AP Statistics Quiz: Sampling Distributions For Sample Means

What this quiz covers

This quiz focuses on Sampling Distributions For Sample Means, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The sampling distribution of $\bar{x}$ is approximately normal with mean $100\ \Omega$. (correct answer)
2. The sampling distribution of $\bar{x}$ has mean $100\ \Omega$, but it is not normal because $n=16$ is too small.
3. The sampling distribution of $\bar{x}$ has the same standard deviation as the population, $8\ \Omega$.
4. The sampling distribution of $\bar{x}$ is centered at the sample size, $16$.
5. The sampling distribution of $\bar{x}$ is centered at $100\ \Omega$ only if the population distribution is uniform.

Explanation: This question examines sampling distributions when the population is already normal. Since individual resistances are approximately normal, the sampling distribution of x̄ is also normal for any sample size (not just n ≥ 30). The mean of the sampling distribution equals the population mean μ = 100 Ω. The standard deviation of x̄ is σ/√n = 8/√16 = 2 Ω, which is less than the population standard deviation. When the population is normal, the sampling distribution of x̄ is always normal, making the Central Limit Theorem unnecessary.

Question 2

1. The sampling distribution of $\bar{x}$ is left-skewed with the same shape as the population because the population is left-skewed.
2. The sampling distribution of $\bar{x}$ is approximately normal and centered at $3.2$ seconds. (correct answer)
3. The sampling distribution of $\bar{x}$ has mean $3.2$ seconds only if the sample is more than 10% of the population.
4. The sampling distribution of $\bar{x}$ has greater variability than the population because it is based on 50 observations.
5. The sampling distribution of $\bar{x}$ is approximately normal only if the population distribution is normal.

Explanation: This question tests the Central Limit Theorem with a strongly skewed population. Despite the left-skewed population distribution, with n = 50 visits (n ≥ 30), the sampling distribution of x̄ becomes approximately normal. The mean of the sampling distribution equals the population mean μ = 3.2 seconds regardless of skewness. The standard deviation of x̄ equals σ/√n, which is less than the population standard deviation. The Central Limit Theorem ensures normality for large samples even when the population is non-normal.

Question 3

1. The sampling distribution of $\bar{x}$ is centered at $520$, and its standard deviation is smaller than the standard deviation of individual exam scores. (correct answer)
2. The sampling distribution of $\bar{x}$ is centered at $520/64$, and its standard deviation is smaller than the standard deviation of individual exam scores.
3. The sampling distribution of $\bar{x}$ is centered at $520$, and its standard deviation is the same as the standard deviation of individual exam scores.
4. The sampling distribution of $\bar{x}$ is centered at $64$, and its standard deviation is $520$.
5. The sampling distribution of $\bar{x}$ must have the same shape as the population distribution of exam scores, regardless of $n$.

Explanation: This AP Statistics question probes understanding of sampling distributions for sample means. The sampling distribution is centered at μ = 520, with standard deviation σ/√64 = σ/8, smaller than individual scores' variability. Distractor choice B divides the center by 64, perhaps mistaking it for a rate or total. Mini-lesson: Generating many samples of size n and plotting their means ar{x} yields a distribution with mean μ, spread σ/√n (less variable due to averaging), and normal approximation by CLT for large n like 64. This makes sample means more reliable estimators. Choice A is accurate.

Question 4

1. The sampling distribution of $\bar{x}$ has mean $42$, and its standard deviation is $\sigma/49$.
2. The sampling distribution of $\bar{x}$ has mean $42$, and its standard deviation is $\sigma/\sqrt{49}$. (correct answer)
3. The sampling distribution of $\bar{x}$ has mean $42/49$, and its standard deviation is $\sigma/\sqrt{49}$.
4. The sampling distribution of $\bar{x}$ has mean $42$, and its standard deviation is the same as $\sigma$.
5. The sampling distribution of $\bar{x}$ is more spread out than the population distribution because it is based on repeated samples.

Explanation: This AP Statistics question evaluates sampling distributions for sample means. The distribution of ar{x} has mean μ = 42 minutes and standard deviation σ/√49 = σ/7. Distractor choice A uses σ/49 without the square root, a common error in recalling the formula. Mini-lesson: Sampling distributions of means show the pattern of ar{x} values from many samples of size n; centered at μ, with spread σ/√n that decreases with n, and normal shape by CLT for sufficient n. With n=49, it's highly normal and precise. Choice B is correct.

