AP Statistics Sampling Distribution of the Sample Mean Practice: Central Limit Theorem
The sampling distribution of x̄ is centered at μ with standard deviation σ/√n, whatever the population's shape. As n grows, the spread shrinks and, by the Central Limit Theorem, the shape becomes approximately normal. The population itself never becomes normal — only the distribution of sample means does.
Try the question
The distribution of individual household incomes in a large city is strongly right-skewed. A researcher plans to take a random sample of n households and compute the sample mean income, x̄.
As the sample size n increases from 10 to 100, what happens to the shape and variability of the sampling distribution of x̄?
- A. The shape becomes more skewed and the variability increases, matching the population more closely as n grows.
- B. The shape stays exactly as skewed as the population, but the variability decreases.
- C. The shape becomes more normal and the variability (standard error) decreases.
- D. The shape becomes more normal, but the variability stays the same regardless of n.
Answer key: why each choice is right or wrong
✕ A — Incorrect
This reverses both effects of the Central Limit Theorem. Larger samples make the sampling distribution of the mean look less like the skewed population, not more.
✕ B — Incorrect
This ignores the Central Limit Theorem's effect on shape. As n increases, the sampling distribution of x̄ becomes more normal, not equally skewed.
✓ C — Correct
As n increases, the sampling distribution of x̄ becomes more approximately normal regardless of population shape, and its standard deviation σ/√n decreases.
✕ D — Incorrect
This describes the shape change correctly but claims the spread is unaffected by sample size. σ/√n decreases as n increases.
The core idea
x̄ is a random variable with its own distribution across samples, not a fixed number. Its center and spread hold regardless of shape; only the shape claim depends on whether the population is normal or n is large.
| Claim | What to write | Depends on population shape? |
|---|---|---|
| Center | μ_x̄ = μ | No |
| Spread | σ_x̄ = σ/√n | No |
| Shape | Normal population → exactly normal; otherwise n ≥ 30 → approximately normal | Yes |
The Central Limit Theorem is a claim about shape only, and it depends on n. A skewed population needs a larger n before x̄ looks normal; a normal population needs no minimum n at all.
How points are earned and lost
Earned: a rubric typically credits center, spread and shape separately. State μx̄ = μ and σx̄ = σ/√n with the right symbols, then justify shape as its own claim tied to n or to the population's known shape.
Lost: treating “n is large” as making the population itself normal, rather than making the sampling distribution of x̄ approximately normal.
The answer move
Write three separate lines before you calculate anything: center (μx̄ = μ), spread (σx̄ = σ/√n), and shape (normal population → exact; otherwise n ≥ 30 → approximate).
Weak answer vs. point-earning answer
A shipping company's package weights are strongly right-skewed with μ = 8 lb. A sample of n = 25 packages is weighed. Is it valid to treat x̄ as approximately normal?
✕ Weak answer
Yes, because n is a decent-sized sample so the average should be normal.
✓ Point-earning answer
No. Because the population is strongly skewed and n = 25 is below the n ≥ 30 guideline used for skewed populations, the sampling distribution of x̄ cannot be assumed approximately normal here; a larger sample would be needed.
Before your next sample-mean question
Take a problem that gives you a population shape and a sample size, and write center, spread and shape as three separate lines first.
Review the topic behind this question: Topic 4.1: Sampling distributions for sample means. More practice: Unbiased estimator practice, Two-way table practice and AP Statistics practice questions.
Frequently asked questions
What does the Central Limit Theorem say?
Does a large sample make the population normal?
What is the standard deviation of x̄?
When can I treat x̄ as normal?
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