Central Limit Theorem - Biostatistics

A nutritionist measures daily caloric intake, which has an unknown distribution with μ = 2200 calories and σ = 400 calories. She takes samples of size 16. If she increases her sample size to 64, how does this change affect the sampling distribution of the sample mean?
Answer: The mean stays the same and the standard error decreases by a factor of 2 When you encounter questions about changing sample sizes, you're dealing with the Central Limit Theorem and properties of sampling distributions. The key insight is understanding how sample size affects the mean and standard error of the sampling distribution. The sampling distribution of the sample mean always has the same mean as the population mean, regardless of sample size. So whether you sample 16 or 64 people, the mean of your sampling distribution remains μ = 2200 calories. However, the standard error (SE) changes dramatically with sample size. The formula is $SE = \frac{σ}{\sqrt{n}}$ . With the original sample size of 16: $SE = \frac{400}{\sqrt{16}} = \frac{400}{4} = 100$ . With the new sample size of 64: $SE = \frac{400}{\sqrt{64}} = \frac{400}{8} = 50$ . The standard error decreased from 100 to 50, which is a reduction by a factor of 2. Looking at the wrong answers: Choice A incorrectly claims the mean increases, but the sampling distribution mean never changes with sample size. Choice C wrongly states the mean decreases and suggests the standard error decreases by a factor of 4. Choice D correctly identifies that the mean stays the same but incorrectly claims the standard error decreases by a factor of 4 rather than 2. Remember this pattern: when sample size quadruples (16 to 64), the standard error halves because you're taking the square root in the denominator. Larger samples give more precise estimates (smaller standard error) but don't change the expected value.
In a clinical trial, patient recovery times follow an exponential distribution with mean 12 days. A researcher plans to compare the effectiveness of two treatments by examining sample means from groups of 36 patients each. Which statement correctly describes the implications of the Central Limit Theorem for this study design?
Answer: The sample means will be approximately normal with mean 12 days and standard error 2 days For an exponential distribution with mean μ = 12, the standard deviation also equals 12. With n = 36, the CLT ensures the sampling distribution of means is approximately normal with mean = 12 and standard error = σ/√n = 12/√36 = 12/6 = 2 days. Choice A incorrectly maintains the exponential shape. Choice C uses the population standard deviation instead of standard error. Choice D is wrong because the CLT applies to any distribution with finite variance when n is sufficiently large.
A market researcher studies consumer spending that follows a chi-square distribution with 4 degrees of freedom (μ = 4, σ = 2√2 ≈ 2.83). She takes samples of size 50. According to CLT, what is the approximate shape and spread of the sampling distribution?
Answer: Approximately normal with standard error 2.83/√50 ≈ 0.40 When you encounter questions about sampling distributions from non-normal populations, the Central Limit Theorem (CLT) is your key framework. The CLT tells us that regardless of the original population's shape, sample means will be approximately normally distributed when the sample size is sufficiently large (typically n ≥ 30). Here's the critical calculation: The CLT transforms our chi-square distribution into a normal sampling distribution with the same mean (μ = 4) but with standard error equal to the population standard deviation divided by the square root of sample size. So we get $SE = \frac{\sigma}{\sqrt{n}} = \frac{2.83}{\sqrt{50}} ≈ 0.40$ . Choice A correctly applies this formula and recognizes that CLT creates an approximately normal distribution. Choice B incorrectly suggests the sampling distribution retains the chi-square shape—this contradicts the CLT, which normalizes the distribution of sample means regardless of the original shape. Choice C makes a calculation error by dividing by n instead of $\sqrt{n}$ , giving 2.83/50 ≈ 0.057, which dramatically underestimates the standard error. Choice D reflects a fundamental misunderstanding of CLT, incorrectly believing that skewed population distributions prevent normal sampling distributions. Study tip: Remember the CLT standard error formula: $\frac{\sigma}{\sqrt{n}}$ , not $\frac{\sigma}{n}$ . This distinction appears frequently on biostatistics exams. Also, CLT works its "magic" on any population distribution when sample sizes are adequate—the original shape becomes irrelevant for the sampling distribution of means.
