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AP Statistics Help: Sampling Distributions For Sample Means

Review real example questions for Sampling Distributions For Sample Means in AP Statistics.

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A manufacturer produces resistors with mean resistance $\mu=100\ \Omega$ and population standard deviation $\sigma=8\ \Omega$ . Individual resistances are approximately normal. A quality engineer repeatedly selects simple random samples of size $n=16$ and records the sample mean resistance $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?

All questions

Question 1

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

A website's page-load times (in seconds) for all visits have mean $\mu=3.2$ and a large standard deviation; the distribution is strongly left-skewed due to a hard lower bound near 0 seconds. A data analyst repeatedly takes random samples of $n=50$ visits and calculates $\bar{x}$ for each sample. Which statement about the sampling distribution of $\bar{x}$ is correct?

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

The mean score on a certain professional exam is $\mu=520$ . A test-prep company repeatedly takes random samples of $n=64$ exam scores and computes the sample mean $\bar{x}$ . Assume the sampling distribution of $\bar{x}$ is approximately normal. Which statement about the sampling distribution is correct?

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)
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.
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.
The sampling distribution of $\bar{x}$ is centered at $64$ , and its standard deviation is $520$ .
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 \bar{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

A call center tracks the length of individual customer calls (in minutes). The population mean is $\mu=8.0$ minutes with population standard deviation $\sigma=5.0$ minutes, and the population distribution is strongly right-skewed. Each day, a supervisor takes a simple random sample of $n=25$ calls and computes the sample mean $\bar{x}$ . Over many days, which statement about the sampling distribution of $\bar{x}$ is correct?

The sampling distribution of $\bar{x}$ has mean $8.0$ and standard deviation $5.0$ because $\bar{x}$ is computed from call times.
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.
The sampling distribution of $\bar{x}$ has mean $8.0$ and standard deviation $5.0/25$ because the sample size is 25.
The sampling distribution of $\bar{x}$ has mean $8.0$ and standard deviation $5.0/\sqrt{25}$ , and it is approximately normal. (correct answer)
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 5

For a large population of car trips, the mean fuel economy is $\mu=27$ mpg with population standard deviation $\sigma=5$ mpg. The distribution of individual mpg values is bimodal because it mixes city and highway driving. A researcher repeatedly takes random samples of size $n=45$ trips and computes the sample mean mpg $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?

Because the population is bimodal, the sampling distribution of $\bar{x}$ must also be bimodal.
The sampling distribution of $\bar{x}$ is approximately normal and centered at 27 mpg. (correct answer)
The sampling distribution of $\bar{x}$ is centered at 27 mpg, but it must be uniform since the population is mixed.
The sampling distribution of $\bar{x}$ has mean $27/45$ mpg because it is an average of 45 trips.
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 6

A university reports that the population mean time to walk between two common campus locations is $\mu=12$ minutes, with population standard deviation $\sigma=4$ minutes. The distribution of individual walking times is somewhat skewed. A student repeatedly selects simple random samples of $n=64$ students and computes the sample mean walking time $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?

The sampling distribution of $\bar{x}$ has mean $12$ and standard deviation $4$ , and it has the same shape as the population.
The sampling distribution of $\bar{x}$ has mean $12$ and standard deviation $4/\sqrt{64}$ , and it is approximately normal. (correct answer)
The sampling distribution of $\bar{x}$ has mean $64$ and standard deviation $4/\sqrt{12}$ .
The sampling distribution of $\bar{x}$ has mean $12/64$ and standard deviation $4/\sqrt{64}$ .
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 7

At a factory, the fill amount (in ounces) of individual bottles has population mean $\mu=20$ ounces and a fixed population standard deviation $\sigma$ (distribution may be skewed). Each hour, a quality engineer randomly selects $n=36$ bottles and records the sample mean fill amount $\bar{x}$ . This process is repeated many times under identical conditions. Which statement about the sampling distribution of $\bar{x}$ is correct?

The sampling distribution of $\bar{x}$ has mean $20$ and standard deviation $\sigma$ because each sample still comes from the same population.
The sampling distribution of $\bar{x}$ has mean $20$ and standard deviation $\sigma/36$ because the mean averages 36 values.
The sampling distribution of $\bar{x}$ has mean $20$ and standard deviation $\sigma/\sqrt{36}$ , and it is approximately normal. (correct answer)
The sampling distribution of $\bar{x}$ has mean $36$ and standard deviation $\sigma/\sqrt{20}$ because $n=36$ and $\mu=20$ .
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 8

A city monitors the concentration of a pollutant (in parts per billion, ppb) at a location each day. The population mean daily concentration is $\mu=18$ ppb with population standard deviation $\sigma=7$ ppb, and the daily values can be quite variable. Each month, an analyst repeatedly selects random samples of $n=30$ days and computes the sample mean concentration $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?

The sampling distribution of $\bar{x}$ has mean $18$ and standard deviation $7/\sqrt{30}$ , and it is approximately normal. (correct answer)
The sampling distribution of $\bar{x}$ has mean $18$ and standard deviation $7$ because $\bar{x}$ is based on the same pollutant measurements.
The sampling distribution of $\bar{x}$ has mean $30$ and standard deviation $7/\sqrt{18}$ .
The sampling distribution of $\bar{x}$ has mean $18/30$ and standard deviation $7/\sqrt{30}$ .
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 9

A hospital records systolic blood pressure for a large population of adults with mean $\mu=122$ mmHg and standard deviation $\sigma=15$ mmHg. The population distribution is approximately normal. A researcher repeatedly takes random samples of size $n=9$ and computes $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?

The sampling distribution of $\bar{x}$ is approximately normal and centered at 122 mmHg. (correct answer)
The sampling distribution of $\bar{x}$ is centered at $122/9$ mmHg because it averages 9 values.
The sampling distribution of $\bar{x}$ is not normal because $n=9$ is less than 30.
The sampling distribution of $\bar{x}$ has the same variability as the population because both measure blood pressure.
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 10

A machine produces bolts with mean length $\mu=5.0$ cm. A quality engineer repeatedly selects random samples of $n=100$ bolts and computes the sample mean length $\bar{x}$ . Assume the sampling distribution of $\bar{x}$ is approximately normal. Which statement about the sampling distribution is correct?

The sampling distribution of $\bar{x}$ is centered at $5.0$ , and its variability is less than the variability of individual bolt lengths. (correct answer)
The sampling distribution of $\bar{x}$ is centered at $5.0/100$ , and its variability is the same as the variability of individual bolt lengths.
The sampling distribution of $\bar{x}$ is centered at $5.0$ , and its variability is greater than the variability of individual bolt lengths.
Because $n=100$ is large, the sampling distribution of $\bar{x}$ must be uniform rather than normal.
The sampling distribution of $\bar{x}$ has the same shape and spread as the distribution of individual bolt lengths.

Explanation: This AP Statistics question evaluates concepts in sampling distributions for sample means. The sampling distribution of \bar{x} is centered at the population mean μ = 5.0 cm, with variability (standard deviation) σ/√100 = σ/10, which is less than that of individual bolt lengths. Choice C is a distractor, incorrectly stating greater variability, perhaps from confusing sample means with sums or misunderstanding the averaging effect. Mini-lesson: Sampling distributions of means arise from repeatedly sampling n items and calculating \bar{x} each time; the resulting distribution has mean μ, standard error σ/√n (decreasing with larger n), and approximates normality via the CLT for n ≥ 30, even if the population isn't normal. Here, with n=100, the distribution is normal with reduced spread. Choice A accurately describes this.