College Statistics Quiz: Sampling Distributions And Clt

20 questions · exam conditions

0:00

Sampling Distributions And CltQuestion 1 of 20

Standard formulas for standard error, such as $\sigma/\sqrt{n}$ , assume that samples are drawn with replacement or from an infinitely large population. When sampling without replacement from a finite population, these formulas are an approximation. In which of the following scenarios would the standard formula provide the least accurate estimate of the true standard error?

A sample of

n=50

is drawn from a population of

N=500,000

A sample of

n=100

is drawn from a population of

N=2,000

A sample of

n=400

is drawn from a population of

N=1,000

A sample of

n=1,000

is drawn from a population of

N=10,000,000

← Back to quizzes

College Statistics Quiz

College Statistics Quiz: Sampling Distributions And Clt

Practice Sampling Distributions And Clt in College Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Sampling Distributions And Clt, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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.

All questions

Question 1

A sample of $n=50$ is drawn from a population of $N=500,000$ .
A sample of $n=100$ is drawn from a population of $N=2,000$ .
A sample of $n=400$ is drawn from a population of $N=1,000$ . (correct answer)
A sample of $n=1,000$ is drawn from a population of $N=10,000,000$ .

Explanation: The accuracy of the standard formula depends on the ratio of sample size to population size (

n/N

). The approximation is poor when this ratio is large. A common rule of thumb is that the approximation is acceptable if

n/N \le 0.05

n/N \le 0.10

. Let's calculate the ratio for each option: A)

50/500,000 = 0.0001

100/2,000 = 0.05

400/1,000 = 0.40

1,000/10,000,000 = 0.0001

The ratio is largest in choice C, where the sample is 40% of the population. In this case, the standard formula would significantly overestimate the true standard error, making it the least accurate.

Question 2

A population of annual household incomes is known to be strongly right-skewed with a mean of $70,000 and a standard deviation of $50,000. A researcher plans to take a random sample of size $n=400$ and calculate the sample mean income, $\bar{x}$ . Which of the following statements gives the most accurate description of the sampling distribution of $\bar{x}$ ?

The distribution will be approximately normal with a mean of $70,000. (correct answer)
The distribution will be strongly right-skewed, just like the population.
The distribution will be approximately normal, and any sample drawn will also be approximately normal.
The mean of the distribution will be $70,000, but its shape cannot be determined from the information given.

Explanation: The Central Limit Theorem (CLT) states that for a sufficiently large sample size, the sampling distribution of the sample mean (

\bar{x}

) will be approximately normal, regardless of the shape of the population distribution. A sample size of

n=400

is considered large. The mean of this sampling distribution (

\mu_{\bar{x}}

) is equal to the population mean (

\mu

), which is $70,000. Therefore, the sampling distribution of

\bar{x}

will be approximately normal with a mean of $70,000.

Question 3

A company's employee performance scores are measured on a 1-to-5 integer scale. The true distribution of scores for all employees is: 10% score 1, 15% score 2, 20% score 3, 40% score 4, and 15% score 5. A manager takes a random sample of 120 employees. Which statement most accurately describes the sampling distribution of the average performance score, $\bar{x}$ ?

It will be approximately normal with a mean of 3.35. (correct answer)
It will have the same discrete, non-symmetric shape as the population distribution.
It will be approximately normal with a mean of 3.0, the median of the possible scores.
It must be analyzed with a t-distribution because the population standard deviation is unknown.

Explanation: First, find the population mean (

\mu

\mu = (1)(0.10) + (2)(0.15) + (3)(0.20) + (4)(0.40) + (5)(0.15) = 0.10 + 0.30 + 0.60 + 1.60 + 0.75 = 3.35

. The mean of the sampling distribution of

\bar{x}

is equal to

\mu

. Since the sample size

n=120

is large, the Central Limit Theorem applies, and the shape of the sampling distribution will be approximately normal, regardless of the population's discrete, non-symmetric shape. Therefore, the sampling distribution is approximately normal with a mean of 3.35.

Question 4

A researcher wants to estimate the proportion, $p$ , of people who own a specific brand of smartphone. They plan to survey a random sample of $n=120$ people. In which of the following scenarios would the normal approximation to the sampling distribution of the sample proportion, $\hat{p}$ , be the LEAST appropriate to use?

The true population proportion is believed to be $p=0.95$ . (correct answer)
The true population proportion is believed to be $p=0.50$ .
The true population proportion is believed to be $p=0.20$ .
The total population size is very large, estimated to be over 200 million.