An epidemiologist studies disease prevalence in different regions. In Region A, prevalence follows a distribution with μ₁ = 0.12 and σ₁ = 0.08. In Region B, μ₂ = 0.15 and σ₂ = 0.06. She takes samples of 64 from each region. What can be concluded about the difference between sample means (X̄₁ - X̄₂)?
Answer: The difference will be approximately normal with mean -0.03 and standard error 0.0125 When you encounter questions about differences between sample means, you're dealing with sampling distribution theory. The key insight is that when you subtract two independent sample means, their sampling distributions combine in a predictable way. Let's work through this step-by-step. First, find the sampling distribution for each region's sample mean. With samples of size 64, the standard error for Region A is $\sigma_{\bar{X}_1} = \frac{0.08}{\sqrt{64}} = 0.01$ , and for Region B it's $\sigma_{\bar{X}_2} = \frac{0.06}{\sqrt{64}} = 0.0075$ . For the difference $\bar{X}_1 - \bar{X}_2$ , the mean equals $\mu_1 - \mu_2 = 0.12 - 0.15 = -0.03$ . The standard error combines as $\sqrt{\sigma_{\bar{X}_1}^2 + \sigma_{\bar{X}_2}^2} = \sqrt{0.01^2 + 0.0075^2} = \sqrt{0.0001 + 0.00005625} = 0.0125$ . This confirms answer A is correct. Answer B uses the wrong standard error calculation—it appears to incorrectly use just one region's standard error rather than properly combining both. Answer C incorrectly assumes we need correlation information; since we're told this involves "different regions," we can reasonably assume independence, making correlation irrelevant. Answer D misunderstands that differences between sample means create their own distinct sampling distribution, not the same as individual sample means. Study tip: For differences between independent sample means, always remember the formula: combine means by subtraction, combine standard errors by taking the square root of the sum of squared standard errors.
A researcher collects data from a population where individual values range from 10 to 90 with μ = 50 and σ = 15. She plans to calculate sample means from samples of size 9. What is the probability that a single sample mean will be less than 45?
Answer: Cannot be calculated because n = 9 is too small for Central Limit Theorem When you encounter questions about sampling distributions and the Central Limit Theorem (CLT), you need to consider whether the conditions for applying the CLT are met. The CLT allows us to treat sample means as normally distributed, but only under certain circumstances. The CLT requires either: (1) the original population to be normally distributed, or (2) a sufficiently large sample size (typically n ≥ 30) when the population distribution is unknown or non-normal. In this problem, we know the population parameters (μ = 50, σ = 15) and the range (10 to 90), but we're not told that the population follows a normal distribution. Answer A is correct because with n = 9, the sample size is too small to invoke the CLT for a population of unknown distribution. Without knowing the population's shape, we cannot assume the sampling distribution of means will be approximately normal, making probability calculations impossible using normal approximation methods. Answer B incorrectly assumes we can use normal approximation. While it correctly calculates the standard error as $\sigma/\sqrt{n} = 15/\sqrt{9} = 5$ , this approach is invalid without confirming normality conditions. Answer C mistakenly uses the population standard deviation (15) instead of the standard error (5) for the sampling distribution. Answer D incorrectly assumes we can calculate exact probabilities and misapplies the standard error concept. Study tip: Always check CLT conditions before applying normal approximations to sampling distributions. When n < 30, you need explicit confirmation that the population is normally distributed to proceed with probability calculations.
A researcher studying enzyme activity levels observes that individual measurements have a uniform distribution between 20 and 80 units. She collects samples of size 25 and calculates sample means. According to the Central Limit Theorem, what are the mean and standard deviation of the sampling distribution of these sample means?
Answer: Mean = 50 units, Standard deviation = 3.46 units, and the distribution is approximately normal For a uniform distribution from 20 to 80: mean = (20+80)/2 = 50, and standard deviation = (80-20)/√12 = 60/√12 = 17.32. With n=25, the sampling distribution has: mean = 50 (unchanged), standard error = 17.32/√25 = 17.32/5 = 3.46, and is approximately normal by CLT. Choice A uses the wrong standard error calculation (σ/√3 instead of σ/√n). Choice C incorrectly uses population standard deviation and claims uniform shape. Choice D has incorrect mean calculation.
