AP Statistics Unit 5 Sampling Distributions

Topic 5.7: Sampling Distributions for Sample Means

From Proportions to Means — Parameters, Shape, & Probability
Video Follow-Along Worksheet
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Learning Objectives: Essential Knowledge:
🖩 Practice on the TI-84 →

Key Vocabulary

Sample Mean (x) The average of all values in a sample; a statistic used to estimate the population mean μ
Sampling Distribution The distribution of a statistic (here, x) across all possible random samples of a given size from a population
Central Limit Theorem For large enough sample sizes (n ≥ 30), the sampling distribution of x is approximately normal regardless of the population shape
10% Condition The sample size must be less than 10% of the population size for the standard deviation formula to be valid
Unbiased Estimator A statistic whose sampling distribution is centered at the true parameter — x is an unbiased estimator of μ

Part 1: What Is the Sampling Distribution of x?

[Video 1: 0:00 – 3:19]
Setup — Sampling Lemons: [0:51] A large tree produces lemons with weights that are approximately normally distributed with μ = 4 oz and σ = 0.5 oz. We randomly select n = 6 lemons, weigh them, and calculate the sample mean x.
1. [1:18] Each dot on the sampling distribution represents the of one random sample of 6 lemons — not an individual lemon's weight.
2. [2:53] After 500 simulated samples, the sampling distribution is approximately , centered at ounces.
3. The center of the simulated distribution (4 oz) matches the population mean μ. This confirms that x is an estimator of μ.

Part 2: Formulas & Conditions

[Video 1: 3:19 – 4:54]

Formulas (on your AP formula sheet)

Mean: μx = μ
SD: σx = σ / √n
4. [3:49] The condition for the standard deviation formula to be valid: the sample size must be less than of the population size.
5. [4:21] Calculate for the lemon example (μ = 4, σ = 0.5, n = 6):
σx = 0.5 / √6 = ounces

Pause & Think

6. The SD of the sampling distribution (0.204 oz) is much than the population SD (0.5 oz). This makes sense: when you average 6 values, extreme highs and lows tend to cancel out, so the result is less spread out.

Part 3: Shape — When Is the Distribution Approximately Normal?

[Video 1: 4:54 – 6:31]
Two Paths to Normality (only one needs to hold):
[5:15] Path 1: The population distribution is approximately normal.
[5:27] Path 2: The sample size is large enough (n ≥ 30) — this is the Central Limit Theorem.
7. In the lemon example, the population is already approximately normal, so the sampling distribution of x is approximately normal even though n = , which is less than 30.
8. If a population were strongly skewed (not normal), we would need a sample size of at least before we could say the sampling distribution of x is approximately normal.

Means vs. Proportions — How Do Normality Conditions Compare?

Sample Proportion (p̂) Sample Mean (x)
Normality Condition Large Counts:
np ≥ 10 and n(1−p) ≥ 10		 Population normal
OR n ≥
SD Formula Condition n < 10% of N n < 10% of N
9. Both sampling distributions share the same condition for the SD formula ( condition), but their normality conditions are different.

Part 4: Interpreting Parameters in Context

[Video 2: 0:00 – 3:19]
10. [1:50] Interpreting the mean: A correct interpretation of μx = 4 must reference " random samples" of size 6 from this population and include .
11. [2:24] Interpreting the SD: σx = 0.204 means the sample mean weights vary by about 0.204 ounces from the population mean. (Use the word "typically" or "on average.")
12. [2:55] Knowing how much x typically varies from μ is one of the most important reasons to study . This idea is the foundation for confidence intervals and significance tests in later units.

Part 5: Calculating & Interpreting Probabilities

[Video 2: 3:19 – 5:28]
Probability Question: [3:19] Would it be unusual to get a sample mean of x = 4.5 oz or greater from a random sample of 6 lemons?
13. [4:22] Calculate the z-score: z = (4.5 − 4) / 0.204 = . This sample mean is about 2.5 standard deviations above the mean of the sampling distribution.
14. [4:40] Using Table A or technology: P(x ≥ 4.5) ≈ .
15. [4:51] This result is unusual: getting a sample mean of 4.5+ oz happens in only about % of all possible samples of size 6.

The Big Idea: Individuals vs. Averages

16. [4:51] A single lemon weighing 4.5 oz is not surprising — it's only 1 SD above the mean of the population. But an average of 6 lemons at 4.5+ oz is very surprising. This is because averages are much variable than individual values: the SD drops from 0.5 oz (individual) to 0.204 oz (sample mean of 6).

Reflection Questions

[AI-Graded]
R1. Compare the sampling distribution of x (sample means) with the sampling distribution of p̂ (sample proportions) from Topic 5.5. How are the formulas for the mean and standard deviation similar? How are the conditions for approximate normality different? Why do you think the normality conditions differ?
R2. The video shows that getting a single lemon weighing 4.5 ounces is not surprising, but getting an average of 6 lemons weighing 4.5 ounces or more IS very surprising (P = 0.0071). Explain in your own words why averages are less variable than individual values. What would happen to the probability P(x ≥ 4.5) if we increased the sample size from 6 to 50? Explain your reasoning.

Exit Ticket

The heights of adult women in a country are normally distributed with a mean of 64 inches and a standard deviation of 3 inches. A researcher takes a random sample of n = 36 women.

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