IB Maths AI HLSample Mean DistributionsPaper 1 & 2~6 min read
The Central Limit Theorem
Last topic, X̄ ∼ N(μ, σ2n) only worked because the population X was normal. The Central Limit Theorem (CLT) removes that condition: take a large enough sample from any distribution — even a weird, skewed, or discrete one — and the sample mean X̄ is approximately normal. In the exam, “large enough” means n > 30. That’s the whole idea — and it’s what lets you do normal-distribution probabilities on populations that aren’t normal at all.
📘 What you need to know
CLT: for a large sample from any distribution X, the sample mean X̄ is approximately normal.
Same distribution as before: X̄ ≈ N(μ, σ2n) where μ, σ2 are the population’s mean and variance.
“Large enough” = n > 30 in IB exams.
Population already normal? CLT is not needed — X̄ is exactly normal for anyn.
Population not normal? CLT is needed, and it only works because n is large.
“Approximately” is the key word with CLT — the result is an approximation, not exact.
What the CLT says
Whatever shape the population X has, if you take a big enough random sample and average it, the distribution of those averages is approximately normal, centred on μ with variance σ2n.
Central Limit Theorem
for large n: X̄ ≈ N(μ, σ2n)NOT in the booklet — know the n > 30 rule ✗
🤔 Why does averaging “normalise” a weird distribution?
A single draw can land anywhere, but an average of many draws smooths out the extremes — high and low values cancel. The more values you average, the more the bumps and skew of the original distribution wash away, leaving the familiar bell shape. The bigger n is, the better the normal approximation.
Any population → normal sample mean
The population can be any shape; averaging large samples gives an approximately normal X̄ centred on μ.
When do I actually need it?
This is the part examiners test in the “explain” questions. The deciding factor is only whether the population is normal — not the sample size on its own.
Population is normal
CLT not needed
X̄ is exactly normal for anyn — even small samples.
Population not normal
CLT needed
X̄ is only approximately normal, and only because n > 30.
🧠 Memory aid — “normal population? skip the theorem”
Ask one question: is the original population normal? If yes → X̄ is normal automatically, no CLT. If no → you lean on the CLT, and you must check the sample is large (n > 30). The maths you do is identical either way; only the justification changes.
🧭 Recipe — a CLT probability question
Read off the population mean μ, variance σ2, and sample size n.
Check n > 30 (if the population isn’t normal) — this licenses the approximation.
ModelX̄ ≈ N(μ, σ2n) and find SD = √σ2n for the GDC.
Run the normal CD with the right limits and round (3 sf).
WE 1
Mean of a sample (Susie’s counters)
Counters numbered 1 to 29 are in a bag. One pick has expected value 15 and variance 70. Susie picks 40 integers (replacing each one) and finds their mean. Find the probability that the mean of her 40 numbers is less than 13.
n large (40 > 30) → CLT appliesX̄ ≈ N(15, 70/40) = N(15, 1.75)SD for the GDCσ = √1.75 ≈ 1.32P(X̄ < 13): lower = −999…, upper = 13= 0.065285…P(X̄ < 13) ≈ 0.0653 (3sf)enter μ = 15, σ = √1.75 — use the SD, not the variance 1.75.
WE 2
Was the CLT necessary?
For the counters in WE 1, explain whether it was necessary to use the Central Limit Theorem.
check: is the population normal?the number on a counter is uniform over 1–29, NOT normal.conclusionYes — CLT was necessarybecause the variable for the number picked is not normally distributed, so X̄ is only approximately normal thanks to n > 30.
WE 3
When the CLT is NOT needed
A population is X ∼ N(15, 70). A sample of n = 8 is taken. Explain whether the CLT is needed to model X̄, and state its distribution.
is the population normal?yes — X is already N(15, 70)so X̄ is exactly normal, any nCLT not neededX̄ ∼ N(15, 70/8) = N(15, 8.75)a normal population gives a normal X̄ even with a small sample (n = 8).
WE 4
A “greater than” CLT probability
Using Susie’s counters (μ = 15, σ2 = 70, n = 40), find the probability that the mean of her 40 numbers is greater than 16.
model the sample meanX̄ ≈ N(15, 1.75), σ = √1.75 ≈ 1.32P(X̄ > 16): lower = 16, upper = 999…= 0.224846…P(X̄ > 16) ≈ 0.225 (3sf)“greater than” → lower limit 16, large upper limit on the GDC.
WE 5
A range for the sample mean
For the same counters (μ = 15, σ2 = 70, n = 40), find the probability that the sample mean lies between 14 and 16.
model the sample meanX̄ ≈ N(15, 1.75), σ = √1.75 ≈ 1.32P(14 < X̄ < 16): lower = 14, upper = 16= 0.550308…P(14 < X̄ < 16) ≈ 0.550 (3sf)a range just uses both limits directly in the normal CD.
💡 Top tips
CLT = “any population → normal X̄“ provided n is large (n > 30 in IB).
Same formula as the normal case: X̄ ≈ N(μ, σ2/n).
Decide on the population, not the sample: normal population ⇒ no CLT; non-normal ⇒ CLT.
Enter the SD√(σ2/n) into the GDC, never the variance.
Say “approximately” when justifying a non-normal case — the result isn’t exact.
For “explain” marks, name the population’s distribution (e.g. “not normally distributed”) rather than just “because n is big”.
⚠ Common mistakes
Claiming the CLT was needed when the population is already normal — it isn’t.
Saying CLT is needed “because n > 30″ alone — the reason is the population isn’t normal.
Using the variance as the SD — square-root σ2/n first.
Applying the CLT with a small sample from a non-normal population — the approximation needs n > 30.
Calling the result exact — for non-normal populations X̄ is only approximately normal.
Forgetting to divide the variance by n — X̄ uses σ2/n, not σ2.
Next up — the Confidence Interval for the Mean. A single sample mean x̄ is only a point estimate; a confidence interval gives a range the true mean μ is likely to sit in. The CLT (or a normal population) is exactly what makes those intervals valid, and you’ll choose a z-interval or t-interval depending on whether σ is known.
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