Sampling Distributions for Sample Means - AP Statistics

A phone battery model has a mean lifetime of $\mu=11$ hours under a standard test. A technician repeatedly selects random samples of $n=25$ batteries and computes the sample mean lifetime $\bar{x}$ . Assume the sampling distribution of $\bar{x}$ is approximately normal. Which statement about the sampling distribution is correct?
Answer: The sampling distribution of $\bar{x}$ has mean $11$ , and its standard deviation is $\sigma/\sqrt{25}$ , where $\sigma$ is the population standard deviation. This question assesses sampling distributions for sample means in AP Statistics. The mean of \bar{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 \bar{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.
Daily commute times for all employees at a large company have mean $\mu=28$ minutes and standard deviation $\sigma=10$ minutes. The distribution of individual commute times is approximately normal. A researcher repeatedly selects many random samples of $n=25$ employees and records the sample mean commute time $\bar{x}$ each time. Which statement about the sampling distribution of $\bar{x}$ is correct?
Answer: The sampling distribution of $\bar{x}$ is approximately normal with mean 28 and standard deviation $10/\sqrt{25}$ . This question examines the sampling distribution of means when the population is already normal. When individual commute times are normally distributed with μ = 28 and σ = 10, the sampling distribution of x̄ for samples of size n = 25 will also be normal (not just approximately normal). The mean of the sampling distribution equals the population mean (28 minutes), while the standard deviation equals σ/√n = 10/√25 = 10/5 = 2 minutes. Choice B incorrectly keeps the standard deviation at 10, failing to account for the reduction in variability that occurs when averaging. The fundamental principle is that sample means have less variability than individual observations because extreme values tend to balance out in the averaging process.
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?
Answer: The sampling distribution of $\bar{x}$ is approximately normal and centered at $3.2$ seconds. 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.
A bottling plant fills 20-oz bottles. The fill amounts for individual bottles have mean $\mu=20$ oz and standard deviation $\sigma=0.6$ oz, with a distribution that is right-skewed. An inspector repeatedly takes many random samples of $n=36$ bottles and computes the sample mean fill amount $\bar{x}$ each time. Which statement about the sampling distribution of $\bar{x}$ is correct?
Answer: The sampling distribution of $\bar{x}$ has mean 20 oz and standard deviation $0.6/\sqrt{36}$ oz, and it is approximately normal. This question tests understanding of the sampling distribution of sample means, specifically how the Central Limit Theorem applies when the population is skewed. The sampling distribution of x̄ has mean equal to the population mean (μ = 20 oz) and standard deviation equal to σ/√n = 0.6/√36 oz. Even though the population distribution is right-skewed, the Central Limit Theorem tells us that for n = 36 (which is sufficiently large), the sampling distribution of x̄ will be approximately normal. Choice C incorrectly claims the standard deviation remains 0.6 oz, but averaging reduces variability by a factor of √n. The key concept is that while individual measurements may be skewed, sample means from large samples follow an approximately normal distribution with reduced spread.
A nutrition researcher studies the sodium content (in milligrams) of individual servings of a packaged soup. The population has mean $\mu=910$ mg and standard deviation $\sigma=120$ mg. The researcher repeatedly takes simple random samples of size $n=9$ servings and calculates the sample mean $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?
Answer: The sampling distribution of $\bar{x}$ has mean $910$ and standard deviation $120/\sqrt{9}$ . For sodium content measurements with population mean μ=910 mg and σ=120 mg, when taking samples of size n=9, the sampling distribution of x̄ has mean 910 (unchanged from population) and standard deviation σ/√n = 120/√9 = 120/3 = 40. The key concept is that sample means have less variability than individual values, with the standard error decreasing by a factor of √n. Choice C incorrectly keeps σ=120, while D divides by n=9 instead of √9. Note that with n=9<30, we cannot assume normality unless the population is approximately normal.
A farmer measures the weight (in pounds) of individual apples from a large orchard. The population mean is $\mu=0.35$ lb with population standard deviation $\sigma=0.08$ lb, and the population distribution is not known. Each week, the farmer repeatedly takes random samples of size $n=16$ apples and calculates the sample mean weight $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?
