Lesson 18: Probability with Sampling Distributions

(a) Warm-Up

Three lessons, three pieces of one machine. In Lesson 15 you learned the sampling distribution of a proportion p̂. In Lesson 16 you learned the sampling distribution of a mean x̄. In Lesson 17 you saw why these distributions are unbiased and how sample size sharpens them. This lesson bolts the pieces together.

Here's the scene the AP exam keeps setting. A problem appears. It does not announce "this is a proportion question" or "use σ/√n." You have to read the wording, decide which statistic is in play, and pick the matching formulas — all before you touch the calculator. Then comes the sneakiest fork of all: is the question about one individual (a single car, one customer, a lone battery) or about a sample statistic (the average of 40 cars, the proportion in a sample of 200)? Those use different standard deviations, and confusing them is one of the most common ways students lose points on this exact topic.

By the end of this lesson you'll have a two-question decision guide that sorts any of these problems in about ten seconds. Let's build it.

(b) Core Concept

The two-question decision guide

Before any arithmetic, ask two questions in order.

Question 1 — Proportion or mean?

Is the variable categorical (each individual is a yes/no, success/failure, defective/not) → you're working with a proportion p̂.

Is the variable quantitative (each individual has a number — weight, time, score) → you're working with a mean x̄.

Question 2 — Individual or sample statistic?

Does the question ask about one randomly chosen individual → use the population σ (no √n, no sampling distribution).

Does it ask about a statistic computed from a sample of size n (an average, a sample proportion) → use the sampling-distribution standard deviation.

Question 1 tells you which formulas to grab. Question 2 tells you which standard deviation goes in the z-score. Get both right and the rest is mechanical.

The full reference table

Everything you need from L15–L17 in one place. Memorize the structure, not just the entries.

Both columns share the 10% condition — it's the price of sampling without replacement, and it justifies the spread formula in both cases. What differs is the shape rule: for p̂ it's about counts (np and n(1−p)); for x̄ it's about sample size and population shape. And note the asymmetry: a mean has a "population was already Normal" shortcut; a proportion never does (categorical data isn't Normal).

The individual-vs-statistic fork, made explicit

This is the heart of the lesson, so let's say it slowly. Suppose individual values have population mean μ and population standard deviation σ.

• Probability about ONE individual: standardize with the population σ.

z = (x − μ) / σ

• Probability about a SAMPLE MEAN of n: standardize with the sampling-distribution σ.

z = (x̄ − μ) / (σ/√n)

Same μ in the center, but the second denominator is smaller (divided by √n), so a sample mean has to travel fewer standard deviations to reach the same value. That makes the probabilities for a sample mean far more extreme. A value that's perfectly ordinary for one individual can be wildly unusual for an average of 40. Read the question stem like a hawk: "a randomly selected car" is an individual; "the mean of 40 cars" is a statistic.

Worked probability #1 — a proportion (p̂)

Problem. A manufacturer claims 40% of customers who try a free sample go on to buy the product, so p = 0.40. A marketer surveys an SRS of n = 200 customers who tried the sample. What is the probability that fewer than 35% of the sample buy, P(p̂ < 0.35)?

Decision guide. "Fewer than 35% of the sample buy" — buy/don't buy is categorical, and we're asked about a sample proportion. → proportion p̂, sample statistic.

Step 1 — Conditions.

• 10% condition: the manufacturer's customer base far exceeds 10 × 200 = 2000, so n ≤ 0.10N. ✓ Spread formula valid.
• Large Counts: np = 200(0.40) = 80 ≥ 10 and n(1−p) = 200(0.60) = 120 ≥ 10. ✓ Shape approximately Normal.

Step 2 — Center and spread.

μ_p̂ = p = 0.40
σ_p̂ = √( p(1−p)/n ) = √( (0.40)(0.60)/200 ) = √( 0.24/200 ) = √0.0012 = 0.034641

Step 3 — z-score.

z = (p̂ − μ_p̂) / σ_p̂ = (0.35 − 0.40) / 0.034641 = −0.05 / 0.034641 = −1.44

Step 4 — Probability. Lower-tail area below z = −1.44.

TI-84: normalcdf(lower: −1E99, upper: 0.35, μ: 0.40, σ: 0.034641)
Output: 0.0745
(Equivalently normalcdf(−1E99, −1.44, 0, 1) ≈ 0.0749 — small rounding difference from rounding z.)

