug in directly.
Fix: Always compute σ = √variance as a separate line before finding z.
Using the wrong gamma parameters, such as mean α × λ.
Different sources use rate or scale parameterisations.
Fix: Check which parameter the question gives. For rate λ the mean is α ÷ λ and variance α ÷ λ². For scale θ the mean is αθ and variance αθ².
Using variance k instead of 2k for the chi-squared distribution.
Students confuse it with the Poisson, where mean equals variance.
Fix: Remember χ²(k) has mean k and variance 2k, because it is a sum of k squared standard normals each with variance 2.
Applying a continuity correction to a gamma or chi-squared variable.
Students treat the correction as part of every normal approximation.
Fix: Only use it when the original variable is discrete.

Worked examples

Example 1

The number of claims X in a year on a portfolio is Poisson with mean 36. Use a normal approximation to estimate P(X ≥ 42).

Show the solution

X ~ Poisson(36), so mean = 36 and variance = 36.
Approximate X by Y ~ N(36, 36), so σ = 6.
X is discrete, so P(X ≥ 42) ≈ P(Y > 41.5).
z = (41.5 − 36) ÷ 6 = 5.5 ÷ 6 = 0.9167, about 0.92.
From tables, Φ(0.92) ≈ 0.8212.
P(X ≥ 42) ≈ 1 − 0.8212 = 0.1788.

Answer: About 0.179

Example 2

The total of 50 independent waiting times, each exponential with rate 0.5 per minute, is T. Use a normal approximation to find P(T > 110).

Show the solution

The sum of 50 independent exponential(0.5) variables is Gamma(α = 50, λ = 0.5).
Mean = α ÷ λ = 50 ÷ 0.5 = 100.
Variance = α ÷ λ² = 50 ÷ 0.25 = 200, so σ = √200 = 14.142.
T is continuous, so no continuity correction is needed.
z = (110 − 100) ÷ 14.142 = 0.7071, about 0.71.
From tables, Φ(0.71) ≈ 0.7611.
P(T > 110) ≈ 1 − 0.7611 = 0.2389.

Answer: About 0.239

Exam tips

Write the mean and variance on their own lines. Method marks are usually given for them even if the final number is wrong.
In multiple-choice questions, the wrong options are often what you get by skipping the continuity correction or dividing by the variance. Do both steps carefully.
State the approximation explicitly, such as 'X is approximately N(36, 36)', in written answers.
For gamma and chi-squared, check whether the question gives rate or scale before you compute the variance.
If asked to comment on accuracy, mention that the exact distributions are skewed and the approximation improves as n, λ or the shape parameter increases.

Practice questions from Central limit theorem

Normal Approximations to Standard Distributions in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Normal Approximations to Standard Distributions: frequently asked questions

When can I use the normal approximation to the binomial?

Use it when n is large and p is not too close to 0 or 1, so the distribution is roughly symmetric. A common guide is that np and n(1 − p) are both reasonably large, but it is only a rule of thumb. Always apply the continuity correction.

Why is the CLT applicable to the Poisson distribution?

A Poisson(λ) variable can be viewed as the sum of many independent Poisson variables with smaller means that add up to λ. The CLT then gives an approximate normal distribution with mean λ and variance λ for large λ.

Do I need a continuity correction for gamma or chi-squared?

No. These are continuous distributions, so you can standardise the stated value directly. The correction is only for discrete variables such as binomial and Poisson.

What are the normal approximation parameters for a chi-squared distribution?

For χ² with k degrees of freedom the approximation is N(k, 2k). This follows from its mean k and variance 2k. It improves as k grows.