Business Statistics Quiz: Sampling Distributions And Clt
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Sampling Distributions And CltQuestion 1 of 20

A financial analyst wants to estimate the average annual return of a particular stock. The standard deviation of annual returns is known to be 15%. The analyst wishes to reduce the standard error of their estimate to be only one-third of what it would be for a sample of size n=25n=25. What sample size should the analyst use?

75
150
225
625
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Business Statistics Quiz

Business Statistics Quiz: Sampling Distributions And Clt

Practice Sampling Distributions And Clt in Business 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 Business 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.

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Question 1

A financial analyst wants to estimate the average annual return of a particular stock. The standard deviation of annual returns is known to be 15%. The analyst wishes to reduce the standard error of their estimate to be only one-third of what it would be for a sample of size n=25n=25. What sample size should the analyst use?

  1. 75
  2. 150
  3. 225 (correct answer)
  4. 625
Explanation: The standard error of the mean is given by σxˉ=σ/n\sigma_{\bar{x}} = \sigma / \sqrt{n}. Let the initial sample size be n1=25n_1 = 25 and the new sample size be n2n_2. The goal is for σxˉ2=(1/3)σxˉ1\sigma_{\bar{x}_2} = (1/3) \sigma_{\bar{x}_1}. Substituting the formula: σ/n2=(1/3)(σ/n1)\sigma / \sqrt{n_2} = (1/3) (\sigma / \sqrt{n_1}). The σ\sigma terms cancel out. 1/n2=(1/3)(1/25)=(1/3)(1/5)=1/151 / \sqrt{n_2} = (1/3) (1 / \sqrt{25}) = (1/3) (1/5) = 1/15. Therefore, n2=15\sqrt{n_2} = 15, which means n2=152=225n_2 = 15^2 = 225.

Question 2

The time required to complete a specific assembly task is uniformly distributed between 20 and 30 minutes. An operations manager takes a random sample of 48 assemblies. What is the approximate probability that the average assembly time for this sample is between 24.5 and 25.5 minutes?

  1. 0.100
  2. 0.683
  3. 0.789 (correct answer)
  4. 0.954
Explanation: First, find the mean and standard deviation of the uniform population distribution. Mean μ=(a+b)/2=(20+30)/2=25\mu = (a+b)/2 = (20+30)/2 = 25 minutes. Variance σ2=(b−a)2/12=(30−20)2/12=100/12\sigma^2 = (b-a)^2/12 = (30-20)^2/12 = 100/12. Standard deviation σ=100/12≈2.887\sigma = \sqrt{100/12} \approx 2.887 minutes. Second, since n=48n=48 is large, apply the CLT. The standard error is σxˉ=σ/n=2.887/48≈0.4167\sigma_{\bar{x}} = \sigma / \sqrt{n} = 2.887 / \sqrt{48} \approx 0.4167. Third, find the Z-scores for the interval boundaries: Z1=(24.5−25)/0.4167=−1.20Z_1 = (24.5 - 25) / 0.4167 = -1.20 and Z2=(25.5−25)/0.4167=1.20Z_2 = (25.5 - 25) / 0.4167 = 1.20. Finally, find the probability P(−1.20<Z<1.20)=P(Z<1.20)−P(Z<−1.20)=0.8849−0.1151=0.7698P(-1.20 < Z < 1.20) = P(Z < 1.20) - P(Z < -1.20) = 0.8849 - 0.1151 = 0.7698, which is approximately 0.789.

Question 3

A pharmaceutical company measures reaction times to a new drug. The population has mean 12.5 minutes and standard deviation 4.2 minutes. Clinical trials use samples of size 64. If a trial reports a sample mean of 11.8 minutes, what is the approximate z-score for this result?

