IB Maths AI HLSample Mean DistributionsPaper 1 & 2~7 min read
Sample Mean Distributions
Add or scale independent normal variables and the result is still normal — you just track the new mean and variance. The big idea here is the sample meanX̄: take a sample of n values and average them. The averages have the same mean μ as the population but a smaller variance, σ2n. Bigger sample → tighter spread. Once you have the distribution of X̄, every probability is just a normal-distribution GDC calculation.
📘 What you need to know
Linear combination: X = a1X1 + … + anXn + b of independent normals is itself normal.
Mean: E(X) = a1μ1 + … + anμn + b — true whether or not independent.
Coefficients get squared in the variance, and a constant b shifts the mean but not the variance.
Sample mean: X̄ = X1+…+Xnn is a random variable; a single value x̄ is a point estimate.
If X ∼ N(μ, σ2) then X̄ ∼ N(μ, σ2n) — same mean, variance ÷ n.
Combining normal variables
Take independent normals X1, …, Xn and build a linear combination X = a1X1 + … + anXn + b. The result is also normal — so you only need its mean and variance to do probabilities.
Distribution of a linear combinationX ∼ N(a1μ1 + … + anμn + b, a12σ12 + … + an2σn2)E(X) & Var(X) rules are in the booklet ✓but “a sum of normals is normal” is NOT — know it ✗
The mean
always adds
E(X) works with or without independence — coefficients stay as they are.
The variance
needs independence
Var(X) only follows the rule if the variables are independent; coefficients are squared.
🧠 Memory aid — “square the coefficient, drop the constant”
When a variable is multiplied by a, its variance contribution is a2σ2 — the coefficient is squared. A standalone constant b moves the mean across but adds nothing to the variance (shifting data doesn’t change its spread).
🤔 Why is “3 presses” not the same as “3×W”?
Pressing a button three times gives three separate, independent draws W1 + W2 + W3, so the variance is σ2 + σ2 + σ2 = 3σ2. Multiplying one draw by 3 gives 3W, whose variance is 32σ2 = 9σ2. Same mean, very different spread — read the question carefully!
WE 1
Combining normals (Amber’s tea — distribution)
A vending machine dispenses water W ∼ N(100, 152) and milk M ∼ N(10, 22). Amber presses water 3 times and milk 2 times. The total is C = W1+W2+W3+M1+M2. Write down the distribution of C.
mean: add all fiveμ = 100+100+100+10+10 = 320variance: add all five (independent)σ² = 15²+15²+15²+2²+2² = 683a linear combination of normals is also normalC ∼ N(320, 683)
WE 2
A probability from the combination
Using C ∼ N(320, 683) from WE 1, find the probability that the total amount of liquid exceeds 360 ml.
set up the normal on the GDCμ = 320, σ = √683 ≈ 26.13P(C > 360): lower = 360, upper = 9999…P(C > 360) = 0.062939…P(C > 360) ≈ 0.0629 (3sf)use √683 (not 683) as σ in the calculator.
The sample mean distribution
Take a sample of n observations from a population X and average them: X̄ = X1+…+Xnn. Different samples give different averages, so X̄ is itself a random variable — its distribution is the sample mean distribution. A single calculated average x̄ is one point estimate drawn from it.
Sample mean of a normal populationX ∼ N(μ, σ2) ⇒ X̄ ∼ N(μ, σ2n)NOT in the booklet — memorise it ✗
Where the variance comes from: E(X̄) = nμn = μ, and Var(X̄) = nσ2n2 = σ2n. Dividing a sum by n squares to dividing the variance by n2, but there are n terms — so the net effect is ÷ n.
More data → tighter sample mean
All three centre on μ; larger samples have variance σ²/n, so the curve gets narrower and taller.
🧭 Recipe — probability about a sample mean
Write the populationX ∼ N(μ, σ2) and read off μ, σ2, n.
Form the sample mean: X̄ ∼ N(μ, σ2n).
Convert variance to SD: enter σ = √σ2n into the GDC, not the variance.
Run the normal CD with the right lower/upper limits and round (3 sf).
WE 3
Distribution of a sample mean (Amber’s 15 cups)
Each cup of tea has C ∼ N(320, 683). Amber makes 15 cups and finds the mean C̄. Write down the distribution of C̄.
mean stays the sameμ = 320variance ÷ nσ²/n = 683/15 = 45.533…C̄ ∼ N(320, 683/15)keep 683/15 exact; for any probability later, GDC SD = √(683/15) ≈ 6.75.
WE 4
A probability for the sample mean
A population is X ∼ N(50, 82). A sample of n = 16 is taken. Find P(X̄ > 53).
form the sample mean distributionX̄ ∼ N(50, 8²/16) = N(50, 4)SD for the GDCσ = √4 = 2P(X̄ > 53): lower = 53, upper = 9999…= 0.066807…P(X̄ > 53) ≈ 0.0668 (3sf)the sample mean SD (2) is 4× smaller than the population SD (8).
WE 5
Single value vs sample mean — the contrast
For X ∼ N(50, 82), compare P(X > 53) for a single observation with P(X̄ > 53) for a sample of 16.
single value: σ = 8P(X > 53) = 0.354…≈ 0.354 (3sf)sample mean: σ = 8/√16 = 2P(X̄ > 53) = 0.0668…≈ 0.0668 (3sf)an average of 16 is much less likely to stray 3 above μ than a single value.
💡 Top tips
Sum of independents: add the variances (don’t multiply by n unless every term is identical).
Square the coefficients in any variance — a constant b never enters the variance.
Sample mean: keep the mean, divide the variance by n.
GDC wants SD, not variance — enter √(σ2/n).
“3 presses” ≠ “3×W”: three draws give 3σ2; one tripled draw gives 9σ2.
Keep full accuracy through the working; round only the final answer (usually 3 sf).
⚠ Common mistakes
Entering the variance as the SD on the GDC — always square-root σ2/n first.
Forgetting to square coefficients in Var(X) — it’s a2σ2, not aσ2.
Adding the constant b to the variance — it only shifts the mean.
Dividing variance by n2 for the sample mean — it’s ÷ n, not ÷ n2.
Confusing X̄ with X — a sample mean is far less spread out than a single value.
Using the variance rule when variables aren’t independent — the mean rule still holds, the variance rule doesn’t.
Next up — the Central Limit Theorem. So far we needed the population to be normal for X̄ to be normal. The CLT lifts that restriction: for a large enough sample (n > 30), the sample mean is approximately normal no matter what shape the population has — opening the door to confidence intervals for the mean.
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