Actuarial Statistics · Central limit theorem

Central Limit Theorem Applications to Aggregate Claims and Sample Sizes

Updated 11 October 2026 · Fact-checked

The central limit theorem says a sum or mean of many independent, identically distributed claims is approximately normal. For total claims, use mean nμ and variance nσ². Standardise, then read the normal table. For sample size, set the required accuracy equal to z × σ ÷ √n and solve for n, rounding up.

Understand Applications to Aggregate Claims and Sample Sizes

An insurer holds many policies. Each policy gives a random claim. The total claim is the sum of these random amounts. Individual claims are usually skewed, with many small claims and a few large ones. The total is much better behaved.

The central limit theorem (CLT) explains why. If X₁, X₂, ..., Xₙ are independent and identically distributed with mean μ and finite variance σ², then the sum S = X₁ + ... + Xₙ is approximately normal for large n. Its mean is nμ and its variance is nσ². The approximation does not need the claims themselves to be normal.

The sample mean X̄ = S ÷ n is also approximately normal, with mean μ and variance σ² ÷ n. The standard deviation of X̄ is σ ÷ √n. It shrinks as n grows. This is what lets you choose a sample size for a given accuracy.

You use the CLT in two ways in exams. First, find a probability, such as the chance that total claims exceed a given amount. Second, find the smallest n so that the sample mean is within a stated distance of μ with a stated probability.

The approximation is better when n is larger and the claim distribution is closer to symmetric. For heavily skewed claims you need a larger n. If the question says to use the CLT, state that you are using a normal approximation and give its parameters.

Key rules to remember

Sum of n iid claims: S = X₁ + ... + Xₙ ≈ N(nμ, nσ²); Needs independence, identical distribution, finite variance and large n.
Sample mean: X̄ ≈ N(μ, σ² ÷ n); The standard deviation of X̄ is σ ÷ √n.
Standardisation: Z = (S − nμ) ÷ (σ√n) ≈ N(0, 1); Use Z to read Φ from the normal table.
Compound distribution moments: E[S] = E[N] E[X]; Var(S) = E[N] Var(X) + Var(N) (E[X])²; Use when the number of claims N is random and independent of the claim amounts.
Compound Poisson with mean λ: E[S] = λ E[X]; Var(S) = λ E[X²]; Use E[X²], not Var(X). The CLT applies when λ is large.
Sample size for a mean: P(|X̄ − μ| ≤ ε) ≈ 2Φ(ε√n ÷ σ) − 1, so n ≥ (zσ ÷ ε)²; z is the standard normal value with P(|Z| ≤ z) equal to the required probability. For 95%, z = 1.96. Round n up.
Sample size for a proportion: n ≥ z² p(1 − p) ÷ ε²; If p is unknown, p = 0.5 gives the largest n and is the safe choice.
Continuity correction: For integer-valued S: P(S ≤ k) ≈ Φ((k + 0.5 − mean) ÷ sd); Use it when S is discrete and takes integer values, for example a count of claims.

How to solve Applications to Aggregate Claims and Sample Sizes questions

Use this method for any question on totals, means or required sample sizes.

1Define the random variable. Is it a total S, a mean X̄, or a count? Write what each Xᵢ represents.
2Check the conditions: independent, identically distributed, finite variance, and n large. State them in a line.
3Find μ and σ² for one claim. If the number of claims is random, use the compound formulas for E[S] and Var(S).
4Write the normal approximation with its parameters, for example S ≈ N(nμ, nσ²). Take the square root of the variance to get the standard deviation.
5For a probability: apply a continuity correction only if the variable is discrete. Standardise and read Φ from the table. For an upper-tail probability use 1 − Φ(z).
6For a sample size: write the probability statement, convert it to a z value, set ε√n ÷ σ equal to z, and solve for n. Round up to the next whole number.
7Check that the answer is sensible, then state it in context with units, such as rupees or number of policies. Note that it is approximate.

Quickest way: Mean, SD, z, table

When to use it: Use this in multiple-choice questions and when time is short. It suits plain sum or mean problems with iid claims.

Total: mean = nμ and SD = σ√n. Write both numbers down first.
Compute z = (limit − mean) ÷ SD.
Read Φ(z). For 'exceeds', answer 1 − Φ(z).
Sample size: compute n = (zσ ÷ ε)² directly. Use z = 1.96 for 95% two-sided and z = 1.645 for 90% two-sided.
Always round n up. Check that z has the right tail: 95% two-sided means 2.5% in each tail.