Question 5

1. The sampling distribution of $\bar{x}$ has mean $8.0$ and standard deviation $5.0$ because $\bar{x}$ is computed from call times.
2. The sampling distribution of $\bar{x}$ has mean $8.0$ and standard deviation $5.0/\sqrt{25}$, but it will still be strongly right-skewed.
3. The sampling distribution of $\bar{x}$ has mean $8.0$ and standard deviation $5.0/25$ because the sample size is 25.
4. The sampling distribution of $\bar{x}$ has mean $8.0$ and standard deviation $5.0/\sqrt{25}$, and it is approximately normal. (correct answer)
5. The sampling distribution of $\bar{x}$ has mean $25$ and standard deviation $5.0/\sqrt{8.0}$.

Explanation: This problem involves the sampling distribution of mean call times. The population has μ=8.0 minutes and σ=5.0 minutes with a right-skewed distribution. For samples of size n=25, the sampling distribution of x̄ has mean 8.0 (same as population) and standard deviation σ/√n = 5.0/√25 = 5.0/5 = 1.0. Despite the population being strongly right-skewed, with n=25 (close to 30), the Central Limit Theorem indicates the sampling distribution will be approximately normal. Choice B incorrectly claims it remains skewed, while C divides by n instead of √n.

Question 6

1. Because the population is bimodal, the sampling distribution of $\bar{x}$ must also be bimodal.
2. The sampling distribution of $\bar{x}$ is approximately normal and centered at 27 mpg. (correct answer)
3. The sampling distribution of $\bar{x}$ is centered at 27 mpg, but it must be uniform since the population is mixed.
4. The sampling distribution of $\bar{x}$ has mean $27/45$ mpg because it is an average of 45 trips.
5. The sampling distribution of $\bar{x}$ has the same variability as individual mpg values because both are measured in mpg.

Explanation: This question tests understanding of sampling distributions from bimodal populations. Despite the bimodal population distribution, with n = 45 trips (n ≥ 30), the Central Limit Theorem ensures the sampling distribution of x̄ is approximately normal, not bimodal. The mean of the sampling distribution equals μ = 27 mpg, not 27/45. The standard deviation of x̄ equals σ/√n = 5/√45 ≈ 0.75 mpg. The averaging process in sampling distributions smooths out unusual shapes like bimodality when n is large.

Question 7

1. The sampling distribution of $\bar{x}$ has mean $12$ and standard deviation $4$, and it has the same shape as the population.
2. The sampling distribution of $\bar{x}$ has mean $12$ and standard deviation $4/\sqrt{64}$, and it is approximately normal. (correct answer)
3. The sampling distribution of $\bar{x}$ has mean $64$ and standard deviation $4/\sqrt{12}$.
4. The sampling distribution of $\bar{x}$ has mean $12/64$ and standard deviation $4/\sqrt{64}$.
5. The sampling distribution of $\bar{x}$ has standard deviation $4/64$ because averaging divides by $n$.

Explanation: This question examines the sampling distribution for walking times with μ=12 minutes and σ=4 minutes. With samples of size n=64, the sampling distribution of x̄ has mean 12 and standard deviation σ/√n = 4/√64 = 4/8 = 0.5. Since n=64 is well above 30, the Central Limit Theorem ensures the sampling distribution is approximately normal despite the somewhat skewed population. Choice A incorrectly maintains the population standard deviation, while E incorrectly divides by n=64 rather than √64.

Question 8

1. The sampling distribution of $\bar{x}$ has mean $20$ and standard deviation $\sigma$ because each sample still comes from the same population.
2. The sampling distribution of $\bar{x}$ has mean $20$ and standard deviation $\sigma/36$ because the mean averages 36 values.
3. The sampling distribution of $\bar{x}$ has mean $20$ and standard deviation $\sigma/\sqrt{36}$, and it is approximately normal. (correct answer)
4. The sampling distribution of $\bar{x}$ has mean $36$ and standard deviation $\sigma/\sqrt{20}$ because $n=36$ and $\mu=20$.
5. The sampling distribution of $\bar{x}$ is the same as the population distribution because the same measurement is being recorded.

Explanation: This question tests understanding of the sampling distribution of sample means. When we take samples of size n=36 from a population with mean μ=20 and standard deviation σ, the sampling distribution of x̄ has mean equal to the population mean (20) and standard deviation equal to σ/√n = σ/√36 = σ/6. Since n=36≥30, the Central Limit Theorem tells us this sampling distribution is approximately normal, even though the original population may be skewed. Choice A incorrectly keeps the population standard deviation, while B incorrectly divides by n instead of √n.

