Lesson 1: Sampling Distributions and the Central Limit TheoremAP Deep Dive

The main course built sampling distributions for one statistic at a time. Real exam questions — especially the investigative task — compare two groups, and everything in this lesson is about differences: x̄₁ − x̄₂ and p̂₁ − p̂₂. The deep-dive insight is that differences have their own sampling distributions with a signature structure: means subtract, but variances add. That single sentence, properly understood, generates every formula below.

The sampling distribution of a difference of means

If two independent samples are drawn, x̄₁ − x̄₂ has mean μ₁ − μ₂ and standard deviation √(σ₁²/n₁ + σ₂²/n₂). The shape is approximately normal if both populations are normal, or both sample sizes are large (CLT applied to each). Note what did *not* happen: the standard deviations did not subtract, and they did not average. Variability in either sample adds uncertainty to the difference — intuitively, two noisy measurements make a noisier comparison.

Worked example 1. Two battery brands: brand A lifetimes have μ = 8.2 h, σ = 0.6; brand B, μ = 7.9 h, σ = 0.5. Samples of n₁ = 50 and n₂ = 40 are taken. The difference x̄_A − x̄_B has mean 0.3 h and standard deviation √(0.6²/50 + 0.5²/40) = √(0.0072 + 0.00625) = √0.01345 = 0.116 h. What is the probability that the sample mean for A exceeds that for B by *any* amount — P(x̄_A − x̄_B > 0)? Standardise: z = (0 − 0.3)/0.116 = −2.59, so P ≈ 0.995. Even though individual batteries overlap enormously in lifetime, the *means* are separated by 2.6 standard errors — a clean illustration that sampling distributions of statistics are far tighter than distributions of individuals, which is the conceptual hinge of this whole unit.

Totals versus means: the √n trap

Worked example 2. An elevator serves 16 passengers. Individual weights have μ = 75 kg, σ = 18 kg. The safe load is 1300 kg. What is the probability a full elevator exceeds it?

The question is about a total, not a mean. The total of 16 independent weights has mean 16 × 75 = 1200 kg and standard deviation 18 × √16 = 72 kg — the standard deviation of a *sum* grows by √n, while the standard error of a *mean* shrinks by √n; both come from the same rule, Var(sum) = nσ². Then z = (1300 − 1200)/72 = 1.39, and P(overload) ≈ 0.082. The trap the question is built around: computing the mean's standard error (18/√16 = 4.5) and comparing 1300/16 = 81.25 kg against μ = 75 — which actually gives the same z if done consistently, but the common corruption is mixing the two scalings (mean 1200 with SD 4.5, or mean 75 with SD 72) and producing nonsense. Decide first: *total or mean?* — then both the centre and the spread follow.

Differences of proportions

For p̂₁ − p̂₂: mean p₁ − p₂, standard deviation √(p₁(1−p₁)/n₁ + p₂(1−p₂)/n₂), approximately normal when all four counts n₁p₁, n₁(1−p₁), n₂p₂, n₂(1−p₂) are at least 10.

Worked example 3. Suppose p₁ = 0.60 and p₂ = 0.50, with n₁ = n₂ = 200. SD of the difference = √(0.24/200 + 0.25/200) = √0.00245 = 0.0495. The probability the observed difference exceeds 0.15: z = (0.15 − 0.10)/0.0495 = 1.01, so P ≈ 0.156. Sampling variability of ±0.05 around a true gap of 0.10 is entirely normal at this sample size — the quantitative seed of every "is this difference real?" question in Lesson 2.

Bias, variability, and what n can and cannot fix

A statistic is unbiased if its sampling distribution is centred on the parameter: x̄ and p̂ are unbiased; the sample maximum is (badly) biased for the population maximum. Increasing n shrinks the *spread* of a sampling distribution toward zero but never moves its centre — a biased method with a million observations is precisely wrong. Simulations make this vivid: sample means from a heavily skewed exponential population (μ = 10, σ = 10) with n = 40 cluster normally around 10 with SD ≈ 10/√40 = 1.58. The CLT repaired the *shape* and the *spread*; the centre was never broken — random sampling had guaranteed that from the start. Exam questions ask you to attribute each property to its source: centre from unbiased design, spread from n, shape from CLT or population normality.

Common mistakes at the top band

• Subtracting standard deviations for a difference (or dividing them). Variances add — always, for independent samples, whether you add or subtract the statistics.
• Applying the CLT to individuals: "n = 40, so the weights themselves are normal." The CLT describes the distribution of x̄, full stop.
• Mixing total and mean scalings (Worked Example 2).
• Forgetting the independence condition: without replacement, check the 10% condition for *each* sample; paired data are not two independent samples at all (Lesson 2).
• Treating an observed difference as the parameter: p̂₁ − p̂₂ = 0.10 in a sample is a statistic with a sampling distribution, not a fact about p₁ − p₂.

