Sampling Distributions & CLT - Business Statistics

1. 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?
   Answer: 0.0228 First, find the population standard deviation from the given variance: $\sigma = \sqrt{225} = 15$. Second, since the sample size is large ($n=36$), the CLT applies. Calculate the standard error of the mean: $\sigma_{\bar{x}} = \sigma / \sqrt{n} = 15 / \sqrt{36} = 15 / 6 = 2.5$. Third, calculate the Z-score for a sample mean of 85: $Z = (\bar{x} - \mu) / \sigma_{\bar{x}} = (85 - 90) / 2.5 = -5 / 2.5 = -2.0$. Finally, find the probability $P(Z < -2.0)$, which from the standard normal table is approximately 0.0228.
2. 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 $n=25$. What sample size should the analyst use?
   Answer: 225 The standard error of the mean is given by $\sigma_{\bar{x}} = \sigma / \sqrt{n}$. Let the initial sample size be $n_1 = 25$ and the new sample size be $n_2$. The goal is for $\sigma_{\bar{x}_2} = (1/3) \sigma_{\bar{x}_1}$. Substituting the formula: $\sigma / \sqrt{n_2} = (1/3) (\sigma / \sqrt{n_1})$. The $\sigma$ terms cancel out. $1 / \sqrt{n_2} = (1/3) (1 / \sqrt{25}) = (1/3) (1/5) = 1/15$. Therefore, $\sqrt{n_2} = 15$, which means $n_2 = 15^2 = 225$.
3. 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?
   Answer: Team B's sample mean has smaller standard error and is more likely to be normally distributed Team A's standard error is $\frac{18000}{\sqrt{25}} = 3600$, while Team B's is $\frac{18000}{\sqrt{100}} = 1800$. 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.
4. A retail chain tracks daily sales at individual stores. Store sales have a right-skewed distribution with mean $8,500 and standard deviation $2,200. A regional manager samples 49 stores randomly to estimate average daily sales. Which statement about the sampling distribution of the sample mean is most accurate?
   Answer: The distribution is approximately normal with standard error approximately $314 By the Central Limit Theorem, with n = 49 ≥ 30, the sampling distribution of the sample mean is approximately normal regardless of the population shape. The standard error is $\sigma_{\bar{x}} = \frac{2200}{\sqrt{49}} = \frac{2200}{7} ≈ 314$. Choice A incorrectly assumes the sampling distribution maintains the population's skewness. Choice C uses the population standard deviation instead of standard error. Choice D ignores the CLT applicability.
5. 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?
   Answer: 0.0548 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 $n \ge 30$ is only required when the population distribution is unknown or not normal. The standard error is $\sigma_{\bar{x}} = \sigma / \sqrt{n} = 5 / \sqrt{4} = 2.5$ ml. The Z-score is $Z = (236 - 240) / 2.5 = -4 / 2.5 = -1.6$. The probability $P(Z < -1.6)$ is approximately 0.0548.
6. 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?
   Answer: 0.789 First, find the mean and standard deviation of the uniform population distribution. Mean $\mu = (a+b)/2 = (20+30)/2 = 25$ minutes. Variance $\sigma^2 = (b-a)^2/12 = (30-20)^2/12 = 100/12$. Standard deviation $\sigma = \sqrt{100/12} \approx 2.887$ minutes. Second, since $n=48$ is large, apply the CLT. The standard error is $\sigma_{\bar{x}} = \sigma / \sqrt{n} = 2.887 / \sqrt{48} \approx 0.4167$. Third, find the Z-scores for the interval boundaries: $Z_1 = (24.5 - 25) / 0.4167 = -1.20$ and $Z_2 = (25.5 - 25) / 0.4167 = 1.20$. Finally, find the probability $P(-1.20 < Z < 1.20) = P(Z < 1.20) - P(Z < -1.20) = 0.8849 - 0.1151 = 0.7698$, which is approximately 0.789.
7. A quality control process at a battery factory measures cell voltage. The population of voltages is known to be slightly skewed with $\mu = 3.7$ volts and $\sigma = 0.2$ 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?
   Answer: $n=40$, because this sample size is large enough for the Central Limit Theorem to apply. 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 $n \ge 30$ 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, $n=40$ 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 $\sigma$ doesn't guarantee normality of the sampling distribution for small samples from a non-normal population.
8. 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?