Question 9

1. The sampling distribution of $\bar{x}$ has mean $18$ and standard deviation $7/\sqrt{30}$, and it is approximately normal. (correct answer)
2. The sampling distribution of $\bar{x}$ has mean $18$ and standard deviation $7$ because $\bar{x}$ is based on the same pollutant measurements.
3. The sampling distribution of $\bar{x}$ has mean $30$ and standard deviation $7/\sqrt{18}$.
4. The sampling distribution of $\bar{x}$ has mean $18/30$ and standard deviation $7/\sqrt{30}$.
5. The sampling distribution of $\bar{x}$ has standard deviation $7/30$ because the sample mean divides by $n$.

Explanation: For pollutant concentrations with μ=18 ppb and σ=7 ppb, samples of n=30 days yield a sampling distribution of x̄ with mean 18 and standard deviation σ/√n = 7/√30 ≈ 1.28. With n=30, we're at the threshold where the Central Limit Theorem begins to provide approximate normality for the sampling distribution, even with variable daily values. Choice B incorrectly maintains the population standard deviation, while E divides by n=30 instead of √30, confusing the formula for standard error with the calculation of the mean.

Question 10

1. The sampling distribution of $\bar{x}$ is approximately normal and centered at 122 mmHg. (correct answer)
2. The sampling distribution of $\bar{x}$ is centered at $122/9$ mmHg because it averages 9 values.
3. The sampling distribution of $\bar{x}$ is not normal because $n=9$ is less than 30.
4. The sampling distribution of $\bar{x}$ has the same variability as the population because both measure blood pressure.
5. The sampling distribution of $\bar{x}$ is centered at the sample mean from the first sample taken.

Explanation: This question involves sampling from a normal population with a small sample size. Since the population distribution is approximately normal, the sampling distribution of x̄ is also normal regardless of sample size (n = 9). The mean of the sampling distribution equals μ = 122 mmHg, not 122/9. The standard deviation of x̄ equals σ/√n = 15/√9 = 5 mmHg, which is less than the population standard deviation. When sampling from normal populations, normality is preserved in the sampling distribution for any n.

Question 11

1. Because the population is bimodal, the sampling distribution of $\bar{x}$ must also be bimodal.
2. The sampling distribution of $\bar{x}$ is approximately normal and centered at 27 mpg. (correct answer)
3. The sampling distribution of $\bar{x}$ is centered at 27 mpg, but it must be uniform since the population is mixed.
4. The sampling distribution of $\bar{x}$ has mean $27/45$ mpg because it is an average of 45 trips.
5. The sampling distribution of $\bar{x}$ has the same variability as individual mpg values because both are measured in mpg.

Explanation: This question tests understanding of sampling distributions from bimodal populations. Despite the bimodal population distribution, with n = 45 trips (n ≥ 30), the Central Limit Theorem ensures the sampling distribution of x̄ is approximately normal, not bimodal. The mean of the sampling distribution equals μ = 27 mpg, not 27/45. The standard deviation of x̄ equals σ/√n = 5/√45 ≈ 0.75 mpg. The averaging process in sampling distributions smooths out unusual shapes like bimodality when n is large.

Question 12

1. The sampling distribution of $\bar{x}$ must be approximately normal because the student is sampling randomly.
2. The mean of the sampling distribution of $\bar{x}$ is $95$ mg. (correct answer)
3. The sampling distribution of $\bar{x}$ has mean $95/4$ mg because the sample mean averages 4 cups.
4. The sampling distribution of $\bar{x}$ has the same spread as the population because $\bar{x}$ is measured in mg.
5. The sampling distribution of $\bar{x}$ is guaranteed to be symmetric even if the population is skewed.

Explanation: This question addresses sampling distributions with small sample sizes. The mean of the sampling distribution of x̄ always equals the population mean μ = 95 mg, not 95/4. With only n = 4 cups and unknown population distribution, we cannot assume the sampling distribution is approximately normal (Central Limit Theorem requires n ≥ 30 for non-normal populations). The standard deviation of x̄ equals σ/√n = 20/√4 = 10 mg, which is less than the population standard deviation. Small samples from non-normal populations generally produce non-normal sampling distributions.

Question 13

1. The sampling distribution of $\bar{x}$ has mean $11$, and its standard deviation is $\sigma/\sqrt{25}$, where $\sigma$ is the population standard deviation. (correct answer)
2. The sampling distribution of $\bar{x}$ has mean $11/25$, and its standard deviation is $\sigma/\sqrt{25}$.
3. The sampling distribution of $\bar{x}$ has mean $11$, and its standard deviation is $\sigma\sqrt{25}$.
4. The sampling distribution of $\bar{x}$ has the same mean and the same standard deviation as the distribution of individual battery lifetimes.
5. Because $n=25$, the sampling distribution of $\bar{x}$ cannot be approximately normal unless the population is exactly normal.