A psychologist studies reaction times that follow a gamma distribution with mean 250 ms and standard deviation 50 ms. She takes samples of size 100. According to CLT, what is the approximate probability that a sample mean exceeds 255 ms?
Answer: 0.16 because 255 is one standard error above the mean When you encounter a problem involving sample means from any population distribution, the Central Limit Theorem (CLT) is your key tool. The CLT tells you that regardless of the original population's shape—whether normal, gamma, or any other distribution—the sampling distribution of sample means will be approximately normal with mean equal to the population mean and standard deviation equal to the standard error. Here, you need the standard error: $SE = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{100}} = \frac{50}{10} = 5$ ms. So sample means are normally distributed with mean 250 ms and standard error 5 ms. To find P(sample mean > 255), you calculate: $z = \frac{255 - 250}{5} = 1$ . Since 255 is exactly one standard error above the mean, and P(Z > 1) ≈ 0.16, the answer is C. Let's examine why the other options miss the mark: A incorrectly uses the population standard deviation (50 ms) instead of the standard error (5 ms) in the reasoning, though it stumbles into the right probability by coincidence. B suggests gamma distributions need special CLT adjustments, but the CLT works for any distribution with finite variance. D claims 255 is a critical value, but critical values depend on chosen significance levels, not inherent to the sample size. Study tip: Always distinguish between population standard deviation (σ) and standard error (σ/√n) in CLT problems. The CLT transforms any population into normally distributed sample means using the standard error—this is the foundation of most inferential statistics.
An environmental scientist measures pollutant concentrations that follow a Poisson distribution with mean λ = 16 particles per sample and σ = √16 = 4. She averages measurements from 64 samples. What is the approximate probability that the sample mean exceeds 16.5 particles?
Answer: 0.16 using normal approximation with standard error 0.5 When you encounter questions about sample means from any distribution, think about the Central Limit Theorem (CLT). The CLT tells us that sample means will be approximately normally distributed, regardless of the original distribution's shape, as long as the sample size is large enough. Here's the key calculation: You have a Poisson distribution with λ = 16, so the population mean is 16 and population standard deviation is √16 = 4. When you take the mean of 64 samples, the standard error becomes $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{64}} = \frac{4}{8} = 0.5$ . The sample mean follows a normal distribution with mean = 16 and standard error = 0.5. To find P(X̄ > 16.5), you calculate the z-score: $z = \frac{16.5 - 16}{0.5} = 1$ . Using the standard normal table, P(Z > 1) ≈ 0.16. Choice A correctly applies this logic. Choice B incorrectly suggests Poisson distributions need special adjustment for the CLT—they don't when sample sizes are adequate. Choice C mistakenly uses the original distribution parameters instead of the sampling distribution parameters (it ignores the standard error calculation). Choice D is wrong because even though individual Poisson values are discrete, the CLT makes sample means approximately continuous and normally distributed. Study tip: For any sampling distribution problem, always calculate the standard error first: divide the population standard deviation by √n. The CLT works for virtually any population distribution when n ≥ 30.
A biostatistician is analyzing data from multiple studies. Study A uses samples of size 16 from a normal population, while Study B uses samples of size 100 from a highly skewed population. Both populations have the same mean and variance. Which statement best compares the sampling distributions of the sample means from these two studies?
Answer: Study A will have a normal sampling distribution while Study B will be approximately normal, and Study B will have a smaller standard error Study A: With n=16 from a normal population, the sampling distribution is exactly normal (not just approximately). Study B: With n=100 from a skewed population, the CLT ensures the sampling distribution is approximately normal. Both have the same mean as the population. Standard errors: Study A has SE = σ/√16 = σ/4, Study B has SE = σ/√100 = σ/10, so Study B has smaller standard error. Choice A incorrectly states identical standard errors. Choice B incorrectly says both are 'approximately' normal (Study A is exactly normal). Choice C incorrectly claims Study B remains skewed and same standard errors.
A researcher studies two populations: Population 1 has μ₁ = 100, σ₁ = 20 (normal distribution), and Population 2 has μ₂ = 100, σ₂ = 20 (exponential distribution). She takes samples of size 30 from each. How do the sampling distributions compare?