   Answer: The probability is smaller for the 100-day sample. For $n=64$, the standard error is $\sigma_{\bar{x}} = 80 / \sqrt{64} = 10$. The Z-score for $\bar{x}=470$ is $Z = (470-450)/10 = 2.0$. For $n=100$, the standard error is $\sigma_{\bar{x}} = 80 / \sqrt{100} = 8$. The Z-score for $\bar{x}=470$ is $Z = (470-450)/8 = 2.5$. Since a Z-score of 2.5 is further in the tail of the distribution than a Z-score of 2.0, the probability $P(Z > 2.5)$ is smaller than $P(Z > 2.0)$. Increasing the sample size makes the sampling distribution tighter around the mean, making large deviations less likely.
9. 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?
   Answer: 68 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: $SE = \frac{2.8\%}{\sqrt{20}} = 0.626\%$. 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 $100 \times 0.68 = 68$ 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.
10. 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?
    Answer: 2 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 $SE = \frac{\sigma}{\sqrt{n}}$, 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: $SE_{36} = \frac{75}{\sqrt{36}} = \frac{75}{6}$. For n = 144: $SE_{144} = \frac{75}{\sqrt{144}} = \frac{75}{12}$. The factor by which standard error decreases is $\frac{SE_{36}}{SE_{144}} = \frac{75/6}{75/12} = \frac{12}{6} = 2$. 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.
11. 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?
    Answer: -1.33 The standard error is $\sigma_{\bar{x}} = \frac{4.2}{\sqrt{64}} = \frac{4.2}{8} = 0.525$. The z-score is $z = \frac{11.8 - 12.5}{0.525} = \frac{-0.7}{0.525} ≈ -1.33$. Choice B uses the population standard deviation instead of standard error: $\frac{-0.7}{4.2} ≈ -0.17$. Choice C makes an arithmetic error in the calculation. Choice D uses an incorrect standard error calculation.
12. A telecommunications company measures call durations with mean 8.2 minutes and standard deviation 3.5 minutes. An analyst samples calls from two different time periods: 25 calls from morning hours and 64 calls from evening hours. If both sample means are calculated, what is the ratio of the standard error for morning calls to the standard error for evening calls?
    Answer: 1.6 When you encounter questions about sampling distributions, focus on how the standard error changes with sample size. The standard error of the mean equals the population standard deviation divided by the square root of the sample size: $SE = \frac{\sigma}{\sqrt{n}}$. For the morning sample (n = 25), the standard error is $SE_{morning} = \frac{3.5}{\sqrt{25}} = \frac{3.5}{5} = 0.7$ minutes. For the evening sample (n = 64), the standard error is $SE_{evening} = \frac{3.5}{\sqrt{64}} = \frac{3.5}{8} = 0.4375$ minutes. The ratio of morning to evening standard errors is $\frac{0.7}{0.4375} = 1.6$, confirming answer D. Choice A (1.28) likely comes from incorrectly using $\sqrt{\frac{64}{25}} = \frac{8}{5} = 1.6$, then taking its square root. Choice B (0.625) represents the inverse of the correct ratio - you might get this by accidentally calculating evening-to-morning instead of morning-to-evening. Choice C (2.56) results from squaring the correct ratio, possibly from confusing variance relationships with standard error relationships. Remember that standard error decreases as sample size increases, so the smaller sample (morning, n = 25) will always have a larger standard error than the bigger sample (evening, n = 64). When comparing standard errors, focus on the square root relationship with sample size - the ratio of standard errors equals the square root of the inverse ratio of sample sizes.
13. 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?
    Answer: 0.8351 First, note that the population variance ($\sigma^2$) is 0.16 mm², so the standard deviation ($\sigma$) is $\sqrt{0.16} = 0.4$ mm. Next, calculate the standard error of the mean for a sample of size 64: $\sigma_{\bar{x}} = \sigma / \sqrt{n} = 0.4 / \sqrt{64} = 0.4 / 8 = 0.05$ mm. Then, calculate the Z-scores for the interval boundaries: $Z_{lower} = (19.875 - 20.00) / 0.05 = -0.125 / 0.05 = -2.5$. $Z_{upper} = (20.05 - 20.00) / 0.05 = 0.05 / 0.05 = 1.0$. Finally, find the probability $P(-2.5 < Z < 1.0) = P(Z < 1.0) - P(Z < -2.5) = 0.8413 - 0.0062 = 0.8351$.
14. The sampling distribution of the mean for samples of size $n=49$ 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?