Explanation: This question assesses sampling distributions for sample means in AP Statistics. The mean of ar{x}'s distribution is μ = 11 hours, and the standard deviation is σ/√25 = σ/5. Choice C is a distractor, multiplying by √25 instead, which would increase spread incorrectly. Mini-lesson: The sampling distribution of the sample mean captures how ar{x} varies across samples of size n; it's centered at μ, with standard error σ/√n that shrinks as n grows, and approximates a normal distribution per the CLT for n ≥ 30 or normal populations. For n=25, it's close to normal with reduced variability. Choice A correctly identifies this.

Question 14

1. The sampling distribution of $\bar{x}$ has mean $66$ and standard deviation $3.5/\sqrt{100}$. (correct answer)
2. The sampling distribution of $\bar{x}$ has mean $66$ and standard deviation $3.5$ because the sample size does not affect variability.
3. The sampling distribution of $\bar{x}$ has mean $66/100$ and standard deviation $3.5/100$.
4. The sampling distribution of $\bar{x}$ has mean $100$ and standard deviation $66/\sqrt{3.5}$.
5. The sampling distribution of $\bar{x}$ is approximately normal only if the population distribution is exactly normal.

Explanation: For student heights with μ=66 inches and σ=3.5 inches, samples of size n=100 yield a sampling distribution of x̄ with mean 66 and standard deviation σ/√n = 3.5/√100 = 3.5/10 = 0.35. The large sample size (n=100) ensures approximate normality through the Central Limit Theorem. Choice B incorrectly claims sample size doesn't affect variability - a fundamental misunderstanding since larger samples produce more precise estimates. Choice E incorrectly states normality requires a normal population; with n≥30, the CLT provides approximate normality.

Question 15

1. The mean of $\bar{x}$ is $27$, and the standard deviation of $\bar{x}$ is larger than the standard deviation of individual commute times.
2. The mean of $\bar{x}$ is $27/9$, and the standard deviation of $\bar{x}$ is $27$.
3. The mean of $\bar{x}$ is $27$, and the standard deviation of $\bar{x}$ is smaller than the standard deviation of individual commute times. (correct answer)
4. The sampling distribution of $\bar{x}$ is exactly the same as the population distribution of individual commute times.
5. The mean of $\bar{x}$ depends on the sample size, so it is not necessarily $27$.

Explanation: This question tests knowledge of sampling distributions for sample means in AP Statistics. The center of the sampling distribution of ar{x} is the population mean μ = 27 minutes, not affected by the sample size n=9. The spread is reduced, with the standard deviation of ar{x} being σ/√9 = σ/3, which is smaller than the standard deviation of individual commute times. A frequent distractor is choice B, which wrongly divides the mean by 9 and sets the standard deviation to 27, mixing up concepts of mean and variability. Mini-lesson: The sampling distribution of the sample mean describes the variability in ar{x} across many random samples of size n; it has mean μ, standard deviation σ/√n (smaller for larger n), and is approximately normal by the CLT when n is sufficiently large or the population is normal. Thus, choice C is correct, emphasizing the unchanged mean and reduced variability.

Question 16

1. The mean of $\bar{x}$ is $150$, and the standard deviation of $\bar{x}$ is the population standard deviation divided by $\sqrt{10}$. (correct answer)
2. The mean of $\bar{x}$ is $150/10$, and the standard deviation of $\bar{x}$ is the population standard deviation divided by $\sqrt{10}$.
3. The mean of $\bar{x}$ is $150$, and the standard deviation of $\bar{x}$ is the population standard deviation multiplied by $\sqrt{10}$.
4. The mean of $\bar{x}$ is $150$, and the standard deviation of $\bar{x}$ is the same as the population standard deviation.
5. The sampling distribution of $\bar{x}$ is centered at the sample size $10$ because it averages 10 apples.

Explanation: This question tests sampling distributions for sample means in AP Statistics. The mean of ar{x} is μ = 150 grams, with standard deviation σ/√10, reduced from σ. Choice C is a distractor, multiplying by √10, which would inflate the spread erroneously. Mini-lesson: The distribution of sample means from repeated sampling of size n has center μ, standard deviation σ/√n (demonstrating the benefit of larger samples for precision), and is approximately normal via the CLT, especially for n ≥ 30. For n=10, normality is assumed here. Choice A correctly describes the mean and standard deviation.

