IB Maths AI HL Random Variables Paper 1 & 2 ~6 min read

Unbiased Estimates

You rarely measure a whole population — you take a sample and use it to estimate the population’s mean and variance. The sample mean does this job perfectly: it’s an unbiased estimate of the population mean. The sample variance sn2 does not — it systematically undershoots, so you scale it up by nn − 1 to get the unbiased estimate sn−12. That one correction is the whole topic.

📘 What you need to know

Estimators & bias

An estimator is a recipe (a random variable) for guessing a population parameter from a sample. Run it on a sample and out comes an estimate. An estimator is unbiased if, on average across all possible samples, it lands exactly on the true parameter.

Sample mean
unbiased
E() = μ. On average it hits the population mean dead on.
Sample variance sₙ²
biased
E(Sn2) = n−1nσ2 < σ2 — it undershoots.

🤔 Why does the sample variance undershoot?

The sample variance measures spread around the sample mean , not the true mean μ. The sample mean sits right in the middle of your data, so the data is always a little closer to than to μ. That makes sn2 a touch too small — and the bigger the sample, the smaller the gap.

The formulae

A sample of n values gives unbiased estimates of both the population mean and variance.

Unbiased estimate of the mean = xn the sample mean — already unbiased ✓
Unbiased estimate of the variance sn−12 = nn − 1 sn2   =   ∑(x)2n − 1 in the formula booklet ✓

🧭 Recipe — finding the unbiased estimates

  1. Mean: divide the data total by n  →  = xn. That’s it — no correction needed.
  2. Variance: if you’re given the sample variance sn2, multiply by nn − 1.
  3. Read the wording first — make sure what you’re given is sn2, not σ2 or sn−12 already.

🧠 Memory aid — “n−1 on the bottom = unbiased”

Whenever the divisor is n − 1 you have the unbiased estimate; a plain n gives the biased one. The nn − 1 multiplier is always bigger than 1, which makes sense — you’re boosting the too-small sample variance up to size.

Which quantity has the question given you?
σ² population variance sₙ² sample variance (biased) sₙ₋₁² unbiased estimate × n/(n−1)Already given σ²? No work. Given sₙ²? Multiply up. Given sₙ₋₁²? You’re done.
Always check the wording to see which of the three you’ve been handed.

Standard deviation & notation

One catch: even though sn−12 is an unbiased estimate of the variance, its square root sn−1 is not an unbiased estimate of the standard deviation. There’s no single formula that fixes this for all populations, so the exam never asks for it — always work with the unbiased variance.

Calculator notation warning: different calculators label the unbiased estimator differently — you may see sn−12, σn−12, s2 or ŝ2 (and their square roots). Know which button on your calculator gives the n − 1 version.
Two proofs (understanding only — not for memorising): using the linear-combination rules, E() = μ+μ+…+μn = μ, so the sample mean is unbiased. And E(Sn2) works out to n−1nσ2σ2, which is exactly why multiplying by nn−1 repairs it.

Worked examples

WE 1

Unbiased mean and variance from a sample

The times, X minutes, of daily revision by 50 IB students are summarised by n = 50, ∑x = 6174, sn2 = 1384.3. Find unbiased estimates of the population mean and variance.

mean: x̄ = ∑x / n = 6174 / 50 = 123.48 x̄ ≈ 123 minutes (3sf) variance: sₙ₋₁² = n/(n−1) × sₙ² = 50/49 × 1384.3 = 1412.55… sₙ₋₁² ≈ 1410 minutes² (3sf) mean needs no correction; variance is scaled up by 50/49.
WE 2

Correcting a sample variance

A sample of n = 20 has sample variance sn2 = 38. Find the unbiased estimate of the population variance.

sₙ₋₁² = n/(n−1) × sₙ² = 20/19 × 38 = 1.0526… × 38 = 40 sₙ₋₁² = 40 the unbiased estimate is always a little larger than sₙ².
WE 3

Estimating from raw data

A sample of 5 values is 4, 7, 8, 10, 11. Find unbiased estimates of the population mean and variance.

mean: x̄ = ∑x / n = (4+7+8+10+11)/5 = 40/5 = 8 x̄ = 8 variance: sₙ₋₁² = ∑(x−x̄)² / (n−1) deviations²: 16, 1, 0, 4, 9 → sum = 30 = 30 / (5−1) = 30/4 = 7.5 sₙ₋₁² = 7.5 from raw data, divide the squared deviations by n−1 directly.
WE 4

Reading the wording carefully

A question states “the population variance is σ2 = 64.” Another states “the sample variance is sn2 = 64 for n = 16.” For each, give the unbiased estimate of the population variance.

case 1: already population variance σ² = 64 → no estimate needed, it’s exact 64 case 2: sample variance → correct it sₙ₋₁² = 16/15 × 64 = 68.27… ≈ 68.3 (3sf) only the SAMPLE variance needs the n/(n−1) boost.
WE 5

Working from ∑x and ∑x²

For a sample of n = 8, ∑x = 96 and ∑x2 = 1228. Find unbiased estimates of the mean and variance.

mean: x̄ = ∑x / n = 96 / 8 = 12 x̄ = 12 sₙ² = ∑x²/n − x̄² = 1228/8 − 12² = 153.5 − 144 = 9.5 sₙ₋₁² = n/(n−1) × sₙ² = 8/7 × 9.5 = 10.857… sₙ₋₁² ≈ 10.9 (3sf) find the biased sₙ² first, then scale it up by 8/7.

💡 Top tips

⚠ Common mistakes

That wraps up the Random Variables unit! You can now transform and combine variables with the linear-combination rules, and use those same rules to produce unbiased estimates of a population’s mean and variance. Next this thinking feeds into hypothesis testing and confidence intervals, where the unbiased variance becomes the engine behind t-tests and interval estimates.

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