    Answer: 28 The standard deviation of the sampling distribution of the mean is the standard error, $\sigma_{\bar{x}}$. The formula relating it to the population standard deviation $\sigma$ is $\sigma_{\bar{x}} = \sigma / \sqrt{n}$. We are given $\sigma_{\bar{x}} = 4$ and $n=49$. We need to solve for $\sigma$. Rearranging the formula gives $\sigma = \sigma_{\bar{x}} \times \sqrt{n}$. Plugging in the values: $\sigma = 4 \times \sqrt{49} = 4 \times 7 = 28$.
15. 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?
    Answer: 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. According to the Central Limit Theorem, for a large sample size (n=100 is large), the sampling distribution of the sample mean ($\bar{x}$) will be approximately normal, regardless of the population's shape. The mean of this sampling distribution is equal to the population mean ($\mu = 18$ kg), and its standard deviation (the standard error) is $\sigma_{\bar{x}} = \sigma / \sqrt{n} = 6 / \sqrt{100} = 0.6$ 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.
16. 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?
    Answer: No, because the z-score is -1.5, which does not exceed the threshold of more than 1.5 standard errors The standard error is $\sigma_{\bar{x}} = \frac{12}{\sqrt{36}} = 2$. The z-score is $z = \frac{42-45}{2} = -1.5$. 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.
17. 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 $P(\bar{X} > 5.08) \approx 0.0548$. Which of the following is the most accurate interpretation of this result?
    Answer: 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. When you encounter probability statements involving sample means (denoted as $\bar{X}$), 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 $P(\bar{X} > 5.08) \approx 0.0548$ 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 $P(X > 5.08)$ 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 $\bar{X}$ has its own distribution (the sampling distribution) with mean μ = 5 and standard error $\sigma_{\bar{X}} = \frac{0.1}{\sqrt{4}} = 0.05$. This distribution describes the behavior of sample means across repeated sampling. Remember: When you see $\bar{X}$ in a probability statement, think "repeated sampling" and "long-run behavior of sample means," not individual observations or single samples.
18. 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?
    Answer: n = 59 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 $\mu = 58$ and standard error $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{14}{\sqrt{n}}$. You want $P(|\bar{x} - 58| \leq 3) = 0.90$, 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: $1.645 \times \frac{14}{\sqrt{n}} = 3$ Solving for n: $\frac{14 \times 1.645}{3} = \sqrt{n}$, which gives $\sqrt{n} = 7.677$, so $n = 58.9 \approx 59$. 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.
19. The fill weights of cereal boxes from a production line are normally distributed with a mean of 450 grams and a standard deviation of 8 grams. A quality control manager takes a random sample of 16 boxes. The manager wants to identify a lower threshold for the sample mean weight, below which only 5% of all possible sample means would fall. What is this threshold weight?
    Answer: 446.71 grams This is an inverse probability problem. First, find the standard error of the mean: $\sigma_{\bar{x}} = \sigma / \sqrt{n} = 8 / \sqrt{16} = 8 / 4 = 2$ grams. Next, find the Z-score corresponding to the 5th percentile (a cumulative probability of 0.05). This Z-score is approximately -1.645. Finally, use the Z-score formula and solve for $\bar{x}$: $Z = (\bar{x} - \mu) / \sigma_{\bar{x}} \Rightarrow -1.645 = (\bar{x} - 450) / 2 \Rightarrow \bar{x} = 450 + 2(-1.645) = 450 - 3.29 = 446.71$ grams.
20. Let $X$ be a random variable for the weekly sales at a retail store, with $\mu = \$25,000$ and $\sigma = \$4,500$. Let $\bar{X}_{36}$ be the sample mean of weekly sales from a random sample of 36 weeks. Which statement correctly compares $P(X > \$26,500)$ and $P(\bar{X}_{36} > \$26,500)$, assuming $X$ is not normally distributed?
    Answer: $P(\bar{X}_{36} > \$26,500)$ is smaller than $P(X > \$26,500)$ because the sampling distribution has less variability. The sampling distribution of the mean $\bar{X}_{36}$ has a standard deviation (standard error) of $\sigma_{\bar{x}} = \sigma / \sqrt{n} = 4500 / \sqrt{36} = \$750$. This is much smaller than the population standard deviation of $\sigma = \$4,500$. A value of $26,500 is $\$1500/\$4500 = 1/3$ of a standard deviation above the mean for an individual observation. For the sample mean, $26,500 is $\$1500/\$750 = 2$ 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.

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