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Central Limit Theorem

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Edexcel A Level Further Maths — Further Statistics 1

The Central Limit Theorem

If you take a big enough sample from literally any population and average it, that average behaves like a normal distribution — even if the original population doesn't look normal at all.

Quick Summary

A random sample is a set of independent, identically distributed copies of a population variable X.
The sample mean, X̄, is not a fixed number — it changes depending on which sample you happen to draw.
If the population itself is normal, then X̄ ~ N(μ, σ²/n) exactly, for any sample size n.
The Central Limit Theorem (CLT) says that for any population distribution, as long as n is large, X̄ is approximately N(μ, σ²/n).
You always need to find μ = E(X) and σ² = Var(X) of the population first, then divide the variance by n.
Bigger samples → sample means cluster tighter around μ → smaller spread.

1. Samples and the Sample Mean

1.1 What is a "random sample" of a random variable?

Imagine you flip a fair coin 10 times and count the heads. That's one trial of X ~ B(10, 1/2), and here E(X) = np = 5, Var(X) = np(1-p) = 2.5.

Now suppose you repeat this whole experiment (10 flips → count heads) once a day, for 7 days. Each day gives you a new random variable:

X₁ ~ B(10, ½) X₂ ~ B(10, ½) … X₇ ~ B(10, ½)

Think of it like this: X is the "recipe" (the distribution). X₁, X₂, ..., X₇ are seven separate meals cooked from the exact same recipe. They're all made the same way (identical distributions), but that doesn't mean each meal tastes exactly alike (they're still independent, random outcomes). One day you might get 4 heads, the next day 7, the next day 5 — even though every day used the same underlying B(10, ½) rule.

X here is called the population distribution, with population mean μ and population variance σ². The collection X₁, X₂, ..., Xₙ is called a random sample of size n — each Xᵢ is an independent copy of X.

1.2 The sample mean, X̄

Definition

X̄ = (X₁ + X₂ + ... + Xₙ) / n

Plain English: add up all n observations in your sample, then divide by how many there are — exactly like an ordinary average, except here the whole thing is treated as a random variable in its own right.

Crucial idea: X̄ is not a fixed number. Every time you draw a fresh sample of size n, you get a slightly different X̄. Do this 7-day coin experiment once and you might get a sample like 4, 6, 5, 5, 3, 5, 6 → mean 4.857. Do it again next week and you'll get a different set of 7 numbers, and a different mean. If you did this hundreds of times, all those different sample means would themselves form a distribution — and that distribution is what the rest of this chapter is about.

Why this matters Statisticians almost never know the true population mean μ. What they usually have is one sample, and they use its mean X̄ as an estimate of μ. To judge how trustworthy that estimate is, we need to know how X̄ itself is distributed — how spread out sample means tend to be around the true μ.

1.3 The distribution of X̄ when the population IS normal

If your random sample X₁, X₂, ..., Xₙ comes from a normal population, X ~ N(μ, σ²), then the sample mean has an exact distribution too:

Distribution of the sample mean (normal population)

X̄ ~ N( μ, σ²/n )

Same centre as the population (mean = μ), but a squeezed-in spread (variance = σ² divided by n). This holds exactly, for any n, with no approximation needed — because it only works this cleanly when X is already normal.

Mean of X̄ = population mean μ (no shift — averaging doesn't change where it's centred).
Variance of X̄ = population variance ÷ n (bigger samples shrink the spread).
Larger n → σ²/n gets smaller → X̄ clusters tighter around μ → your sample mean becomes a more reliable estimate of the true mean.

The catch This exact result only holds when the population distribution X is itself normal. If X is binomial, Poisson, geometric, or anything else — this formula doesn't strictly apply. That's exactly the gap the Central Limit Theorem fills.

Practice Question 1.1

A population is known to follow X ~ N(40, 36). A random sample of size 9 is taken. State the exact distribution of the sample mean X̄, and calculate P(X̄ > 42).

Practice Question 1.2

Explain, in your own words, why X̄ is described as a "random variable" rather than a single fixed value.

2. The Central Limit Theorem (CLT)

2.1 What does the CLT actually say?

Section 1.3 only worked because we assumed the population was already normal. But real-world populations are often not normal — number of heads in coin flips (binomial), number of customers arriving per hour (Poisson), number of attempts until success (geometric), or something with a completely irregular shape. The Central Limit Theorem is the result that rescues us in every one of these cases.