Answer: The sampling distribution of $\bar{x}$ has mean $0.35$ and standard deviation $0.08/\sqrt{16}$ . 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.
A company tracks the number of defects on individual circuit boards. The population mean is $\mu=1.6$ defects per board with population standard deviation $\sigma=1.2$ . Each day, an inspector randomly selects $n=49$ boards and computes the sample mean number of defects $\bar{x}$ . This is repeated many times. Which statement about the sampling distribution of $\bar{x}$ is correct?
Answer: The sampling distribution of $\bar{x}$ has mean $1.6$ and standard deviation $1.2/\sqrt{49}$ , and it is approximately normal. This problem involves defect counts with μ=1.6 and σ=1.2. For samples of size n=49, the sampling distribution of x̄ has mean 1.6 and standard deviation σ/√n = 1.2/√49 = 1.2/7 ≈ 0.171. With n=49>30, the Central Limit Theorem guarantees the sampling distribution is approximately normal regardless of the population shape. Choice A incorrectly divides by n=49 instead of √49, producing 1.2/49 ≈ 0.024 rather than the correct 1.2/7.
The weights of apples from an orchard have population mean $\mu=180$ grams. The distribution of individual apple weights is moderately skewed. A grocer repeatedly selects random samples of size $n=40$ apples and computes the sample mean weight $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?
Answer: The sampling distribution of $\bar{x}$ has mean 180 grams. This question tests understanding of unbiased estimators with moderate skewness. The mean of the sampling distribution of x̄ always equals the population mean μ = 180 grams, regardless of population skewness. With n = 40 apples (n ≥ 30), the Central Limit Theorem ensures the sampling distribution is approximately normal despite the moderate skew. The standard deviation of x̄ equals σ/√n, not σ. The sampling distribution's center is determined by the population mean, not the sample size or skewness.
A nutrition label claims the mean sodium content of a certain soup is $\mu=690$ mg per serving. A student repeatedly takes random samples of $n=16$ servings and computes the sample mean sodium content $\bar{x}$ . Assume the sampling distribution of $\bar{x}$ is approximately normal. Which statement about the sampling distribution is correct?
Answer: The mean of $\bar{x}$ is $690$ , and the standard deviation of $\bar{x}$ is the population standard deviation divided by $\sqrt{16}$ . This question focuses on sampling distributions for sample means in AP Statistics. The mean of the sampling distribution of \bar{x} is μ = 690 mg, and its standard deviation is σ/√16 = σ/4, smaller than σ. A common distractor is choice A, which uses σ/16 instead of σ/√16, likely from forgetting the square root in the standard error formula. Mini-lesson: The sampling distribution of \bar{x} is the collection of all possible sample means from samples of size n; it centers at the population mean μ, has standard deviation σ/√n (reflecting how averaging reduces variability), and is approximately normal by the CLT for large n. For n=16, the spread is quartered compared to σ. Choice B correctly states this.
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?
Answer: The sampling distribution of $\bar{x}$ has mean $8.0$ and standard deviation $5.0/\sqrt{25}$ , and it is approximately normal. 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.
A company tracks the number of minutes employees spend commuting. For all employees, the population mean commute time is $\mu=28$ minutes and the distribution is strongly right-skewed. An HR analyst repeatedly selects simple random samples of size $n=100$ employees and computes the sample mean commute time $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?
Answer: The sampling distribution of $\bar{x}$ is approximately normal and has mean 28 minutes. This question demonstrates the Central Limit Theorem with a large sample from a skewed population. With n = 100 employees (well above 30), the sampling distribution of x̄ is approximately normal despite the right-skewed population. The mean of the sampling distribution equals μ = 28 minutes; skewness doesn't affect the center of the sampling distribution. The standard deviation of x̄ equals σ/√n, which is much smaller than the population standard deviation. Large samples effectively "wash out" population skewness through the averaging process.
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?
Answer: The sampling distribution of $\bar{x}$ has mean $20$ and standard deviation $\sigma/\sqrt{36}$ , and it is approximately normal. 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.