  1. -1.33 (correct answer)
  2. -0.17
  3. -1.67
  4. -0.70
Explanation: The standard error is σxˉ=4.264=4.28=0.525\sigma_{\bar{x}} = \frac{4.2}{\sqrt{64}} = \frac{4.2}{8} = 0.525. The z-score is z=11.8−12.50.525=−0.70.525≈−1.33z = \frac{11.8 - 12.5}{0.525} = \frac{-0.7}{0.525} ≈ -1.33. Choice B uses the population standard deviation instead of standard error: −0.74.2≈−0.17\frac{-0.7}{4.2} ≈ -0.17. Choice C makes an arithmetic error in the calculation. Choice D uses an incorrect standard error calculation.

Question 4

A logistics company tracks delivery times with population mean 45 minutes and standard deviation 12 minutes. A dispatcher samples 36 deliveries and calculates that the sample mean is 42 minutes. The company policy requires investigation if a sample mean is more than 1.5 standard errors below the population mean. Should this sample trigger an investigation?

  1. No, because the z-score is -1.5, which equals but does not exceed the threshold
  2. Yes, because the z-score is -1.5, which meets the 'more than 1.5' criterion when considering absolute value
  3. No, because the z-score is -1.5, which does not exceed the threshold of more than 1.5 standard errors (correct answer)
  4. Yes, because the sample mean of 42 is significantly different from the population mean of 45
Explanation: The standard error is σxˉ=1236=2\sigma_{\bar{x}} = \frac{12}{\sqrt{36}} = 2. The z-score is z=42−452=−1.5z = \frac{42-45}{2} = -1.5. The policy requires investigation if the sample mean is MORE THAN 1.5 standard errors below the mean. Since |-1.5| = 1.5 exactly, it does not exceed the threshold. Choice A correctly calculates but misinterprets the 'more than' criterion. Choice B incorrectly applies absolute value logic. Choice D ignores the specific policy threshold.

Question 5

An economist studies household incomes in a metropolitan area where the population mean is $67,000 with standard deviation $18,000. Two research teams independently collect samples: Team A samples 25 households, Team B samples 100 households. If both teams calculate sample means, which comparison of their sampling distributions is correct?

  1. Team B's sample mean has smaller standard error and is more likely to be normally distributed (correct answer)
  2. Team A's sample mean has larger standard error but both are equally likely to be normally distributed
  3. Both teams have identical standard errors since they sample from the same population
  4. Team A's sample mean is more reliable because smaller samples reduce bias
Explanation: Team A's standard error is 1800025=3600\frac{18000}{\sqrt{25}} = 3600, while Team B's is 18000100=1800\frac{18000}{\sqrt{100}} = 1800. Team B has smaller standard error (more precise). For normality, Team B's larger sample size (n=100 vs n=25) makes the CLT more effective, so Team B's sampling distribution is more likely to be normal. Choice B incorrectly states equal normality likelihood. Choice C ignores the sample size effect on standard error. Choice D incorrectly suggests smaller samples are more reliable.

Question 6

An agricultural researcher studies wheat yields per acre. The population distribution is unknown but has mean 58 bushels and standard deviation 14 bushels. She plans to sample n farms and wants the probability that the sample mean falls within 3 bushels of the population mean to be approximately 0.90. What minimum sample size should she use?

  1. n = 34
  2. n = 47
  3. n = 59 (correct answer)
  4. n = 65
Explanation: When you encounter a problem asking for sample size to achieve a specific probability about the sample mean, you're working with the Central Limit Theorem and the sampling distribution of the mean. The key insight is that sample means follow a normal distribution with mean μ=58\mu = 58 and standard error σxˉ=σn=14n\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{14}{\sqrt{n}}. You want P(∣xˉ−58∣≤3)=0.90P(|\bar{x} - 58| \leq 3) = 0.90, which means the sample mean should fall within 3 bushels of 58 with 90% probability. For a 90% probability, you need the interval to extend 1.645 standard errors on each side of the mean (using the z-table). So: 1.645×14n=31.645 \times \frac{14}{\sqrt{n}} = 3 Solving for n: 14×1.6453=n\frac{14 \times 1.645}{3} = \sqrt{n}, which gives n=7.677\sqrt{n} = 7.677, so n=58.9≈59n = 58.9 \approx 59. Option A (n = 34) is too small—this would give a standard error of about 2.4, making the margin of error about 3.95 bushels, which exceeds the desired 3 bushels. Option B (n = 47) is closer but still insufficient, yielding a margin of error around 3.36 bushels. Option D (n = 65) would work but represents oversampling—it's larger than necessary and wastes resources. Therefore, C (n = 59) is correct. Remember: when determining sample size for a given margin of error, always round up to ensure you meet the minimum probability requirement, and use the appropriate z-score for your desired confidence level.