Explanation: The normal approximation for the sampling distribution of a sample proportion is considered appropriate when the Large Counts Condition is met:

np \ge 10

and

n(1-p) \ge 10

. We check this for each choice with

n=120

. A)

n(1-p) = 120(1-0.95) = 120(0.05) = 6

. Since

6 < 10

, the condition is not met. B)

np = 120(0.50) = 60

and

n(1-p) = 120(0.50) = 60

. Both are

\ge 10

. C)

np = 120(0.20) = 24

and

n(1-p) = 120(0.80) = 96

. Both are

\ge 10

. The condition is violated when the proportion

p

is very close to 0 or 1, making choice A the least appropriate scenario.

Question 5

A statistics student claims, 'For the sampling distribution of the sample mean to be approximately normal, the sample size must be at least 30.' Which of the following scenarios presents a valid reason why this statement is not universally correct?

If the population distribution is itself normal, the sampling distribution of the mean is exactly normal for any sample size. (correct answer)
If the population distribution is strongly skewed, a sample size of 30 may not be sufficiently large for the normal approximation to be accurate.
If the population standard deviation is unknown, the t-distribution must be used, which negates the Central Limit Theorem.
The statement is a fundamental theorem of statistics and is correct in all situations involving sample means.

Explanation: The rule of thumb that

n \ge 30

is a guideline for applying the Central Limit Theorem when the population distribution is not normal. However, if the population from which samples are drawn is already normally distributed, the sampling distribution of the sample mean will be exactly normal for any sample size,

n \ge 1

. This scenario demonstrates that

n \ge 30

is a sufficient condition in many cases but not a universally necessary one.

Question 6

A political analyst calculates a 95% confidence interval for voter turnout and reports that the standard error of the sample proportion was 0.02. What does this value of 0.02 represent?

The estimated standard deviation of the voter turnout data within the single sample collected by the analyst.
The approximate average distance between the sample proportion and the true population proportion over many hypothetical samples. (correct answer)
The margin of error for the 95% confidence interval, indicating the maximum likely error.
The probability that the true population proportion is not contained within the calculated confidence interval.

Explanation: The standard error of a statistic (in this case, the sample proportion) is the standard deviation of its sampling distribution. Conceptually, it quantifies the typical amount of error, or the average distance, between the values of the statistic from all possible samples of a given size and the actual population parameter. It is not the standard deviation of a single sample's data (A), nor is it the full margin of error (C), which is typically

z^* \times SE

. It is also not a probability like alpha (D).

Question 7

The formula for the standard error of a sample mean, $\sigma/\sqrt{n}$ , relies on the assumption that individual observations are independent. This assumption is technically violated when sampling without replacement. Why is it still considered acceptable to use this formula when the sample size $n$ is less than 10% of the population size $N$ ?

Because for a small sample fraction, the sampling distribution remains perfectly normal.
Because the sample mean $\bar{x}$ becomes an unbiased estimator only when the 10% condition is met.
Because the 10% condition ensures that the sample size is large enough for the Central Limit Theorem to apply.
Because removing a small fraction of the population does not meaningfully change the remaining probabilities, so the observations are 'nearly' independent. (correct answer)

Explanation: When you encounter questions about sampling without replacement, you need to understand why we can still use independence-based formulas despite technically violating the independence assumption. The key insight is that independence in sampling is about whether selecting one item affects the probability of selecting another. When you sample without replacement from a large population, removing a few observations barely changes the composition of the remaining population. If you remove 50 people from a population of 10,000, the characteristics of the remaining 9,950 people are virtually unchanged. This makes successive draws "nearly independent" even though they're technically dependent. The correct answer is D because it captures this practical reality. When your sample is less than 10% of the population, the finite population correction factor

\sqrt{\frac{N-n}{N-1}}

is so close to 1 that it's negligible, meaning the standard error formula

\sigma/\sqrt{n}

remains highly accurate. Answer A is wrong because normality of the sampling distribution isn't the issue here—we're discussing independence, not distribution shape. Answer B incorrectly suggests bias is the concern; the sample mean is unbiased regardless of the 10% condition. Answer C confuses the 10% rule with sample size requirements for the Central Limit Theorem—these are separate concepts. Remember this pattern: The 10% rule is about maintaining "practical independence" when sampling without replacement. It's not about normality, bias, or CLT requirements—it's purely about keeping dependence effects negligible.