Interpretation. About a 7.5% chance a random sample of 200 customers has fewer than 35% buyers. Uncommon but not shocking.

Worked probability #2 — a mean (x̄)

Problem. A coffee shop's espresso shots have population mean weight μ = 64 grams with σ = 8 grams, and the individual weights are Normally distributed. A barista pulls a random sample of n = 36 shots. What is the probability the sample mean weight exceeds 66 grams, P(x̄ > 66)?

Decision guide. Weight is quantitative, and we want the average of 36 shots. → mean x̄, sample statistic.

Step 1 — Center. μ_x̄ = μ = 64 g.

Step 2 — Spread. σ_x̄ = σ/√n = 8/√36 = 8/6 = 1.3333 g.

(10% condition: 36 shots is well under 10% of all shots the shop pulls, so σ/√n is valid.)

Step 3 — Shape. The population is Normal, so x̄ is exactly Normal for any n — no CLT needed.

Step 4 — z-score.

z = (x̄ − μ_x̄) / σ_x̄ = (66 − 64) / 1.3333 = 2 / 1.3333 = 1.50

Step 5 — Probability. Upper-tail area above z = 1.50.

TI-84: normalcdf(lower: 66, upper: 1E99, μ: 64, σ: 1.3333)
Output: 0.0668

Interpretation. About a 6.7% chance the mean of 36 shots tops 66 grams.

Hold that thought — the individual contrast

Same espresso context, but now: what is the probability one randomly chosen shot weighs more than 66 grams? That's an individual, so use σ = 8, not σ/√n:

z = (66 − 64) / 8 = 0.25  →  P(x > 66) = normalcdf(66, 1E99, 64, 8) = 0.4013

Look at the gap: 40.1% for one shot, but only 6.7% for the mean of 36. A single shot over 66 g is routine; an average of 36 over 66 g is rare, because averaging squeezes the spread to σ/√n. Same number, same center — different question, different σ, wildly different answer. That contrast is exactly what the AP exam is testing.

(c) Worked Examples

Example 1 (easy) — a proportion

A polling firm believes 25% of teens use a certain app daily, p = 0.25. They survey an SRS of n = 120 teens. Find P(p̂ > 0.30).

Decision guide. Uses-app/doesn't is categorical; we want a sample proportion. → p̂.

Solution.

• Conditions: 10% — teen population ≫ 1200 ✓; Large Counts — np = 120(0.25) = 30 ≥ 10, n(1−p) = 120(0.75) = 90 ≥ 10 ✓. Shape ≈ Normal.
• μ_p̂ = 0.25; σ_p̂ = √(0.25·0.75/120) = √(0.001563) = 0.039528.
• z = (0.30 − 0.25)/0.039528 = 1.27.
• P(p̂ > 0.30) = P(z > 1.27) ≈ 0.103.

TI-84: normalcdf(0.30, 1E99, 0.25, 0.039528) = 0.103.

Interpretation. About a 10.3% chance more than 30% of 120 teens use the app daily.

Example 2 (medium) — a mean with a skewed population

Time spent on a checkout page is right-skewed with population mean μ = 500 seconds and σ = 100 seconds. A site samples n = 50 sessions. Find P(480 < x̄ < 520).

Decision guide. Time is quantitative; we want the average of 50. → x̄.

Solution.

• Shape first: the population is skewed, but n = 50 ≥ 30, so by the CLT x̄ is approximately Normal. (Without the CLT we'd be stuck — a skewed population alone would not justify Normal.)
• μ_x̄ = 500; σ_x̄ = 100/√50 = 14.1421. (10% condition: 50 ≪ 10% of all sessions ✓.)
• z₁ = (480 − 500)/14.1421 = −1.41; z₂ = (520 − 500)/14.1421 = +1.41.
• P(−1.41 < z < 1.41) ≈ 0.8427.

TI-84: normalcdf(480, 520, 500, 14.1421) = 0.8427.

Interpretation. About 84% of samples of 50 sessions have a mean within 20 seconds of 500.

Example 3 (AP-style) — individual vs. sample mean on one context

Resting heart rates for a population of adults are Normal with μ = 70 bpm and σ = 12 bpm.

(a) Find the probability one randomly selected adult has a resting heart rate above 74 bpm.