Question 7

A financial analyst examines daily stock returns for a volatile stock. Returns have mean 0.12% and standard deviation 2.8%. She calculates the average return over 20 trading days and repeats this process 100 times, creating 100 sample means. Approximately how many of these 100 sample means should fall within one standard error of the population mean?

  1. 34
  2. 68 (correct answer)
  3. 95
  4. 47
Explanation: When you encounter questions about sample means and their distribution, you're working with the Central Limit Theorem and sampling distributions. The key insight is that sample means follow a normal distribution centered on the population mean, with their own standard deviation called the standard error. Here's how to solve this: The standard error equals the population standard deviation divided by the square root of sample size: SE=2.8%20=0.626%SE = \frac{2.8\%}{\sqrt{20}} = 0.626\%. Since we have 100 sample means that are normally distributed around the population mean of 0.12%, we can apply the empirical rule (68-95-99.7 rule). This rule states that approximately 68% of values in a normal distribution fall within one standard deviation of the mean. In this case, about 68% of the 100 sample means should fall within one standard error of 0.12%, which equals 100×0.68=68100 \times 0.68 = 68 sample means. Answer A (34) represents half of 68%, suggesting confusion about one-sided versus two-sided intervals. Answer C (95) incorrectly applies the two standard deviation rule (95% within two standard deviations) instead of the one standard deviation rule. Answer D (47) has no clear statistical basis and likely represents a calculation error. Remember that the empirical rule is your friend: 68% within one standard deviation, 95% within two, and 99.7% within three. When you see "within one standard error" for sample means, immediately think 68% of your samples.

Question 8

A manufacturing process creates bolts with a specified mean diameter of 20 mm and a population variance of 0.16 mm². A quality check involves sampling 64 bolts. What is the probability that the sample mean diameter will be between 19.875 mm and 20.05 mm?

  1. 0.1056
  2. 0.4938
  3. 0.8351 (correct answer)
  4. 0.9938
Explanation: First, note that the population variance (σ2\sigma^2) is 0.16 mm², so the standard deviation (σ\sigma) is 0.16=0.4\sqrt{0.16} = 0.4 mm. Next, calculate the standard error of the mean for a sample of size 64: σxˉ=σ/n=0.4/64=0.4/8=0.05\sigma_{\bar{x}} = \sigma / \sqrt{n} = 0.4 / \sqrt{64} = 0.4 / 8 = 0.05 mm. Then, calculate the Z-scores for the interval boundaries: Zlower=(19.875−20.00)/0.05=−0.125/0.05=−2.5Z_{lower} = (19.875 - 20.00) / 0.05 = -0.125 / 0.05 = -2.5. Zupper=(20.05−20.00)/0.05=0.05/0.05=1.0Z_{upper} = (20.05 - 20.00) / 0.05 = 0.05 / 0.05 = 1.0. Finally, find the probability P(−2.5<Z<1.0)=P(Z<1.0)−P(Z<−2.5)=0.8413−0.0062=0.8351P(-2.5 < Z < 1.0) = P(Z < 1.0) - P(Z < -2.5) = 0.8413 - 0.0062 = 0.8351.

Question 9

The monthly data usage of a population of smartphone users follows a distribution with a mean of 12 GB and a standard deviation of 5 GB. A random sample of 100 users is taken. There is an approximately 95% probability that the sample mean will fall between which two values?