Question 8

Two different populations of component lifetimes are being studied. Population X has a normal distribution with mean $\mu_X = 1000$ hours and standard deviation $\sigma_X = 100$ hours. Population Y has an exponential distribution (which is highly right-skewed) with mean $\mu_Y = 1000$ hours and standard deviation $\sigma_Y = 1000$ hours. For which of the following sample sizes $n$ would the sampling distribution of $\bar{x}$ from Population X be most similar in shape to the sampling distribution of $\bar{y}$ from Population Y?

$n=5$ for both
$n=50$ for both (correct answer)
$n=5$ for X and $n=50$ for Y
$n=50$ for X and $n=5$ for Y

Explanation: The sampling distribution from Population X will be exactly normal for any sample size because the population is normal. The sampling distribution from Population Y will become approximately normal for a large sample size due to the CLT. For the two sampling distributions to be most similar in shape (i.e., both approximately normal), we need a large enough sample size for the CLT to work its effect on Population Y. With

n=5

, the sampling distribution from Y would still be quite skewed. With

n=50

for both, the sampling distribution from X is exactly normal, and the sampling distribution from Y is approximately normal, making them very similar in shape.

Question 9

A quality control specialist is examining the standard deviation of the sampling distribution of the mean weight of a product. For a sample of size $n$ , the standard error is $\sigma_{\bar{x}}$ . If the specialist wants to reduce this standard error to one-third of its original value (i.e., to $\frac{1}{3}\sigma_{\bar{x}}$ ), what new sample size must be taken?

$3n$
$9n$ (correct answer)
$n/3$
$n/9$

Explanation: The standard error of the mean is given by the formula

\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}

, where

\sigma

is the population standard deviation and

n

is the sample size. To reduce the standard error to

\frac{1}{3}

of its original value, we need to solve for the new sample size,

n'

\frac{\sigma}{\sqrt{n'}} = \frac{1}{3} \left( \frac{\sigma}{\sqrt{n}} \right)

. This simplifies to

\sqrt{n'} = 3\sqrt{n}

. Squaring both sides gives

n' = 9n

. Therefore, the sample size must be multiplied by 9.

Question 10

It will be approximately normal with a mean of 3.35. (correct answer)
It will have the same discrete, non-symmetric shape as the population distribution.
It will be approximately normal with a mean of 3.0, the median of the possible scores.
It must be analyzed with a t-distribution because the population standard deviation is unknown.

Explanation: First, find the population mean (

\mu

\mu = (1)(0.10) + (2)(0.15) + (3)(0.20) + (4)(0.40) + (5)(0.15) = 0.10 + 0.30 + 0.60 + 1.60 + 0.75 = 3.35

. The mean of the sampling distribution of

\bar{x}

is equal to

\mu

. Since the sample size

n=120

Question 11

A population's distribution is strongly bimodal, with peaks at 10 and 90. A researcher draws a random sample of size $n=4$ from this population and computes the sample mean. If this process were repeated many times, what would be the most likely shape of the resulting distribution of sample means?

Strongly bimodal, with peaks at 10 and 90, identical to the population.
Approximately normal and centered near 50, due to the Central Limit Theorem.
Unimodal and roughly symmetric, but likely not perfectly normal, centered near 50. (correct answer)
A uniform distribution spanning the range between 10 and 90.

Explanation: With a very small sample size like

n=4

, the Central Limit Theorem will not have a strong enough effect to make the sampling distribution fully normal, especially from a strongly bimodal population. However, the process of averaging will still tend to pull the means away from the extremes and toward the center. For example, a sample might contain values from both modes, resulting in a mean near 50. The resulting distribution will therefore lose the strong bimodality of the parent population and tend towards a central peak, but it will not yet be bell-shaped. It will be more peaked and less bimodal than the population.

Question 12

The distribution will be approximately normal with a mean of $70,000. (correct answer)
The distribution will be strongly right-skewed, just like the population.
The distribution will be approximately normal, and any sample drawn will also be approximately normal.
The mean of the distribution will be $70,000, but its shape cannot be determined from the information given.

Explanation: The Central Limit Theorem (CLT) states that for a sufficiently large sample size, the sampling distribution of the sample mean (

\bar{x}

) will be approximately normal, regardless of the shape of the population distribution. A sample size of

n=400

is considered large. The mean of this sampling distribution (

\mu_{\bar{x}}

) is equal to the population mean (

\mu

), which is $70,000. Therefore, the sampling distribution of

\bar{x}

will be approximately normal with a mean of $70,000.