The Central Limit Theorem

If X₁, X₂, ..., Xₙ is a random sample of n independent observations from any population distribution X, with mean μ and variance σ², then provided n is large:

X̄ ≈ N( μ, σ²/n )

The "≈∼" symbol means "is approximately modelled by." Unlike the exact result for normal populations, this is only an approximation — but it gets better and better as n increases, and it works no matter what shape the original population distribution has.

Think of it like this: Imagine the population distribution is a weirdly-shaped, lumpy, skewed pile of sand — nothing like a neat bell curve. Now scoop up handfuls (samples) of that sand repeatedly, and mark the "average height" of each handful on a chart. Astonishingly, no matter how strange the original sand-pile shape was, that chart of averages will start to look like a smooth, symmetric bell curve once your handfuls (n) are big enough. That's the magic of the CLT — it doesn't care what X looks like, it only cares that n is large.

Case	Distribution of X̄	Exact or approximate?
Population X is normal	N(μ, σ²/n)	Exact, for any n
Population X is anything else (binomial, Poisson, geometric, uniform...)	N(μ, σ²/n)	Approximate, only valid when n is large

2.2 Worked mini-example: Poisson population

Suppose X₁ to X₅₀ is a random sample of size 50 taken from a Po(λ) distribution. We know that for a Poisson distribution, μ = λ and σ² = λ.

Since n = 50 counts as "large," the CLT tells us:

X̄ ≈ N( λ, λ/50 )

Even though the original population is Poisson (discrete, skewed for small λ, definitely not normal-shaped), the sample mean of 50 observations behaves approximately like a normal distribution.

2.3 Finding μ and σ² from the population

Before you can apply the CLT, you must find the population mean μ and population variance σ² of whatever distribution X follows. There are two routes:

Route A — X is a discrete random variable given as a table:

μ = E(X) σ² = Var(X) = E(X²) − [E(X)]²

Route B — X follows a standard named distribution: use the known formulae. For example:

Distribution	μ = E(X)	σ² = Var(X)
B(n, p)	np	np(1−p)
Po(λ)	λ	λ
Geo(p)	1/p	(1−p)/p²

Watch the letter n! A binomial distribution B(n, p) already uses the letter "n" for its number of trials. But your sample size is also usually called n. If you're taking samples from a binomial population, rename the binomial's parameter to avoid confusing yourself — e.g. write X ~ B(m, p), so that σ²/n = mp(1−p)/n stays unambiguous.

2.4 Full worked example (exactly how exam questions are phrased)

Let X represent the value shown when a four-sided spinner is spun:

x	0	2	4	8
P(X = x)	0.5	0.2	0.1	0.2

The spinner is spun 60 times, and the values are recorded. Estimate the probability that the mean of the values recorded is less than 2.8.

Step-by-step reasoning:

Step 1 — Recognise it's a CLT question. We're asked about the probability of a mean, and we have a large sample (n = 60). That combination is the signal: use X̄ ≈ N(μ, σ²/n).

Step 2 — Find μ = E(X):
μ = 0(0.5) + 2(0.2) + 4(0.1) + 8(0.2) = 0 + 0.4 + 0.4 + 1.6 = 2.4

Step 3 — Find σ² = Var(X):
E(X²) = 0²(0.5) + 2²(0.2) + 4²(0.1) + 8²(0.2) = 0 + 0.8 + 1.6 + 12.8 = 15.2
σ² = E(X²) − [E(X)]² = 15.2 − 2.4² = 15.2 − 5.76 = 9.44

Step 4 — Apply the CLT with n = 60:
X̄ ≈ N( 2.4, 9.44/60 ) = N( 2.4, 0.1573... )

Step 5 — Standardise and use normal tables / calculator:
P(X̄ < 2.8) = 0.8433... ≈ 0.843 (3 s.f.)

Practice Question 2.1

A random variable X has population mean μ = 15 and population variance σ² = 40. A random sample of size 80 is taken. Use the CLT to state the approximate distribution of X̄, and find its standard deviation.

Practice Question 2.2

X ~ Geo(0.25). A random sample of size 100 is taken. Find μ, σ², and use the CLT to write down the approximate distribution of X̄.