Answer: Population 1 gives exact normality; Population 2 gives approximate normality with same parameters When you encounter questions about sampling distributions from different population types, think about how the Central Limit Theorem applies differently depending on the original distribution. For Population 1 (normal distribution), any sample size will produce a perfectly normal sampling distribution of the mean. Since the population is already normal with $\mu_1 = 100$ and $\sigma_1 = 20$ , the sampling distribution will be normal with mean 100 and standard error $\frac{20}{\sqrt{30}} = 3.65$ . For Population 2 (exponential distribution), the Central Limit Theorem kicks in. Even though the population is exponential (highly skewed), a sample size of 30 is generally sufficient for the sampling distribution of the mean to become approximately normal. It will have the same mean (100) and standard error (3.65) as Population 1's sampling distribution. Choice A is wrong because it suggests Population 2's sampling distribution remains exponential, ignoring the Central Limit Theorem's effect. Choice B misses the important distinction between exact and approximate normality - while both sampling distributions will be practically equivalent, their theoretical foundations differ. Choice D incorrectly suggests n=30 is inadequate for exponential distributions; while exponential distributions do take longer to approach normality than symmetric distributions, n=30 is typically sufficient for reasonable approximation. The key insight is that Population 1 gives exact normality (since it starts normal), while Population 2 gives approximate normality (due to CLT), but both end up with essentially the same sampling distribution parameters. Remember: Normal populations always yield normal sampling distributions, while non-normal populations yield approximately normal sampling distributions when n≥30.
A pharmaceutical company tests drug absorption times, which have an unknown distribution with μ = 45 minutes and σ = 12 minutes. They take samples of size 36. If they repeat this sampling process many times, what percentage of sample means will fall between 43 and 47 minutes?
Answer: Approximately 68% because this represents one standard error from the sample mean When you encounter questions about sample means from repeated sampling, you're dealing with the Central Limit Theorem and sampling distributions. The key insight is that even when the original population distribution is unknown, the distribution of sample means becomes approximately normal with sufficient sample size (n ≥ 30). Here's how to solve this: The sampling distribution of sample means has the same mean as the population (μ = 45 minutes) but a smaller standard deviation called the standard error: $SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2$ minutes. Now you can standardize: 43 and 47 minutes are each 2 minutes from the mean of 45, which equals exactly one standard error (2 minutes) in each direction. In a normal distribution, approximately 68% of values fall within one standard deviation (or standard error) of the mean. Option A incorrectly uses the population standard deviation (12) instead of the standard error (2). Option B miscalculates the distance as two standard errors when it's actually one, and incorrectly references "sample mean" instead of the sampling distribution mean. Option D reflects a common misconception—while we don't know the original distribution's shape, the Central Limit Theorem tells us the sampling distribution will be approximately normal with n = 36. Study tip: Always calculate the standard error when dealing with sample means, and remember that 68% corresponds to one standard error, while 95% corresponds to two standard errors in sampling distribution problems.
In a large epidemiological study, researchers measure biomarker levels in blood samples. The population distribution of this biomarker is known to be heavily right-skewed with mean 75 units and standard deviation 30 units. The research team debates whether they need samples of size 30 or 100 for their analysis. Which statement best describes the practical implications of this choice for applying the Central Limit Theorem?
Answer: Both sample sizes will produce normally distributed sample means, but n=100 provides better approximation to normality and more precise estimates The CLT typically applies for n≥30 regardless of population skewness, so both sample sizes should yield approximately normal sampling distributions. However, n=100 provides: (1) better approximation to normality (especially for heavily skewed populations), and (2) smaller standard error (30/√100 = 3 vs. 30/√30 = 5.48). Choice B incorrectly claims n=30 is insufficient for the CLT. Choice C wrongly states identical sampling distributions (they have different standard errors). Choice D incorrectly claims the CLT requires symmetric populations.
A medical device measures glucose levels with known measurement error. Individual readings have mean μ and standard deviation σ, where the population distribution is unknown. A clinician takes n measurements and uses the sample mean as the best estimate. If the clinician wants to reduce the standard error of this estimate by a factor of 3, and currently uses n = 16 measurements, what should the new sample size be?
