Lesson 16: Sampling Distributions — Means

(a) Warm-Up

In Lesson 15 you watched what happens to a sample proportion p̂ as you take sample after sample: it bounces around the true proportion p, with predictable spread. Now we swap categorical for quantitative. Instead of "What fraction of the sample voted yes?" we ask "What is the average value in the sample?"

Picture a city where the typical home uses 24 miles per gallon... no wait, let's make it cleaner. A coffee chain says the typical wait at the drive-through averages 8.5 minutes. You time 50 customers and get a sample mean of x̄ = 9.1 minutes. Is that surprising, or just the ordinary wobble you'd expect from one sample of 50?

You can't answer that until you know how x̄ itself behaves across many samples — its center, its spread, and its shape. That pattern is the sampling distribution of the sample mean, and it is the engine behind every confidence interval and test for a mean you'll meet in Unit 4. Here's the headline you'll earn by the end of this lesson: even when the population is wildly skewed, the distribution of x̄ becomes beautifully Normal once n is large enough. That's the Central Limit Theorem, and it's one of the most important ideas in all of statistics.

(b) Core Concept

From one sample to a distribution of means

Imagine a population with mean μ (mu) and standard deviation σ (sigma). Take a random sample of size n and compute its mean x̄. Take another sample — you get a slightly different x̄. Do this over and over, and the collection of all possible x̄ values forms the sampling distribution of the sample mean. Three questions describe it: Where is it centered? How spread out is it? What shape does it have?

Center: x̄ is unbiased

The mean of the sampling distribution of x̄ equals the population mean:

μ_x̄ = μ

This says the sample mean is an unbiased estimator of μ. Across many samples, x̄ neither systematically overshoots nor undershoots the truth — it is correct on average. There's no "n" anywhere in this formula. A sample of 10 and a sample of 10,000 both center on μ. What changes with n is not where the distribution sits, but how tightly it clusters.

Spread: bigger samples, smaller wobble

The standard deviation of the sampling distribution of x̄ is:

σ_x̄ = σ / √n

This is sometimes called the standard error of the mean. Read it carefully: dividing by √n means larger samples produce less variable means. Quadruple the sample size and you halve the spread (because √4 = 2). This matches intuition — averaging more observations smooths out the extremes, so x̄ lands closer to μ.

The 10% condition. The formula σ / √n assumes the observations are independent. When you sample without replacement from a finite population, that's only approximately true. The fix: require the sample to be no more than 10% of the population, n ≤ 0.10N. As long as you're not scooping up a huge fraction of the population, σ / √n is a valid approximation.

Shape: the Central Limit Theorem

Shape is the subtle one, and there are exactly two ways to justify that x̄ is Normal:

1. The population is Normal. If the population itself is Normally distributed, then x̄ is exactly Normal for any sample size n — even n = 2.

1. The Central Limit Theorem (CLT). If the population is not Normal (skewed, bimodal, whatever), then x̄ is still approximately Normal when n is large. The common rule of thumb on the AP exam is n ≥ 30.

This second case is the magic. The shape of the population can be a mess, but the shape of the distribution of sample means heals toward Normal as n grows. You must state which justification you're using — this is a frequent place to lose points.

[GRAPH: Two panels side by side. LEFT panel titled "Population (n = 1)": a strongly right-skewed distribution, tall bar near 0 and a long tail stretching right, mean μ marked. X-axis "Value", a lump on the left with a long right tail. RIGHT panel titled "Sampling distribution of x̄ (n = 40)": a smooth, symmetric, bell-shaped Normal curve centered at the SAME mean μ but much narrower, centered tick at μ. Caption: "The population is skewed, yet the distribution of sample means is nearly Normal and far less spread out — the Central Limit Theorem in action."]

Putting it together: a fully worked probability

A car model's highway fuel economy has population mean μ = 24 mpg with σ = 6 mpg. The population of individual cars is right-skewed. A reviewer tests a random sample of n = 40 cars. What is the probability the sample mean exceeds 26 mpg, P(x̄ > 26)?

Step 1 — Center. μ_x̄ = μ = 24 mpg.

Step 2 — Spread. σ_x̄ = σ / √n = 6 / √40 = 6 / 6.3246 = 0.9487 mpg.

(10% condition: 40 cars is well under 10% of all cars of this model, so σ_x̄ is valid.)

Step 3 — Shape. The population is skewed, so we lean on the CLT. Since n = 40 ≥ 30, the sampling distribution of x̄ is approximately Normal. (We must say this — skewed population alone would NOT let us use Normal calculations for a single car.)

Step 4 — z-score.

z = (x̄ - μ_x̄) / σ_x̄ = (26 - 24) / 0.9487 = 2.11

Step 5 — Probability. Using the z-score with a Normal table, P(z > 2.11) ≈ 0.0175.