  1. 11.02 GB and 12.98 GB (correct answer)
  2. 7.00 GB and 17.00 GB
  3. 11.50 GB and 12.50 GB
  4. 2.20 GB and 21.80 GB
Explanation: For a 95% probability interval around the mean, we use a Z-score of approximately ±1.96\pm 1.96. First, calculate the standard error: σxˉ=σ/n=5/100=0.5\sigma_{\bar{x}} = \sigma / \sqrt{n} = 5 / \sqrt{100} = 0.5 GB. The margin of error is Z×σxˉ=1.96×0.5=0.98Z \times \sigma_{\bar{x}} = 1.96 \times 0.5 = 0.98 GB. The interval is μ±margin of error\mu \pm \text{margin of error}, which is 12±0.9812 \pm 0.98. This gives a lower bound of 12−0.98=11.0212 - 0.98 = 11.02 GB and an upper bound of 12+0.98=12.9812 + 0.98 = 12.98 GB.

Question 10

Let XX be a random variable for the weekly sales at a retail store, with \mu = \25,000andand\sigma = $4,500.Let. Let \bar{X}{36}bethesamplemeanofweeklysalesfromarandomsampleof36weeks.Whichstatementcorrectlycomparesbe the sample mean of weekly sales from a random sample of 36 weeks. Which statement correctly comparesP(X > $26,500)andandP(\bar{X}{36} > $26,500),assuming, assuming X$ is not normally distributed?

  1. P(\bar{X}_{36} > \26,500)isgreaterthanis greater thanP(X > $26,500)$ because averaging smooths out low values.
  2. P(\bar{X}_{36} > \26,500)issmallerthanis smaller thanP(X > $26,500)$ because the sampling distribution has less variability. (correct answer)
  3. The two probabilities are approximately equal because they are centered on the same mean.
  4. The relationship cannot be determined because the distribution of XX is unknown.
Explanation: The sampling distribution of the mean Xˉ36\bar{X}_{36} has a standard deviation (standard error) of \sigma_{\bar{x}} = \sigma / \sqrt{n} = 4500 / \sqrt{36} = \750.Thisismuchsmallerthanthepopulationstandarddeviationof. This is much smaller than the population standard deviation of \sigma = $4,500.Avalueof$26,500is. A value of $26,500 is $1500/$4500 = 1/3ofastandarddeviationabovethemeanforanindividualobservation.Forthesamplemean,$26,500isof a standard deviation above the mean for an individual observation. For the sample mean, $26,500 is$1500/$750 = 2standarderrorsabovethemean.Anoutcome2standardunitsfromthemeanismuchlesslikelythananoutcome1/3ofastandardunitaway.Therefore,standard errors above the mean. An outcome 2 standard units from the mean is much less likely than an outcome 1/3 of a standard unit away. Therefore,P(\bar{X}_{36} > $26,500)$ is significantly smaller.

Question 11

The sampling distribution of the mean for samples of size n=49n=49 taken from a population is found to be approximately normal with a mean of 150 and a standard deviation of 4. What is the standard deviation of the original population?

  1. 4
  2. 0.57
  3. 28 (correct answer)
  4. 196
Explanation: The standard deviation of the sampling distribution of the mean is the standard error, σxˉ\sigma_{\bar{x}}. The formula relating it to the population standard deviation σ\sigma is σxˉ=σ/n\sigma_{\bar{x}} = \sigma / \sqrt{n}. We are given σxˉ=4\sigma_{\bar{x}} = 4 and n=49n=49. We need to solve for σ\sigma. Rearranging the formula gives σ=σxˉ×n\sigma = \sigma_{\bar{x}} \times \sqrt{n}. Plugging in the values: σ=4×49=4×7=28\sigma = 4 \times \sqrt{49} = 4 \times 7 = 28.

Question 12

A quality control process at a battery factory measures cell voltage. The population of voltages is known to be slightly skewed with μ=3.7\mu = 3.7 volts and σ=0.2\sigma = 0.2 volts. An inspector will sample a number of batteries and calculate the mean voltage. For which of the following sample sizes and justifications is the use of a normal distribution to calculate probabilities about the sample mean most appropriate?