Answer: n = 144, because reducing standard error by factor of 3 requires increasing sample size by factor of 9 When you encounter questions about reducing standard error, you're dealing with the relationship between sample size and sampling distribution precision. The key insight is understanding how the standard error of the sample mean behaves mathematically. The standard error of the sample mean is $SE = \frac{\sigma}{\sqrt{n}}$ , where σ is the population standard deviation and n is the sample size. To reduce the standard error by a factor of 3, you need the new standard error to be one-third of the original: $\frac{SE_{new}}{SE_{old}} = \frac{1}{3}$ . Setting up the equation: $\frac{\sigma/\sqrt{n_{new}}}{\sigma/\sqrt{n_{old}}} = \frac{1}{3}$ . This simplifies to $\frac{\sqrt{n_{old}}}{\sqrt{n_{new}}} = \frac{1}{3}$ , which means $\sqrt{n_{new}} = 3\sqrt{n_{old}}$ . Squaring both sides gives $n_{new} = 9n_{old}$ . With the current sample size of 16, the new sample size must be $9 \times 16 = 144$ . Option A incorrectly assumes a linear relationship between sample size and standard error reduction—tripling sample size only reduces standard error by $\sqrt{3} \approx 1.73$ , not by 3. Option B presents an arbitrary number without mathematical justification; the Central Limit Theorem doesn't change the standard error formula. Option D is a distractor—the standard error formula applies regardless of population distribution shape, especially since we're given that individual readings have known standard deviation σ. Remember this key relationship: to reduce standard error by factor k, you must increase sample size by factor k². The square root in the denominator creates this quadratic relationship.
A public health researcher studies daily calorie intake in a population where individual intakes follow a distribution with μ = 2000 calories and σ = 400 calories. The researcher wants to estimate the population mean using sample means from groups of participants. If the researcher requires that the probability of the sample mean being within 50 calories of the true population mean is at least 0.95, and plans to use the Central Limit Theorem for inference, what is the minimum sample size needed?
Answer: n = 246, since this ensures CLT application and provides the required precision with 95% confidence We need P(|X̄ - 2000| ≤ 50) ≥ 0.95. This translates to P(-50 ≤ X̄ - 2000 ≤ 50) ≥ 0.95, or P(-50/(400/√n) ≤ Z ≤ 50/(400/√n)) ≥ 0.95. For 95% probability, we need 50/(400/√n) ≥ 1.96, which gives √n ≥ 1.96×400/50 = 15.68, so n ≥ 245.86. Rounding up gives n = 246. Choice A only considers CLT requirements, not precision. Choices B and D use incorrect calculations for the required sample size.
A quality engineer samples products from a process where defect rates follow a highly skewed distribution with μ = 0.05 and σ = 0.02. She wants the sampling distribution of the sample mean to have a standard error no greater than 0.004. What minimum sample size satisfies this requirement?
Answer: 25 items because this ensures CLT applies and gives standard error 0.004 When you encounter problems about sampling distributions and standard error, you need to apply the relationship between sample size and sampling variability, along with the Central Limit Theorem (CLT). The standard error of the sample mean is calculated as $SE = \frac{\sigma}{\sqrt{n}}$ , where σ is the population standard deviation and n is the sample size. To find the minimum sample size for a standard error no greater than 0.004, you set up: $0.004 \geq \frac{0.02}{\sqrt{n}}$ Solving this inequality: $\sqrt{n} \geq \frac{0.02}{0.004} = 5$ , so $n \geq 25$ The CLT tells us that for n ≥ 25-30, the sampling distribution becomes approximately normal regardless of the original distribution's shape, making this sample size adequate even for highly skewed data. Choice A is correct because n = 25 gives exactly the required standard error of 0.004 and satisfies CLT requirements. Choice B incorrectly assumes that highly skewed distributions require sample sizes beyond the standard CLT threshold. While more extreme skewness might benefit from larger samples, n = 25 is sufficient for practical normality. Choice C miscalculates the standard error—100 items would give $SE = \frac{0.02}{\sqrt{100}} = 0.002$ , which exceeds the requirement but isn't the minimum needed. Choice D also miscalculates, giving $SE = \frac{0.02}{\sqrt{625}} = 0.0008$ , which is unnecessarily small and wastes resources. Study tip: Remember that standard error problems require you to find the minimum sample size that meets the constraint—don't choose larger samples unless specifically required.
