Means combine; variances always add: Pack a box: its mass is the box plus the contents, two separate random quantities. What's the mean and spread of the total?
The means just combine: E(X + Y) = E(X) + E(Y), and for a difference E(X − Y) = E(X) − E(Y).
The surprise is the spread. When X and Y are independent, the variances ADD — even for a difference: Var(X + Y) = Var(X) + Var(Y) and Var(X − Y) = Var(X) + Var(Y). Two sources of randomness both add uncertainty, whether you're combining or comparing them, so you never subtract variances.
Don't add the standard deviations: It's the variances that add, not the standard deviations. To combine spreads you must square the SDs first, add, then square-root:
SD(X ± Y) = √(SD(X)² + SD(Y)²).
Adding the SDs directly is the single most common error in these questions. And this whole rule needs the variables to be independent — if they influence each other, it doesn't apply.
IB-style question — total mass of a packed box
An empty box has mass with mean 200 g and standard deviation 6 g. Independently, its contents have mean 850 g and standard deviation 15 g.
Find the mean and standard deviation of the total mass of a packed box.
Step by step
- Means add directly.
- Variances add (square each SD first).
- Standard deviation is the square root.
Final answer
Mean 1050 g, standard deviation ≈ 16.2 g. Notice 16.2 is less than 6 + 15 = 21 — adding the SDs would overstate the spread.
IB-style question — a difference of two times
An old commute time has mean 40 min, SD 5 min; a new route has mean 32 min, SD 4 min, independent of the old.
Find the mean and standard deviation of the time saved, D = old − new.
Step by step
- Mean of the difference subtracts.
- Variances still ADD, even though we subtracted the means.
- Standard deviation.
Final answer
On average the new route saves 8 minutes, with SD ≈ 6.4 min. The spread of the saving is larger than either route's alone because both uncertainties pile in.
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Adding n copies vs scaling one copy: Buy a bag of n separate apples, each with mean μ and variance σ². The bag's total is X₁ + X₂ + … + Xₙ of independent copies, so:
E(total) = nμ and Var(total) = nσ², giving SD = σ√n.
This is not the same as weighing one apple and multiplying by n. There, nX is a single variable scaled up, so Var(nX) = n²σ² (much bigger). Six different apples partly cancel each other's randomness; one apple counted six times doesn't. The variances add (n·σ²) only when the copies are genuinely separate and independent.
IB-style question — a bag of apples
Apples have mass with mean 110 g and standard deviation 8 g, and apples are independent of one another. A bag holds 6 apples.
Find the mean and standard deviation of the total mass of the bag.
Step by step
- Mean of the total: n times the single mean.
- Variance of the total: n times the single variance (NOT n²).
- Standard deviation = σ√n.
Final answer
Mean 660 g, SD ≈ 19.6 g. (Scaling one apple by 6 would wrongly give SD = 6 × 8 = 48 g — far too big.)
Estimating a population from a sample: We rarely know the true population mean μ or variance σ² — we estimate them from a sample. An estimator is unbiased if, averaged over all possible samples, it lands on the true value.
The sample mean x̄ is an unbiased estimate of μ — straightforward.
For the variance there's a catch. Dividing the squared deviations by n underestimates σ² (the data hugs its own mean too closely). Dividing by n − 1 corrects this. So the unbiased estimate is sₙ₋₁² = Σ(x − x̄)² / (n − 1) — the value your GDC labels Sx² (with Sx the unbiased SD).
IB-style question — unbiased estimate from data
A sample of 5 reaction times (seconds) is recorded: 4, 7, 5, 9, 5.
Find an unbiased estimate of the population mean and of the population variance.
Step by step
- Unbiased mean = the sample mean.
- Sum the squared deviations from x̄ = 6.
- Divide by n − 1 = 4 (NOT 5) for the unbiased variance.
Final answer
Unbiased mean estimate = 6 s; unbiased variance estimate = 4 s² (so estimated SD = 2 s). On a GDC this is Sx² — dividing by 5 would give the biased 3.2 and lose marks.