  1. n=10n=10, because the sampling distribution is always normal regardless of sample size.
  2. n=20n=20, because the standard error will be small enough to ensure normality.
  3. n=40n=40, because this sample size is large enough for the Central Limit Theorem to apply. (correct answer)
  4. Any sample size is appropriate, since the population standard deviation is known.
Explanation: The Central Limit Theorem (CLT) states that the sampling distribution of the mean becomes approximately normal as the sample size increases. A common rule of thumb is that a sample size of n≥30n \ge 30 is sufficient for the CLT to apply, even if the population is not normal. For a skewed population, a larger sample size is better. Among the choices, n=40n=40 is the only one that safely meets this condition. Choice A is false. Choice B's justification is incorrect; a small standard error doesn't cause normality. Choice D is incorrect; knowing σ\sigma doesn't guarantee normality of the sampling distribution for small samples from a non-normal population.

Question 13

The number of customers entering a store per hour follows a distribution with a mean of 90 and a variance of 225. During a 36-hour sales event, what is the probability that the average number of customers per hour is less than 85?

  1. 0.0228 (correct answer)
  2. 0.1587
  3. 0.3694
  4. 0.4772
Explanation: First, find the population standard deviation from the given variance: σ=225=15\sigma = \sqrt{225} = 15. Second, since the sample size is large (n=36n=36), the CLT applies. Calculate the standard error of the mean: σxˉ=σ/n=15/36=15/6=2.5\sigma_{\bar{x}} = \sigma / \sqrt{n} = 15 / \sqrt{36} = 15 / 6 = 2.5. Third, calculate the Z-score for a sample mean of 85: Z=(xˉ−μ)/σxˉ=(85−90)/2.5=−5/2.5=−2.0Z = (\bar{x} - \mu) / \sigma_{\bar{x}} = (85 - 90) / 2.5 = -5 / 2.5 = -2.0. Finally, find the probability P(Z<−2.0)P(Z < -2.0), which from the standard normal table is approximately 0.0228.

Question 14

A psychologist studies reaction times to visual stimuli. The population has mean 480 milliseconds and standard deviation 75 milliseconds. She observes that when she takes samples of size 9, the sample means vary considerably, but when she takes samples of size 36, they vary much less. If she increases her sample size to 144, by what factor will the standard error decrease compared to samples of size 36?

  1. 1.5
  2. 4
  3. 0.5
  4. 2 (correct answer)
Explanation: When you encounter questions about how sample size affects the variability of sample means, you're dealing with the standard error of the mean, which measures how much sample means vary from the true population mean. The standard error formula is SE=σnSE = \frac{\sigma}{\sqrt{n}}, where σ is the population standard deviation and n is the sample size. Notice that sample size appears in the denominator under a square root, which is crucial for solving this problem. To find how the standard error changes when sample size increases from 36 to 144, you need to compare the ratios. For n = 36: SE36=7536=756SE_{36} = \frac{75}{\sqrt{36}} = \frac{75}{6}. For n = 144: SE144=75144=7512SE_{144} = \frac{75}{\sqrt{144}} = \frac{75}{12}. The factor by which standard error decreases is SE36SE144=75/675/12=126=2\frac{SE_{36}}{SE_{144}} = \frac{75/6}{75/12} = \frac{12}{6} = 2. So the standard error decreases by a factor of 2, making D correct. Looking at the wrong answers: A (1.5) might come from incorrectly calculating the square root relationship. B (4) represents the ratio of sample sizes (144/36 = 4), but this ignores the square root in the formula. C (0.5) is the reciprocal of the correct answer – you might get this if you calculated the ratio backwards. Remember this pattern: when sample size increases by a factor of k², the standard error decreases by a factor of k. Since 144 = 4² × 9 and 36 = 4 × 9, the sample size increased by 4, so standard error decreases by 2.

Question 15

The amount of coffee dispensed by a machine is normally distributed with a mean of 240 ml and a standard deviation of 5 ml. An office manager takes a small sample of 4 cups. What is the probability that the average amount of coffee in these 4 cups is less than 236 ml?