AP Exam Experience

1. A population has mean 80 and standard deviation 16. The standard deviation of the sampling distribution of x̄ for samples of size 64 is

• (A) 16
• (B) 4
• (C) 2
• (D) 0.25

Answer: (C). σ_x̄ = σ/√n = 16/8 = 2. If the sample size is then quadrupled to 256, the standard deviation halves again to 1 — sample size buys precision at a √n rate.

2. Two independent populations have proportions p₁ = 0.30 and p₂ = 0.45. The mean of the sampling distribution of p̂₁ − p̂₂ is

• (A) 0.15
• (B) −0.15
• (C) 0.375
• (D) it cannot be determined from the information given

Answer: (B). The sampling distribution of a difference is centred on the difference of the parameters, in the order written: 0.30 − 0.45 = −0.15. Sign discipline in "first minus second" is a perennial source of lost points.

3. Which of the following is approximately normally distributed?

• (A) The weights of individual tomatoes from a right-skewed population
• (B) The mean weight of samples of 40 tomatoes from a right-skewed population
• (C) The sample mean of 5 tomatoes from a right-skewed population
• (D) The population of tomato weights after collecting a large sample

Answer: (B). By the Central Limit Theorem, x̄ for n = 40 is approximately normal even though the population is skewed. (A) and (D) describe the population itself, which stays skewed; (C) is too small for the CLT to rescue from a strongly skewed parent.

Free-response question. Two schools compare their mathematics programmes. School A: population mean μ_A = 72, σ_A = 10. School B: μ_B = 68, σ_B = 8. Random samples of 30 students are tested at each school.

(a) Describe the sampling distribution of x̄_A − x̄_B (centre, spread, shape), justifying the shape claim. (b) Calculate the probability that the sample mean for School A exceeds the sample mean for School B by more than 6 points. (c) A commentator says: "A's programme is clearly better — a 4-point gap in population means guarantees A wins any comparison." Using your answer to (b), evaluate the word "guarantees."

Scoring guideline.

• (a) 3 points. 1 point for the centre: μ = 72 − 68 = 4 points. 1 point for the spread: √(10²/30 + 8²/30) = √5.467 = 2.34 points — variances added, not standard deviations (an answer of √(100/30) − √(64/30) or (10+8)/30 earns nothing). 1 point for shape: approximately normal, justified by the Central Limit Theorem with n = 30 from each school (accept "both samples are large"); asserting normality with no justification earns no shape point.
• (b) 2 points. 1 point for standardising: z = (6 − 4)/2.34 = 0.86. 1 point for the probability P(x̄_A − x̄_B > 6) ≈ 0.20 (accept 0.19–0.21). A probability reported without the standardisation earns the second point only if the normal calculation is evident.
• (c) 2 points. 1 point for the quantitative rebuttal: there is roughly a 20% chance the observed gap exceeds 6 — and, symmetrically, a substantial chance (about 4%) the observed gap is negative (z = (0−4)/2.34 = −1.71, P ≈ 0.044) even though B's true mean is lower. 1 point for the conclusion in context: sample comparisons are probabilistic, not guarantees; a true 4-point advantage makes A *likely* to post the higher sample mean (with probability ≈ 0.96), but individual samples can and do reverse. Answers that simply repeat "correlation is not causation" or attack the sampling earn neither point — the rebuttal must use the sampling distribution.

Why variances add — even when the statistics subtract

Students often accept "variances add" as a rule to memorise. It deserves one level deeper, because the rule has a condition attached. For independent random variables, Var(X − Y) = Var(X) + Var(Y): the minus sign is squared away, since Var(−Y) = Var(Y). Intuition: two independent sources of randomness cannot cancel each other *systematically* — each wobble in Y adds uncertainty to X − Y regardless of direction, just as each wobble in X does. But independence is doing real work in that sentence. If X and Y are paired — the same student measured before and after, say — then Var(X − Y) = Var(X) + Var(Y) − 2·Cov(X, Y), and because paired measurements are usually positively correlated (strong students score well both times), the covariance term *subtracts*, shrinking the variance of the difference. That is the deep reason pairing is powerful, and it previews Lesson 2: analysing paired data as if the samples were independent throws away the covariance discount and overstates the standard error. The formula you choose is a claim about how the data were produced — which is why the exam describes the design in such detail and expects you to read it.