AP Statistics Unit 5 Sampling Distributions

Topics 5.1–5.2: Sampling Distributions & The Normal Distribution, Revisited

Why Samples Vary, Normal Probability, Linear Combinations & Appropriateness
Video Follow-Along Worksheet
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Learning Objectives:

Key Vocabulary

Sampling Variability The natural variation in statistics produced by different random samples from the same population
Sampling Distribution The distribution of values of a statistic from all possible samples of a given size from a population
Normal Distribution A symmetric, bell-shaped continuous probability distribution described by mean (μ) and standard deviation (σ)
Z-Score The number of standard deviations a value is from the mean: z = (x − μ) / σ
Inverse Normal Given an area (probability), find the boundary value — the reverse of a typical normal problem
Linear Combination A sum or difference of random variables (e.g., X − Y or X + Y); if independent & normal, the result is also normal
Empirical Rule For normal distributions: ~68% within 1 SD, ~95% within 2 SD, ~99.7% within 3 SD of the mean

Video 1: Why Is My Sample Not Like Yours? (Topic 5.1)

[Video: ~5 min]
1. [0:37] Three different samples of five dice rolls produced means of 7, 5.2, and 8.4. This illustrates that different samples produce different .
Koala Example: [2:21] The average weight of adult koalas is about 15 pounds with SD = 1.8 pounds. Zoo researchers randomly select 12 koalas each month to check their health.
2. [3:00] If we calculate the means of many samples of the same size, the sample means will cluster around the true .
3. [3:23] In a close election, because sample proportions , different polls may predict different outcomes.
4. [5:00] Key Takeaway: If the population distribution is known, then the distribution of statistics from many samples of the same size is .

Video 2: The Normal Distribution, Revisited — Probability (Topic 5.2)

[Video: ~6 min]

Z-Score Formula

z = (x − μ) / σ
5. [0:16] The standard normal distribution has a mean of and a standard deviation of .
Giraffe Neck Example: [1:30] The lengths of adult giraffe necks are approximately normal with μ = 5.9 feet and σ = 0.3 feet.
6. [2:00] Calculate P(neck > 6 feet):
z = (6 − 5.9) / 0.3 =  →  P(X > 6) = (Over a third of giraffes have necks longer than 6 feet!)
AP Exam Checklist: [2:42] When solving normal probability problems, graders look for: (1) Define the random variable in context, (2) Show you're using a normal distribution, (3) Identify parameters μ and σ, (4) Show the boundary value and direction, (5) Give the correct probability.
7. [5:03] An normal problem gives you the area/probability and asks you to find the boundary value.
8. [5:03] Find the cutoff for the shortest 25% of giraffe necks:
Z-score for bottom 25% ≈ −0.67  →  Cutoff = 5.9 + (−0.67)(0.3) = feet

Video 3: Linear Combinations of Normal Random Variables (Topic 5.2)

[Video: ~7 min]

Formulas for Linear Combinations of Independent Random Variables

Mean of X − Y: μX−Y = μX − μY
SD of X − Y: σX−Y = √(σX² + σY²)   (always ADD variances!)
Capitol 10K Race (Austin, TX): [0:26] Female and male completion times are independent and approximately normally distributed.
Females (F)Males (M)
Mean (μ)81 min70 min
SD (σ)15 min14 min
9. [2:08] Mean of F − M: μF−M = 81 − 70 = minutes. (On average, males finish 11 min faster)
10. [2:36] Key insight: You cannot subtract standard deviations! You must convert to and them.
σF−M = √(15² + 14²) = minutes.
11. [3:37] If a female finishes faster, her time is lower, so F − M is a number.
12. [4:28] P(female finishes faster than male) = P(F − M < 0) = (Almost a 1-in-3 chance!)
Practice Problem — More Competitive Race: [5:21] For a different, more competitive 10K: μF−M = 9 min, σF−M = 5.19 min
13. [6:00] P(female finishes at least 15 min slower)?
"Slower" means F − M is positive.   z = (15 − 9) / 5.19 = 1.16  →  P =

Video 4: Appropriateness of the Normal Distribution (Topic 5.2)

[Video: ~5 min]
Survey Example: [0:25] 14% of young adults (ages 18–29) claim their parents are doing too little for them. Let X = number in a random sample who make this claim. This is a binomial situation (two outcomes, fixed n, independence, constant p = 0.14).
14. [2:29] With n = 10: The distribution is skewed right. A normal approximation is .
15. [2:45] With n = 75: The distribution is unimodal, roughly symmetric, and -shaped. A normal approximation is appropriate.
16. [3:25] Empirical Rule Check (n = 75): Within 1 SD: ~68%, Within 2 SD: ~95%, Within 3 SD: ~% — matches normal distribution!
17. [4:10] When the sample size is enough, the normal distribution is a good approximation for a binomial distribution.
Normality Checklist: [4:30] To determine if normal approximation is appropriate, check if the distribution is: (1) Unimodal, (2) Roughly symmetric, (3) Bell-shaped. Verify with the 68-95-99.7 empirical rule.

Reflection Questions

[AI-Graded]
R1. Explain in your own words what "sampling variability" means and why it matters in statistics. Use the koala weight example or the election polling example from Video 1 to support your explanation. Why is a single sample statistic not enough to know the "truth" about a population?
R2. A factory produces bolts with lengths that are normally distributed with μ = 5.0 cm and σ = 0.1 cm. A bolt is rejected if it is shorter than 4.8 cm or longer than 5.2 cm. Calculate the probability that a randomly selected bolt is rejected. Show your work with Z-scores and interpret your result in context.

Exit Ticket

The weights of bags of chips from Brand X are normally distributed with μ = 10.0 oz and σ = 0.3 oz. The weights of bags from Brand Y are independently and normally distributed with μ = 9.5 oz and σ = 0.2 oz.

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