  1. 0.0548 (correct answer)
  2. 0.1587
  3. 0.2119
  4. This cannot be calculated as the sample size is less than 30.
Explanation: Because the original population is stated to be normally distributed, the sampling distribution of the mean will also be exactly normal, regardless of the sample size. The condition n≥30n \ge 30 is only required when the population distribution is unknown or not normal. The standard error is σxˉ=σ/n=5/4=2.5\sigma_{\bar{x}} = \sigma / \sqrt{n} = 5 / \sqrt{4} = 2.5 ml. The Z-score is Z=(236−240)/2.5=−4/2.5=−1.6Z = (236 - 240) / 2.5 = -4 / 2.5 = -1.6. The probability P(Z<−1.6)P(Z < -1.6) is approximately 0.0548.

Question 16

The average number of daily transactions at a bank branch is 450 with a standard deviation of 80. An audit is conducted by sampling 64 randomly chosen days. How does the probability of the sample mean being greater than 470 compare for the sample of 64 days versus a larger sample of 100 days?

  1. The probability is larger for the 100-day sample.
  2. The probability is smaller for the 100-day sample. (correct answer)
  3. The probability is the same for both sample sizes.
  4. The relationship cannot be determined without knowing the population distribution.
Explanation: For n=64n=64, the standard error is σxˉ=80/64=10\sigma_{\bar{x}} = 80 / \sqrt{64} = 10. The Z-score for xˉ=470\bar{x}=470 is Z=(470−450)/10=2.0Z = (470-450)/10 = 2.0. For n=100n=100, the standard error is σxˉ=80/100=8\sigma_{\bar{x}} = 80 / \sqrt{100} = 8. The Z-score for xˉ=470\bar{x}=470 is Z=(470−450)/8=2.5Z = (470-450)/8 = 2.5. Since a Z-score of 2.5 is further in the tail of the distribution than a Z-score of 2.0, the probability P(Z>2.5)P(Z > 2.5) is smaller than P(Z>2.0)P(Z > 2.0). Increasing the sample size makes the sampling distribution tighter around the mean, making large deviations less likely.

Question 17

A manager of a fast-food restaurant finds that the average service time per customer is 150 seconds, with a standard deviation of 30 seconds. The distribution of service times is skewed to the right. The manager wants to estimate the probability that the average service time for a sample of customers is between 145 and 155 seconds. Which of the following sample sizes would provide the most reliable estimate using a normal approximation?

  1. n=5n=5, because the population mean is known.
  2. n=15n=15, because this is a moderately sized sample.
  3. n=25n=25, because it is close to the threshold of 30.
  4. n=50n=50, because a larger sample better compensates for population skewness. (correct answer)
Explanation: The Central Limit Theorem allows us to use a normal distribution to approximate the sampling distribution of the mean. However, the quality of this approximation depends on the sample size and the skewness of the population. For a highly skewed population, a larger sample size is required for the sampling distribution to be sufficiently bell-shaped. While n≥30n \ge 30 is a general guideline, a sample size of 50 would provide a more reliable and accurate approximation than smaller sample sizes like 15 or 25, especially given the population is skewed.

Question 18

A machine fills bags with flour. The weight of the flour is normally distributed with a mean of 5 lbs and a standard deviation of 0.1 lbs. A sample of 4 bags is taken. An engineer calculates P(Xˉ>5.08)≈0.0548P(\bar{X} > 5.08) \approx 0.0548. Which of the following is the most accurate interpretation of this result?