TI-84:

normalcdf(lower: 26, upper: 1E99, μ: 24, σ: 0.9487)
Output: 0.0175

(Or, in z-units: normalcdf(2.11, 1E99, 0, 1) ≈ 0.0174. Tiny rounding difference — entering μ_x̄ and σ_x̄ directly avoids rounding the z-score.)

Interpretation. About a 1.75% chance a random sample of 40 cars averages above 26 mpg. That's unusual — if the brochure claims μ = 24 and you keep seeing sample means this high, you'd start to doubt the claim.

Contrast with the proportion case (L15)

Same three questions, different formulas. Keep them straight:

For proportions there's no "the population was already Normal" shortcut — categorical data isn't Normal. For means, that shortcut exists. And the shape rule is about counts (np, n(1−p)) for p̂ but about sample size and population shape for x̄.

(c) Worked Examples

Example 1 (easy) — population already Normal

Battery lifetimes for a phone model are Normally distributed with μ = 18 hours and σ = 4 hours. A store samples n = 25 phones. Find P(x̄ < 17).

Strategy. Identify center, spread, shape; then z-score.

Solution.

• Center: μ_x̄ = 18.
• Spread: σ_x̄ = 4 / √25 = 4 / 5 = 0.8.
• Shape: the population is Normal, so x̄ is exactly Normal for any n — we don't even need the CLT or n ≥ 30 here.
• z-score: z = (17 − 18) / 0.8 = −1.25.
• P(z < −1.25) ≈ 0.1056.

TI-84: normalcdf(−1E99, 17, 18, 0.8) = 0.1056.

Interpretation. About a 10.6% chance the average of 25 phones falls below 17 hours.

Example 2 (medium) — a range, justify with population shape

Adult female heights in a country are approximately Normal with μ = 64 inches and σ = 2.7 inches. A random sample of n = 36 women is selected. Find the probability the sample mean is between 63 and 65 inches, P(63 < x̄ < 65).

Strategy. Two z-scores, subtract.

Solution.

• μ_x̄ = 64; σ_x̄ = 2.7 / √36 = 2.7 / 6 = 0.45.
• Shape: population is approximately Normal, so x̄ is Normal. (n = 36 also satisfies the CLT, but the population-Normal justification is cleaner.)
• z₁ = (63 − 64) / 0.45 = −2.22; z₂ = (65 − 64) / 0.45 = +2.22.
• P(−2.22 < z < 2.22) ≈ 0.9737.

TI-84: normalcdf(63, 65, 64, 0.45) = 0.9737.

Interpretation. About 97.4% of samples of 36 women have a mean height within one inch of 64. Notice how tight that is — because σ_x̄ = 0.45 is small, sample means cluster hard around μ.

Example 3 (AP-style) — the effect of n on spread

A factory fills bags with a target mean weight μ = 50 grams and σ = 15 grams. The fill weights are right-skewed (occasional heavy overfills).

(a) For a sample of n = 25, can you find P(x̄ > 56)?

(b) For a sample of n = 100, find P(x̄ > 53), and comment on how n changed the spread.

Solution.

(a) Shape check first: population is skewed and n = 25 < 30. The CLT does not apply, the population isn't Normal, so we cannot model x̄ as Normal and cannot compute this probability with a Normal calculation. (Stating this is the answer — recognizing when you can't use the tool is an AP skill.)

(b) Now n = 100 ≥ 30, so by the CLT x̄ is approximately Normal.

• μ_x̄ = 50; σ_x̄ = 15 / √100 = 15 / 10 = 1.5.
• z = (53 − 50) / 1.5 = 2.00.
• P(z > 2.00) ≈ 0.0228. TI-84: normalcdf(53, 1E99, 50, 1.5) = 0.0228.

Effect of n. With n = 25, σ_x̄ = 15/√25 = 3.0; with n = 100, σ_x̄ = 15/√100 = 1.5. Multiplying the sample size by 4 cut the spread in half (because √4 = 2). Larger samples give means that hug μ more tightly — and push the shape toward Normal.

(d) Common Mistakes

1. Using σ instead of σ/√n. The single most common error. A question about one individual uses σ; a question about a sample mean uses σ_x̄ = σ/√n. If the problem says "sample of n =" and asks about x̄, you must divide by √n. Forgetting this makes your spread far too big and your probabilities wrong.

2. Invoking the CLT when n is small and the population isn't Normal. The CLT needs large n (rule of thumb n ≥ 30). If the population is skewed and n = 12, you cannot claim x̄ is Normal. The correct response is often "we can't do this calculation," not a number.

3. Forgetting to justify the Normal shape. You must state why x̄ is Normal: either "the population is Normal, so x̄ is Normal for any n," or "n ≥ 30, so by the CLT x̄ is approximately Normal." No justification = lost points, even if your arithmetic is perfect.