2.5 How do I know it's a CLT question?

Exam questions rarely say "use the Central Limit Theorem" outright. Instead, look for these giveaway phrases:

"Estimate the probability that the mean number of... is greater than / less than / between..."
"A random sample X₁, ..., X₅₀ is taken..." — especially with a fairly large subscript like 30, 50, 60, 100.
The population is described using a non-normal distribution (binomial, Poisson, geometric, or "the following probability distribution" as a table) — that's your cue that this must be an approximation via CLT rather than an exact normal result.

What to Memorise

Population distribution, X

The underlying random variable being sampled, with population mean μ and population variance σ².

Random sample of size n

X₁, X₂, ..., Xₙ — independent random variables, each with the same distribution as X.

Sample mean, X̄

X̄ = (X₁ + X₂ + ... + Xₙ) / n. Itself a random variable — varies from sample to sample.

Exact result (normal population)

If X ~ N(μ, σ²), then X̄ ~ N(μ, σ²/n) — exact, for any n.

Central Limit Theorem

For ANY population X with mean μ, variance σ², if n is large: X̄ ≈ N(μ, σ²/n).

"≈∼" symbol

Means "is approximately modelled by" — used only in the CLT approximate case, never in the exact normal case.

Binomial B(n,p) formulae

μ = np, σ² = np(1−p). Rename the "n" if it clashes with sample size.

Poisson Po(λ) formulae

μ = λ, σ² = λ.

Geometric Geo(p) formulae

μ = 1/p, σ² = (1−p)/p².

Effect of sample size n

As n increases, σ²/n decreases → X̄ becomes less spread out and a more reliable estimate of μ.

Concepts Checklist

I can explain what a "random sample of size n" means, and why identical distributions don't give identical outcomes.
I can write down the formula for the sample mean X̄ and explain why it's a random variable, not a fixed number.
I know the exact distribution of X̄ when the population is normal: X̄ ~ N(μ, σ²/n).
I can state the Central Limit Theorem precisely, including the condition that n must be large.
I understand the difference between the exact result (~) and the CLT approximation (≈∼).
I can find μ = E(X) and σ² = Var(X) for a discrete distribution given as a probability table.
I know the mean/variance formulae for B(n,p), Po(λ), and Geo(p), and can apply them under the CLT.
I can spot the exam-question phrases that signal a CLT problem ("estimate the probability the mean is...").
I can fully solve a CLT problem: find μ and σ², apply X̄ ≈ N(μ, σ²/n), standardise, and read off a probability.
I remember to rename clashing letters (e.g. binomial's n) so my sample size stays unambiguous.

Exam Tips & Common Mistakes

Examiner Tip Don't forget that n must be large for the Central Limit Theorem to hold! If a question gives a small sample size (say, n = 3 or n = 5) from a non-normal population, the CLT approximation is not valid, and you cannot assume X̄ is approximately normal.

Forgetting to divide the variance by n The most common slip in the whole chapter. Students correctly find σ² but then plug σ² itself into the normal distribution instead of σ²/n. Always double check: the variance used in X̄'s distribution is population variance divided by sample size, never the raw population variance.

Using the CLT when the population is already normal If X is already stated as normal, the result X̄ ~ N(μ, σ²/n) is exact — you don't need "n large" and you shouldn't write the approximation symbol ≈∼. Reserve the CLT language and symbol only for non-normal populations.

Confusing sample size n with a distribution's own parameter n If your population is B(n, p) and you're also taking a sample of size n, you now have two different things both called "n." Rename one (e.g. call the binomial parameter m) before writing your final answer, or you'll create real confusion in your working.

Mixing up E(X²) with [E(X)]² When calculating σ² = Var(X) = E(X²) − [E(X)]² from a probability table, students sometimes square the probabilities instead of the x-values, or forget to subtract [E(X)]² at the end. Always compute E(X) first, then E(X²) separately, then subtract.

Standardising with the wrong standard deviation When converting X̄ to a Z-score, you must use the standard deviation of X̄ itself — that's √(σ²/n), not √σ² (the population's own standard deviation). This is an easy way to lose marks even when every earlier step was correct.

What examiners look for Full marks usually require: (1) correctly identifying μ and σ² of the population, often via a formula for a named distribution, (2) correctly writing X̄ ≈ N(μ, σ²/n) — or the exact ~ version if the population is normal — with the right numbers substituted, (3) standardising using the correct standard deviation, and (4) reading the final probability to the accuracy asked for (commonly 3 significant figures).

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