Question 13

Let Population A be a normal distribution with mean 50 and standard deviation 12. Let Population B be a uniform distribution on the interval [0, 100], which has a mean of 50. A researcher creates sampling distribution $S_A$ by taking samples of size $n=4$ from Population A, and sampling distribution $S_B$ by taking samples of size $n=36$ from Population B. Which statement best compares the characteristics of $S_A$ and $S_B$ ?

Both $S_A$ and $S_B$ will be exactly normal and will have the same mean.
$S_A$ will be exactly normal, while $S_B$ will be approximately normal, and both will have the same mean. (correct answer)
Both $S_A$ and $S_B$ will be only approximately normal, but $S_A$ will have a smaller standard error.
$S_A$ will be exactly normal, but its mean will be smaller than the mean of $S_B$ .

Explanation: The mean of any sampling distribution of the mean is equal to the population mean, so both

S_A

and

S_B

will have a mean of 50. Since Population A is normal, the sampling distribution

S_A

will be exactly normal, even with a small sample size of

n=4

. Population B is not normal (it's uniform), but since the sample size

n=36

is large, the Central Limit Theorem states that the sampling distribution

S_B

will be approximately normal. Therefore,

S_A

is exactly normal,

S_B

is approximately normal, and they share the same mean.

Question 14

The sample mean ( $\bar{x}$ ) is often described as an 'unbiased estimator' of the population mean ( $\mu$ ). What is the precise statistical meaning of this property?

For any single random sample, the calculated sample mean $\bar{x}$ will be equal to the true population mean $\mu$ .
As the sample size gets infinitely large, the value of the sample mean $\bar{x}$ will get progressively closer to the population mean $\mu$ .
The mean of the theoretical sampling distribution of $\bar{x}\, taken over all possible samples of a given size, is equal to \(\mu$ . (correct answer)
The sampling distribution of $\bar{x}$ is guaranteed to be symmetric, with the population mean $\mu$ at its center.

Explanation: Unbiasedness of an estimator means that the expected value of the estimator is equal to the population parameter it is estimating. For the sample mean, this translates to the idea that the mean of its sampling distribution is equal to the population mean. It does not mean any particular sample mean will equal the population mean (Choice A), nor does it describe convergence as

n

increases (Choice B, which is the Law of Large Numbers). It also doesn't guarantee symmetry (Choice D), although the distribution may be symmetric.

Question 15

A researcher takes a single random sample of $n=40$ exam scores from a large population of scores, which is known to be left-skewed. Which statement correctly distinguishes between the 'distribution of the sample' and the 'sampling distribution of the sample mean' in this context?

Both the distribution of the sample and the sampling distribution will be approximately normal due to the Central Limit Theorem.
The distribution of the sample will likely be left-skewed, while the sampling distribution of the sample mean will be approximately normal. (correct answer)
The distribution of the sample will be approximately normal, while the sampling distribution of the sample mean will be left-skewed.
There is no difference; the two terms describe the same concept of the distribution of 40 exam scores.

Explanation: These are two distinct concepts. The 'distribution of the sample' refers to the distribution of the 40 actual data points collected. As it's a random sample from a left-skewed population, it is also expected to be left-skewed. The 'sampling distribution of the sample mean' is a theoretical distribution of all possible sample means of size 40. Since the sample size

n=40

is large, the Central Limit Theorem applies, and this theoretical distribution will be approximately normal.

Question 16

The number of typos per page in a book manuscript follows a discrete, right-skewed distribution with a mean of 2.5 typos and a standard deviation of 1.5 typos. An editor randomly selects 50 pages and calculates the average number of typos per page, $\bar{x}$ . Which statement best describes the sampling distribution of $\bar{x}$ ?

It is a discrete, right-skewed distribution with a mean of 2.5.
It is approximately normal with a mean of 2.5 and a standard deviation of 1.5.
It is approximately normal with a mean of 2.5 and a standard deviation of $1.5/\sqrt{50}$ . (correct answer)
Its shape cannot be determined because the Central Limit Theorem does not apply to discrete populations.