(b) Find the probability the mean of a random sample of n = 36 adults is above 74 bpm.

(c) Explain why the two answers differ so much.

Solution.

(a) Individual → use σ = 12. z = (74 − 70)/12 = 0.33. P(x > 74) = normalcdf(74, 1E99, 70, 12) = 0.3694. About 36.9%.

(b) Sample mean → use σ_x̄ = 12/√36 = 2. The population is Normal, so x̄ is Normal for any n. z = (74 − 70)/2 = 2.00. P(x̄ > 74) = normalcdf(74, 1E99, 70, 2) = 0.0228. About 2.3%.

(c) The center is the same (70 bpm), but the standard deviation collapses from σ = 12 for one person to σ_x̄ = 2 for an average of 36. The same value, 74, sits only 0.33 SD above the mean for an individual but a full 2.00 SD above for the sample mean. Averaging cancels out high and low values, so sample means cluster tightly around μ — a mean of 36 people above 74 is genuinely unusual, while one person above 74 is common.

Example 4 (AP-style) — pick the distribution from the wording

A school says 18% of its students walk to school, p = 0.18. Separately, the time students take to get to school is right-skewed with μ = 22 minutes, σ = 9 minutes. For each part, name the sampling distribution and find the probability.

(a) In an SRS of n = 250 students, find P(p̂ > 0.18 + 0.03) = P(p̂ > 0.21).

(b) In an SRS of n = 81 students, find P(x̄ < 21) for the commute time.

Solution.

(a) "Walk / don't walk" is categorical → proportion p̂. 10% condition ✓ (school ≫ 2500... if the school is large; assume so). Large Counts: np = 250(0.18) = 45 ≥ 10, n(1−p) = 250(0.82) = 205 ≥ 10 ✓. σ_p̂ = √(0.18·0.82/250) = √(0.0005904) = 0.024298. z = (0.21 − 0.18)/0.024298 = 1.23. P(p̂ > 0.21) = normalcdf(0.21, 1E99, 0.18, 0.024298) = 0.108.

(b) Time is quantitative → mean x̄. Skewed population but n = 81 ≥ 30, so CLT → approximately Normal. σ_x̄ = 9/√81 = 9/9 = 1. z = (21 − 22)/1 = −1.00. P(x̄ < 21) = normalcdf(−1E99, 21, 22, 1) = 0.1587.

Takeaway. Identical lesson, two different machines — the only thing that changed was reading "walk" (categorical) versus "minutes" (quantitative).

(d) Common Mistakes

1. Using population σ instead of σ/√n for a sample mean. The marquee error. "The mean of n = 36 cars" is a statistic — divide by √n. Using the raw σ makes your spread far too big, your z far too small, and your probability badly off. (In Example 3, this turns a 2.3% answer into a 36.9% answer.) Always ask Question 2: individual or statistic?

2. Skipping the shape justification. You must state why the statistic is approximately Normal: for p̂ cite Large Counts (np ≥ 10, n(1−p) ≥ 10); for x̄ cite either population Normal (any n) or the CLT (n ≥ 30). Perfect arithmetic with no shape justification still loses the shape point.

3. Mixing up p̂ and x̄ procedures. Plugging √(p(1−p)/n) into a mean problem, or σ/√n into a proportion problem, is a guaranteed wrong answer. Run Question 1 first: categorical → p̂ formulas; quantitative → x̄ formulas. They are not interchangeable.

4. Using the CLT to rescue a small skewed-population mean. The CLT needs large n (n ≥ 30). If the population is skewed and n = 12, you cannot model x̄ as Normal — the right response is "this calculation isn't justified," not a number.

5. Forgetting the 10% condition. It justifies the spread formula in both the p̂ and x̄ cases. Omitting it is an easy lost point on the FRQ even when every number is right.

(e) Practice Problems

(in context) A streaming service finds that p = 0.15 of trials convert to paid subscriptions. In an SRS of n = 250 trials, find P(p̂ > 0.18) and state your shape justification.

(in context) Daily steps for a fitness-app population are right-skewed with μ = 3.2 (thousand) and σ = 1.1 (thousand). For a random sample of n = 64 users, find P(x̄ > 3.4) and justify the shape.

True or false: For the same context, the probability that one individual exceeds a value is always less than the probability that the sample mean of n individuals exceeds that value. Explain.