In a city, the mean one-way commute time for all workers is $\mu=27$ minutes. A researcher repeatedly takes random samples of $n=9$ workers and records the sample mean commute time $\bar{x}$ . Assume the sampling distribution of $\bar{x}$ is approximately normal. Which statement about the sampling distribution is correct?
Answer: The mean of $\bar{x}$ is $27$ , and the standard deviation of $\bar{x}$ is smaller than the standard deviation of individual commute times. This question tests knowledge of sampling distributions for sample means in AP Statistics. The center of the sampling distribution of \bar{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 \bar{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 \bar{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.
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?
Answer: The sampling distribution of $\bar{x}$ is approximately normal and centered at 122 mmHg. 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.
A cereal company states that the mean amount of cereal in its boxes is $\mu=18$ ounces, with individual box weights varying around this mean. Suppose many random samples of $n=36$ boxes are taken, and the sample mean $\bar{x}$ is computed each time. The sampling distribution of $\bar{x}$ is approximately normal (by the CLT). Which statement about the sampling distribution is correct?
Answer: The mean of $\bar{x}$ is $18$ , and the standard deviation of $\bar{x}$ is smaller than the standard deviation of individual box weights. This question assesses understanding of sampling distributions for sample means in AP Statistics. The sampling distribution of the sample mean \bar{x} has a center (mean) equal to the population mean μ = 18 ounces, so it remains 18 regardless of sample size. The spread, measured by the standard deviation of \bar{x}, is the population standard deviation σ divided by √n, which for n=36 is σ/6, making it smaller than the standard deviation of individual box weights. A common distractor is choice C, which incorrectly divides the mean by 36, confusing the mean of the sampling distribution with something like a proportion or average per unit. In a mini-lesson on sampling distributions of means: when you take many samples of size n from a population with mean μ and standard deviation σ, the sample means form a distribution that is approximately normal (by the Central Limit Theorem for large n), centered at μ, with standard error σ/√n, which decreases as n increases, meaning larger samples give more precise estimates of μ. Choice B correctly captures this by stating the mean is 18 and the standard deviation is smaller.
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?
Answer: The sampling distribution of $\bar{x}$ is approximately normal with mean $100\ \Omega$ . 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.
A delivery service has mean delivery time $\mu=42$ minutes for a certain route. A manager repeatedly takes random samples of $n=49$ deliveries and calculates the sample mean time $\bar{x}$ . Assume the sampling distribution of $\bar{x}$ is approximately normal. Which statement about the sampling distribution is correct?
Answer: The sampling distribution of $\bar{x}$ has mean $42$ , and its standard deviation is $\sigma/\sqrt{49}$ . This AP Statistics question evaluates sampling distributions for sample means. The distribution of \bar{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 \bar{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.
A streaming service records the data usage per hour for individual users. For a typical hour, data usage has mean $\mu=1.8$ GB and standard deviation $\sigma=0.9$ GB, with a distribution that is moderately right-skewed. An analyst repeatedly selects many random samples of $n=15$ users and computes the sample mean data usage $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?
Answer: The sampling distribution of $\bar{x}$ has mean 1.8 GB and standard deviation $0.9/\sqrt{15}$ GB, and it may or may not be approximately normal depending on how skewed the population is. This question addresses uncertainty about normality with moderate skewness and borderline sample size. The sampling distribution of x̄ has mean μ = 1.8 GB and standard deviation σ/√n = 0.9/√15 ≈ 0.232 GB. However, with n = 15 and a moderately right-skewed population, we cannot definitively say whether the sampling distribution is approximately normal - it depends on the degree of skewness. Choice A incorrectly claims n = 15 guarantees normality, but this sample size is borderline for skewed populations. The critical insight is that n = 15 falls in a gray area: it might be sufficient for mild skewness but insufficient for strong skewness, highlighting that the required sample size for approximate normality depends on how far the population deviates from normality.
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?
Answer: The sampling distribution of $\bar{x}$ is centered at $520$ , and its standard deviation is smaller than the standard deviation of individual exam scores. 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.
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?
Answer: The sampling distribution of $\bar{x}$ is approximately normal and centered at 27 mpg. 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.

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