Explanation: The Central Limit Theorem applies to samples from any population distribution, including discrete and skewed ones, provided the sample size is sufficiently large. Here,

n=50

is large enough. The CLT states that the sampling distribution of

\bar{x}

will be approximately normal. Its mean,

\mu_{\bar{x}}

, is equal to the population mean,

\mu = 2.5

. Its standard deviation (the standard error),

\sigma_{\bar{x}}

, is the population standard deviation divided by the square root of the sample size,

\sigma/\sqrt{n} = 1.5/\sqrt{50}

Question 17

Strongly bimodal, with peaks at 10 and 90, identical to the population.
Approximately normal and centered near 50, due to the Central Limit Theorem.
Unimodal and roughly symmetric, but likely not perfectly normal, centered near 50. (correct answer)
A uniform distribution spanning the range between 10 and 90.

Explanation: With a very small sample size like

n=4

Question 18

A simulation is conducted by taking 5,000 separate random samples of size $n=36$ from a population with a mean $\mu=100$ and standard deviation $\sigma=18$ . The mean of each of the 5,000 samples is calculated and stored. If a histogram is created from these 5,000 sample means, what would be the best estimate for the standard deviation of the values shown in this histogram?

18
3 (correct answer)
0.36
0.25

Explanation: The histogram described is an empirical model of the sampling distribution of the sample mean (

\bar{x}

). The standard deviation of this distribution is the standard error,

\sigma_{\bar{x}}

. The formula is

\sigma_{\bar{x}} = \sigma / \sqrt{n}

. Using the given values,

\sigma_{\bar{x}} = 18 / \sqrt{36} = 18 / 6 = 3

. The number of samples in the simulation (5,000) is irrelevant to the calculation of the theoretical standard error.

Question 19

A machine fills bottles with a mean of 500 ml and a standard deviation of 2 ml. The distribution of fill volumes is symmetric and unimodal. An inspector takes a random sample of 16 bottles. What is the mean of the sampling distribution of the sample mean fill volume ( $\bar{x}$ )?

$500 / 16 = 31.25$ ml
$500 / \sqrt{16} = 125$ ml
500 ml (correct answer)
It cannot be determined without knowing the exact shape of the population distribution.

Explanation: A fundamental property of sampling distributions is that the mean of the sampling distribution of the sample mean (

\mu_{\bar{x}}

) is always equal to the mean of the population (

\mu

) from which the samples are drawn. This is true regardless of the sample size or the shape of the population distribution. Given the population mean is 500 ml, the mean of the sampling distribution of

\bar{x}

is also 500 ml.

Question 20

The service time at a bank teller is a random variable with a mean of 3.5 minutes and a standard deviation of 1.2 minutes. Assume the service times for different customers are independent. A manager observes a random sample of 64 customers. According to the Central Limit Theorem, the distribution of the total service time for these 64 customers can be approximated by which of the following?

A normal distribution with a mean of 3.5 minutes and a standard deviation of 0.15 minutes.
A normal distribution with a mean of 224 minutes and a standard deviation of 1.2 minutes.
A distribution with the same shape as the original service time distribution, but with a mean of 224 minutes.
A normal distribution with a mean of 224 minutes and a standard deviation of 9.6 minutes. (correct answer)

Explanation: When you encounter problems about the sum or total of independent random variables, you need to apply the properties of sums along with the Central Limit Theorem. The key insight is that we're looking at the distribution of the total service time, not the sample mean. For the sum of n independent random variables, each with mean μ and standard deviation σ, the sum has mean nμ and standard deviation σ√n. Here, with 64 customers each having mean service time 3.5 minutes, the total has mean 64 × 3.5 = 224 minutes. For the standard deviation of the total: σ√n = 1.2√64 = 1.2 × 8 = 9.6 minutes. The Central Limit Theorem tells us this sum will be approximately normally distributed regardless of the original distribution's shape. Choice A incorrectly uses the standard error for a sample mean (σ/√n = 1.2/8 = 0.15) rather than the standard deviation of a sum, and uses the wrong mean. Choice B has the correct mean but uses the original standard deviation (1.2) instead of the sum's standard deviation. Choice C correctly identifies the mean as 224 but fails to recognize that the Central Limit Theorem guarantees the sum will be normally distributed, not maintain the original shape. Therefore, D is correct: a normal distribution with mean 224 minutes and standard deviation 9.6 minutes. Remember: for sums of random variables, multiply the mean by n and multiply the standard deviation by √n. Don't confuse this with sample means, where you divide the standard deviation by √n.