(in context) Test scores in a population are Normal with μ = 70 and σ = 12.

(a) Find P(one student > 74).

(b) Find P(x̄ > 74) for a sample of n = 36.

(c) Why do the answers differ?

Bottle fills are Normal with μ = 2.0 L and σ = 0.3 L. For a sample of n = 25 bottles, find the value c such that P(x̄ < c) = 0.90. (Hint: z = 1.282.)

(f) FRQ Practice (10 points)

Statistical Practice: 3 — Analyze Data

An airline states that the weight of a checked bag has a population mean of μ = 23 kilograms with population standard deviation σ = 5 kilograms. The distribution of individual bag weights is strongly right-skewed (most bags are light, a few are very heavy). For one flight, a gate agent weighs a random sample of n = 64 checked bags.

(a) The airline first asks about a single bag: explain why the agent cannot reliably compute the probability that one randomly selected bag weighs more than 24 kg using a Normal model, but can compute a probability about the sample mean of 64 bags. (2 points)

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

(c) Find the probability that the sample mean weight exceeds 24 kilograms, P(x̄ > 24). Show your work. (2 points)

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

Model Response

(a) A single bag's weight follows the population distribution, which is strongly right-skewed, not Normal. So we cannot use a Normal model to find the probability for one bag. The sample mean of 64 bags is different: because n = 64 ≥ 30, the Central Limit Theorem makes the sampling distribution of x̄ approximately Normal even though the population is skewed. We can model x̄ as Normal, so a probability about the mean is justified.

(b)

• Center: μ_x̄ = μ = 23 kg. The sample mean is an unbiased estimator of μ.
• Spread: σ_x̄ = σ/√n = 5/√64 = 5/8 = 0.625 kg. This requires the 10% condition: 64 bags is clearly less than 10% of all checked bags the airline handles, so the weights 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 = 64 ≥ 30, the sampling distribution of x̄ is approximately Normal.

(c) Standardize using μ_x̄ = 23 and σ_x̄ = 0.625:

z = (24 − 23) / 0.625 = 1 / 0.625 = 1.60
P(x̄ > 24) = P(z > 1.60) ≈ 0.0548

TI-84: normalcdf(24, 1E99, 23, 0.625) = 0.0548.

(d) If the airline repeatedly took random samples of 64 checked bags, about 5.5% of those samples would have a mean weight greater than 24 kg. A sample mean above 24 kg is therefore somewhat unusual but not extreme; it is broadly consistent with the airline's claimed mean of 23 kg.

Scoring Rubric (10 points)

Part (a) — 2 points

• 1 pt — Explains that one bag follows the skewed population distribution, so a Normal model is not justified for an individual.
• 1 pt — Explains that the sample mean is approximately Normal by the CLT (n ≥ 30), so a probability about x̄ is justified.

Part (b) — 4 points

• 1 pt — Center: μ_x̄ = 23 (states it equals μ).
• 1 pt — Spread: correct formula and value σ_x̄ = 5/√64 = 0.625.
• 1 pt — Shape stated as approximately Normal with CLT justification referencing n ≥ 30 (because the population is skewed).
• 1 pt — 10% condition checked (64 ≤ 10% of all checked bags) to justify the spread formula.

Part (c) — 2 points

• 1 pt — Correct z-score using σ_x̄ = 0.625 (NOT σ = 5): z = 1.60.
• 1 pt — Correct probability ≈ 0.0548 (accept 0.054–0.055) set up as an upper-tail area.

Part (d) — 2 points

• 1 pt — Interpretation framed as a long-run proportion of samples (not individuals): "about 5.5% of samples of 64 would have a mean above 24 kg."
• 1 pt — Connected to context / the airline's claim (somewhat unusual but consistent with μ = 23).

Where students lose points:

• Part (a): saying the individual can be modeled as Normal — it can't, the population is skewed. The whole point is that the CLT rescues the mean, not the individual.
• Part (b): using σ = 5 instead of σ_x̄ = 0.625; saying "the population is Normal" when it is skewed (only valid justification is CLT with n ≥ 30); omitting the 10% condition (automatic −1).
• Part (c): using σ = 5 in the z-score (gives z = 0.20 and the wrong probability ≈ 0.4207); a "naked" answer with no shown work.
• Part (d): interpreting as "the probability one bag weighs more than 24 kg" — that confuses the individual with the sample mean.