4. Confusing the distribution of x̄ with the population. "P(a single car gets > 26 mpg)" uses σ = 6. "P(the mean of 40 cars > 26 mpg)" uses σ_x̄ = 0.9487. The sample-mean distribution is much narrower. Read whether the question asks about one observation or about an average.

5. Treating x̄ as Normal because the sample looks Normal. The shape condition is about the population shape (and n), not the histogram of your one sample. Don't justify normality from a sample dotplot when the population is known to be skewed.

(e) Practice Problems

(in context) A bakery's loaf weights are Normal with μ = 800 g, σ = 25 g. A random sample of n = 16 loaves is weighed. Find P(x̄ > 810) and state your shape justification.

(in context) Daily app-session lengths for a social platform are right-skewed with μ = 28 minutes and σ = 12 minutes. A random sample of n = 50 users is drawn. Find P(x̄ < 25) and justify the shape.

True or false: If the population is Normal, the sampling distribution of x̄ is Normal even for n = 3. Explain.

A delivery service's package weights have μ = 4.2 lb and σ = 1.8 lb (right-skewed). For a sample of n = 9, a student computes P(x̄ > 4.8). What's wrong with attempting this calculation?

(in context) A coffee machine dispenses cups with mean μ = 240 mL and σ = 8 mL; output is Normal. For n = 25 cups, find the value c such that P(x̄ < c) = 0.975. (Hint: z = 1.96.)

(f) FRQ Practice (10 points)

Statistical Practice: 3 — Analyze Data

A regional supermarket chain states that the time customers spend at self-checkout has a population mean of μ = 8.5 minutes with population standard deviation σ = 3.5 minutes. The distribution of individual checkout times is strongly right-skewed (most customers are quick, but a few take a long time). A store manager records the checkout times of a random sample of n = 50 customers during one week.

(a) Describe the shape, center, and spread of the sampling distribution of the sample mean checkout time x̄. Justify the shape, and verify any condition needed for the spread. (4 points)

(b) Find the probability that the sample mean checkout time exceeds 9 minutes, P(x̄ > 9). Show your work. (4 points)

(c) Interpret your probability from part (b) in context. (2 points)

Model Response

(a)

- Center: μ_x̄ = μ = 8.5 minutes. The sample mean is an unbiased estimator of the population mean.

- Spread: σ_x̄ = σ / √n = 3.5 / √50 = 3.5 / 7.0711 = 0.495 minutes. This requires the 10% condition: the 50 sampled customers are clearly less than 10% of all customers the store serves, so the observations are approximately independent and σ/√n is valid.

- Shape: The population is strongly right-skewed, so we rely on the Central Limit Theorem. Because n = 50 ≥ 30, the sampling distribution of x̄ is approximately Normal.

(b) Standardize using μ_x̄ = 8.5 and σ_x̄ = 0.495:

z = (9 − 8.5) / 0.495 = 0.5 / 0.495 = 1.01

P(x̄ > 9) = P(z > 1.01) ≈ 0.1562

TI-84: normalcdf(9, 1E99, 8.5, 0.495) = 0.1562.

(c) If we repeatedly took random samples of 50 customers, about 15.6% of those samples would have a mean checkout time greater than 9 minutes. A sample mean above 9 minutes is therefore not unusual; it is consistent with the chain's claimed mean of 8.5 minutes.

Scoring Rubric (10 points)

Part (a) — 4 points

• 1 pt — Center: μ_x̄ = 8.5 (with correct symbol/idea that it equals μ).
• 1 pt — Spread: correct formula and value σ_x̄ = 3.5/√50 = 0.495.
• 1 pt — Shape stated as approximately Normal with CLT justification referencing n ≥ 30 (because the population is skewed).
• 1 pt — 10% condition checked (50 ≤ 10% of all customers) to justify the spread formula.

Part (b) — 4 points

• 1 pt — Correct z-score using σ_x̄ = 0.495 (NOT σ = 3.5): z ≈ 1.01.
• 1 pt — Correct probability ≈ 0.1562 (accept 0.155–0.157).
• 1 pt — Work shown: either the z-calculation or a correct normalcdf call with μ_x̄ and σ_x̄.
• 1 pt — Probability is set up as an upper-tail area P(x̄ > 9) (correct direction).

Part (c) — 2 points

• 1 pt — Interpretation framed as a long-run proportion of samples (not individuals): "about 15.6% of samples of 50 would have a mean above 9 minutes."
• 1 pt — Connected to context / the chain's claim (not unusual, consistent with μ = 8.5).

Where students lose points:

• Using σ = 3.5 instead of σ_x̄ = 0.495 in the z-score (loses the z and likely the probability point).
• Saying "the population is Normal" — it is skewed; the only valid shape justification here is the CLT with n ≥ 30.
• Omitting the 10% condition entirely (automatic −1 in part a).
• Interpreting part (c) as "the probability one customer takes more than 9 minutes" — that confuses the individual with the sample mean.
• A "naked" probability in (b) with no shown work (loses the work point even if the number is right).