  1. There is a 5.48% chance that any single bag of flour will weigh more than 5.08 lbs.
  2. 94.52% of all bags of flour produced by the machine weigh less than 5.08 lbs.
  3. In a single sample of 4 bags, exactly 5.48% of the bags will weigh more than 5.08 lbs.
  4. If many random samples of 4 bags are taken, about 5.48% of them will have a sample mean weight greater than 5.08 lbs. (correct answer)
Explanation: When you encounter probability statements involving sample means (denoted as Xˉ\bar{X}), you're dealing with sampling distribution concepts, not individual observations. The key distinction is whether the probability refers to a single sample's characteristics or the long-run behavior across many samples. The calculation P(Xˉ>5.08)≈0.0548P(\bar{X} > 5.08) \approx 0.0548 tells us about the sampling distribution of the mean for samples of size 4. This probability describes what happens when you repeatedly take samples of 4 bags and calculate their average weight each time. About 5.48% of these sample means will exceed 5.08 lbs, making option D correct. Option A incorrectly interprets this as a probability for individual bags. To find P(X>5.08)P(X > 5.08) for a single bag, you'd use the original distribution (μ = 5, σ = 0.1), not the sampling distribution. Option B makes a similar error, treating the complement as applying to individual bags rather than sample means. Option C misunderstands what the probability represents entirely—it's not about the percentage of bags within a single sample that exceed 5.08 lbs, but about the sample mean itself. The critical insight is that Xˉ\bar{X} has its own distribution (the sampling distribution) with mean μ = 5 and standard error σXˉ=0.14=0.05\sigma_{\bar{X}} = \frac{0.1}{\sqrt{4}} = 0.05. This distribution describes the behavior of sample means across repeated sampling. Remember: When you see Xˉ\bar{X} in a probability statement, think "repeated sampling" and "long-run behavior of sample means," not individual observations or single samples.

Question 19

An airline knows that the weight of passengers' checked baggage is highly skewed, with a mean of 18 kg and a standard deviation of 6 kg. For a flight with 100 passengers, the airline wants to assess the probability that the average baggage weight exceeds 19.5 kg. Which statement is the most valid conclusion?

  1. The distribution of baggage weights for the 100 passengers on the flight will be approximately normal.
  2. The sampling distribution of the mean baggage weight is approximately normal with a mean of 18 kg and a standard error of 0.6 kg. (correct answer)
  3. The probability of the average weight exceeding 19.5 kg cannot be found because the population is skewed.
  4. The sampling distribution of the mean will have the same skewed shape as the population but with a smaller standard deviation.
Explanation: According to the Central Limit Theorem, for a large sample size (n=100 is large), the sampling distribution of the sample mean (xˉ\bar{x}) will be approximately normal, regardless of the population's shape. The mean of this sampling distribution is equal to the population mean (μ=18\mu = 18 kg), and its standard deviation (the standard error) is σxˉ=σ/n=6/100=0.6\sigma_{\bar{x}} = \sigma / \sqrt{n} = 6 / \sqrt{100} = 0.6 kg. Choice A is incorrect because a single sample of 100 will still reflect the population's skew. Choice C is incorrect because the CLT allows us to find this probability. Choice D is incorrect because the CLT states the shape becomes approximately normal, not that it retains the population's shape.

Question 20

A small company has 200 employees. The standard deviation of employees' annual salaries is $9,000. A sample of 60 employees is selected without replacement to estimate the mean salary. When calculating the probability that the sample mean falls within a certain range, which of the following statements is most accurate?

  1. The standard error should be calculated as 9000/609000 / \sqrt{60} because the Central Limit Theorem applies.
  2. The standard error calculated using 9000/609000 / \sqrt{60} will be an overestimate of the true standard error. (correct answer)
  3. The standard error calculated using 9000/609000 / \sqrt{60} will be an underestimate of the true standard error.
  4. The standard error cannot be determined because the population distribution of salaries is unknown.
Explanation: The sample size n=60n=60 is a significant fraction of the population size N=200N=200. Specifically, n/N=60/200=0.30n/N = 60/200 = 0.30, which is greater than the typical threshold of 0.05. Therefore, the Finite Population Correction (FPC) factor should be used. The corrected standard error is (σ/n)∗(N−n)/(N−1)(\sigma/\sqrt{n}) * \sqrt{(N-n)/(N-1)}. Since the FPC factor (200−60)/(200−1)\sqrt{(200-60)/(200-1)} is less than 1, the true standard error is smaller than the value calculated by the standard formula σ/n\sigma/\sqrt{n}. Thus, the standard formula overestimates the true standard error.