Question 17

1. The sampling distribution of $\bar{x}$ has mean $0.35$ and standard deviation $0.08/\sqrt{16}$. (correct answer)
2. The sampling distribution of $\bar{x}$ has mean $0.35$ and standard deviation $0.08$ because the same orchard is sampled.
3. The sampling distribution of $\bar{x}$ has mean $0.35/16$ and standard deviation $0.08/\sqrt{16}$.
4. The sampling distribution of $\bar{x}$ has mean $16$ and standard deviation $0.08/\sqrt{0.35}$.
5. The sampling distribution of $\bar{x}$ must be uniform because each sample is random.

Explanation: For apple weights with population mean μ=0.35 lb and σ=0.08 lb, samples of size n=16 produce a sampling distribution of x̄ with mean 0.35 and standard deviation σ/√n = 0.08/√16 = 0.08/4 = 0.02. The sampling distribution always has the same mean as the population, but reduced standard deviation. Choice B incorrectly keeps the population standard deviation, while C incorrectly divides the mean by n. With n=16<30 and unknown population distribution, we cannot assume the sampling distribution is normal.

Question 18

1. The sampling distribution of $\bar{x}$ has mean $2.4$ and standard deviation $1.8/\sqrt{40}$, and it is approximately normal. (correct answer)
2. The sampling distribution of $\bar{x}$ has mean $2.4$ and standard deviation $1.8$, and it remains left-skewed.
3. The sampling distribution of $\bar{x}$ has mean $2.4/40$ and standard deviation $1.8/\sqrt{40}$.
4. The sampling distribution of $\bar{x}$ has mean $40$ and standard deviation $1.8/\sqrt{2.4}$.
5. The sampling distribution of $\bar{x}$ has standard deviation $1.8/40$ because the mean divides by $n$.

Explanation: This streaming data problem has μ=2.4 GB and σ=1.8 GB with left-skewed individual usage. For samples of n=40 users, the sampling distribution of x̄ has mean 2.4 and standard deviation σ/√n = 1.8/√40 ≈ 0.285. Despite the left-skewed population, n=40>30 ensures the sampling distribution is approximately normal by the Central Limit Theorem. Choice B incorrectly maintains both the population standard deviation and skewness, while E divides by n instead of √n.

Question 19

1. The sampling distribution of $\bar{x}$ is right-skewed because the population is right-skewed, regardless of $n$.
2. The mean of the sampling distribution of $\bar{x}$ is $20.0$ oz, and the distribution of $\bar{x}$ is approximately normal. (correct answer)
3. The standard deviation of $\bar{x}$ is $0.6$ oz because sample means vary about as much as individual observations.
4. The mean of the sampling distribution of $\bar{x}$ is greater than $20.0$ oz because of the right skew.
5. The sampling distribution of $\bar{x}$ is approximately uniform since the samples are random.

Explanation: This question tests understanding of sampling distributions for sample means. The sampling distribution of x̄ has mean equal to the population mean μ = 20.0 oz (unbiased estimator property). With n = 36 bottles, the Central Limit Theorem applies since n ≥ 30, making the sampling distribution approximately normal regardless of the right-skewed population. The standard deviation of x̄ equals σ/√n = 0.6/√36 = 0.1 oz, not 0.6 oz. Sample means have less variability than individual observations because averaging reduces spread.

Question 20

1. The sampling distribution of $\bar{x}$ must be approximately normal because the student is sampling randomly.
2. The mean of the sampling distribution of $\bar{x}$ is $95$ mg. (correct answer)
3. The sampling distribution of $\bar{x}$ has mean $95/4$ mg because the sample mean averages 4 cups.
4. The sampling distribution of $\bar{x}$ has the same spread as the population because $\bar{x}$ is measured in mg.
5. The sampling distribution of $\bar{x}$ is guaranteed to be symmetric even if the population is skewed.

Explanation: This question addresses sampling distributions with small sample sizes. The mean of the sampling distribution of x̄ always equals the population mean μ = 95 mg, not 95/4. With only n = 4 cups and unknown population distribution, we cannot assume the sampling distribution is approximately normal (Central Limit Theorem requires n ≥ 30 for non-normal populations). The standard deviation of x̄ equals σ/√n = 20/√4 = 10 mg, which is less than the population standard deviation. Small samples from non-normal populations generally produce non-normal sampling distributions.