Common mistakes in Applications to Aggregate Claims and Sample Sizes

Using variance nσ as the variance of the sum, or taking SD as nσ.
Students copy the pattern of the mean, which is nμ.
Fix: Variance of the sum is nσ². SD is σ√n. Always compute SD last by taking a square root.
Using σ² instead of σ in the sample size formula, or forgetting to square at the end.
The formula n = (zσ ÷ ε)² mixes the SD and a square.
Fix: Solve ε√n ÷ σ = z step by step. First √n = zσ ÷ ε, then square to get n.
Rounding the sample size down.
Students round 983.45 to 983 by habit.
Fix: The sample size must be at least the value found. Always round up, here to 984.
Using one-sided z for a two-sided accuracy statement.
The wording 'within ε of μ' means both tails, but students use z = 1.645 for 95%.
Fix: For 'within ε' with probability 95%, put 2.5% in each tail and use z = 1.96.
Forgetting the continuity correction for integer-valued totals, or applying it to continuous claim amounts.
Students treat the correction as always required or never required.
Fix: Apply it only when the variable is discrete and integer-valued. Do not use it for continuous amounts such as claim sizes in rupees.
Using Var(X) instead of E[X²] for the compound Poisson variance.
Students carry over the formula for a fixed number of claims.
Fix: For compound Poisson with mean λ, Var(S) = λE[X²]. If you have Var(X) instead, use E[X²] = Var(X) + (E[X])².

Worked examples

Example 1

A portfolio has 400 independent policies. The annual claim amount per policy has mean ₹5,000 and standard deviation ₹12,000. Use the central limit theorem to estimate the probability that total annual claims exceed ₹22,40,000.

Show the solution

Let S be the total claims, the sum of 400 independent identically distributed claim amounts. n = 400 is large, so the CLT applies.
E[S] = 400 × 5,000 = ₹20,00,000.
Var(S) = 400 × 12,000² = 400 × 14,40,00,000 = 5,76,00,00,00,000.
SD(S) = 12,000 × √400 = 12,000 × 20 = ₹2,40,000.
Approximate S ≈ N(20,00,000, 2,40,000²).
z = (22,40,000 − 20,00,000) ÷ 2,40,000 = 1.
P(S > 22,40,000) ≈ 1 − Φ(1) = 1 − 0.8413 = 0.1587.

Answer: About 0.159 (approximately 15.9%).

Example 2

Claim amounts have a standard deviation of ₹8,000. How many claims must be sampled so that the sample mean is within ₹500 of the true mean claim with probability at least 95%? Use the central limit theorem.

Show the solution

X̄ ≈ N(μ, 8,000² ÷ n), so the standard deviation of X̄ is 8,000 ÷ √n.
We need P(|X̄ − μ| ≤ 500) ≥ 0.95.
Standardise: P(|Z| ≤ 500√n ÷ 8,000) ≥ 0.95.
For 95% two-sided, z = 1.96. So 500√n ÷ 8,000 ≥ 1.96.
√n ≥ 1.96 × 8,000 ÷ 500 = 31.36.
n ≥ 31.36² = 983.45.
Round up to the next whole number.

Answer: n = 984 claims (approximately, by the normal approximation).

Exam tips

Write the normal approximation with both parameters before you touch the table. Examiners give method marks for mean and variance even if the final number is wrong.
State the conditions in one line: independent, identically distributed, finite variance, large n. Written questions often award a mark for this.
Read the wording. 'Within' means two-sided. 'Exceeds' means upper tail. 'At least' for sample size means round up.
In compound questions, check whether you are given Var(X) or E[X²], then pick the right formula. Show the line E[X²] = Var(X) + (E[X])² if you convert.
In computer-based papers, show the R call you would use, for example 1 - pnorm(z), and write the result with the formula you used.

Practice questions from Central limit theorem

Applications to Aggregate Claims and Sample Sizes: frequently asked questions

When can I use the central limit theorem for aggregate claims?

Use it when you add many independent claims with the same distribution and a finite variance. The sum is then approximately normal. It works less well for small n or for very skewed claims.

How do I find the sample size using the central limit theorem?

Write P(|X̄ − μ| ≤ ε) ≥ the required probability. Convert it to z, so ε√n ÷ σ ≥ z. Then n ≥ (zσ ÷ ε)². Round up to a whole number.

Do I need a continuity correction for total claims?

Only if the total is discrete and takes integer values, such as a number of claims. For claim amounts treated as continuous, you do not need one. If the question says to use a continuity correction, apply it with 0.5.

How is the CLT used when the number of claims is random?

Find E[S] and Var(S) using the compound formulas, then approximate S by a normal with those parameters. For compound Poisson, E[S] = λE[X] and Var(S) = λE[X²]. The approximation is better when λ is large.