AP Statistics Unit 5: Sampling Distributions — Free Review Games.
This unit covers sampling distribution of a proportion, sampling distribution of a mean and central limit theorem — essential concepts for AP Statistics. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers sampling distribution of a proportion, sampling distribution of a mean and central limit theorem — essential concepts for AP Statistics. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Sampling Distribution Of A Proportion
The sampling distribution of a sample proportion p̂ describes all possible values of p̂ from samples of size n drawn from a population with true proportion p. Students must know the conditions for normality (Large Counts: np ≥ 10 and n(1−p) ≥ 10) and be able to compute the mean and standard deviation of the distribution. This distribution is central to confidence intervals and significance tests for proportions.
Key Points
- Mean of sampling distribution: μ_p̂ = p
- Standard deviation: σ_p̂ = √(p(1−p)/n) — use this formula, not the sample formula
- Large Counts condition (np ≥ 10 and n(1−p) ≥ 10) must be verified before assuming normality
- Increasing sample size n decreases variability (σ_p̂ shrinks); it does NOT change the center
A large high school reports that 30% of students walk to school. If you take a random sample of 80 students, what is the probability that more than 35% of the sample walks to school?
First verify conditions: np = 80(0.30) = 24 ≥ 10 and n(1−p) = 80(0.70) = 56 ≥ 10, so the distribution of p̂ is approximately normal. Compute σ_p̂ = √(0.30 × 0.70 / 80) ≈ 0.0512. Then find z = (0.35 − 0.30) / 0.0512 ≈ 0.98, and P(p̂ > 0.35) = P(z > 0.98) ≈ 0.1635.
2 Sampling Distribution Of A Mean
The sampling distribution of a sample mean x̄ describes all possible values of x̄ from samples of size n drawn from a population with mean μ and standard deviation σ. Students must know the mean and standard deviation of this distribution and when it is appropriate to use a normal model. The standard deviation of x̄ is called the standard error and equals σ/√n.
Key Points
- Mean of sampling distribution: μ_x̄ = μ (unbiased estimator)
- Standard deviation (standard error): σ_x̄ = σ/√n
- If the population is Normal, x̄ is Normal for ANY sample size n
- If population shape is unknown, use the Central Limit Theorem (n ≥ 30 rule of thumb) to justify normality
The distribution of individual scores on a standardized test is right-skewed with μ = 72 and σ = 15. What is the probability that the mean score of a random sample of 40 students exceeds 75?
Because n = 40 ≥ 30, by the Central Limit Theorem x̄ is approximately normal. The standard error is σ_x̄ = 15/√40 ≈ 2.372. Compute z = (75 − 72) / 2.372 ≈ 1.26, so P(x̄ > 75) = P(z > 1.26) ≈ 0.1038.
3 Central Limit Theorem
The Central Limit Theorem (CLT) states that for a sufficiently large sample size, the sampling distribution of x̄ is approximately normal regardless of the shape of the population distribution. On the AP exam, students must know when to invoke the CLT, how to state it correctly in a free-response justification, and understand that it applies to means (not individual values). The threshold n ≥ 30 is a commonly accepted rule of thumb when the population distribution is unknown.
Key Points
- CLT applies to the distribution of x̄, NOT to individual data values from the population
- The larger the sample size, the closer the sampling distribution is to normal — skewness matters: more skewed populations require larger n
- If the population is already normal, no minimum n is needed for x̄ to be normal
- Exam free-response requires explicit justification: state n ≥ 30 (or that population is normal) before using normal calculations
A bottling machine fills bottles with amounts that follow a strongly right-skewed distribution with mean 16.1 oz and standard deviation 0.4 oz. A quality inspector randomly selects 36 bottles. Is it appropriate to use a normal distribution to model the sample mean fill amount? If so, describe the sampling distribution.
Yes — because n = 36 ≥ 30, the CLT allows us to treat the sampling distribution of x̄ as approximately normal even though the population is skewed. The sampling distribution of x̄ has mean μ_x̄ = 16.1 oz and standard deviation σ_x̄ = 0.4/√36 ≈ 0.067 oz. Note that individual bottles still follow a skewed distribution; only the distribution of the sample mean is approximately normal.
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What is Sampling Distributions?
Sampling Distributions is Unit 5 of AP Statistics, covering sampling distribution of a proportion, sampling distribution of a mean and central limit theorem.
How to study for AP Statistics Unit 5?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 28+ review questions across 5 different game modes.