Two researchers study the same population with mean μ = 100 and σ = 20. Researcher A takes samples of size 25, while Researcher B takes samples of size 100. Both assume their sampling distributions are normal due to CLT. Which statement correctly compares their sampling distributions?
Answer: Both have the same mean and shape, but A's distribution has larger spread When you encounter questions about sampling distributions from the same population, focus on how the Central Limit Theorem affects the three key characteristics: mean, shape, and spread. Both researchers sample from the same population (μ = 100, σ = 20), so their sampling distributions will have identical means equal to the population mean: 100. The CLT also tells us that both sampling distributions will be approximately normal since both researchers assume normality applies to their sample sizes. The crucial difference lies in the standard error (spread). The standard error equals $\frac{\sigma}{\sqrt{n}}$ . For Researcher A: $\frac{20}{\sqrt{25}} = 4$ . For Researcher B: $\frac{20}{\sqrt{100}} = 2$ . Researcher A's sampling distribution has twice the spread of Researcher B's. Looking at the wrong answers: Choice A incorrectly claims different means when both equal the population mean, and wrongly states B has larger spread when B actually has smaller spread. Choice C falsely suggests the distributions have different means and incorrectly implies A's distribution is "less normal" - while larger samples do approach normality faster, both researchers assume CLT applies to their sample sizes. Choice D ignores the fundamental principle that sample size affects standard error - sampling distributions from the same population are not identical when sample sizes differ. Remember this pattern: larger sample sizes always produce sampling distributions with smaller standard errors (less variability), while the mean stays constant at the population mean. The relationship $SE = \frac{\sigma}{\sqrt{n}}$ is essential for comparing sampling distributions.
A psychologist measures IQ scores that are normally distributed with μ = 100 and σ = 15. She takes samples of different sizes and plots the sampling distributions. As sample size increases from 10 to 40 to 160, what changes occur in the sampling distribution characteristics?
Answer: The mean stays constant at 100, while standard error decreases from 4.74 to 2.37 to 1.19 This question tests your understanding of the Central Limit Theorem and how sample size affects sampling distributions. When working with sampling distributions, you need to track what happens to the mean, standard error, and shape as sample size changes. The sampling distribution mean always equals the population mean ( $\mu = 100$ ) regardless of sample size - this is a fundamental property. However, the standard error decreases as sample size increases, following the formula: $SE = \frac{\sigma}{\sqrt{n}}$ . Let's calculate: for n=10, $SE = \frac{15}{\sqrt{10}} = 4.74$ ; for n=40, $SE = \frac{15}{\sqrt{40}} = 2.37$ ; for n=160, $SE = \frac{15}{\sqrt{160}} = 1.19$ . This confirms answer B is correct. Answer A incorrectly claims the mean increases - the sampling distribution mean never changes from the population mean. Answer C suggests both mean and standard error decrease, but again, the mean remains constant at 100. Answer D mentions the distribution becoming "more normal," which is misleading since we started with a normal population, so the sampling distribution is already normal at any sample size. The Central Limit Theorem's normality effect only applies when the original population isn't normal. Remember this pattern: in sampling distributions, the mean stays put while the standard error shrinks by the square root of the sample size. Larger samples give you more precise estimates (smaller standard error) but don't bias your results toward higher or lower values.
A researcher is studying the distribution of systolic blood pressure in a population where the individual measurements follow a right-skewed distribution with mean 140 mmHg and standard deviation 25 mmHg. If the researcher takes random samples of size 64 from this population, which statement best describes the sampling distribution of the sample mean?
Answer: It will be approximately normal with mean 140 mmHg and standard deviation 3.125 mmHg By the Central Limit Theorem, for samples of size 64 (≥30), the sampling distribution of the sample mean will be approximately normal regardless of the population's shape. The mean equals the population mean (140 mmHg), and the standard error equals σ/√n = 25/√64 = 25/8 = 3.125 mmHg. Choice A incorrectly preserves the population's skewness and standard deviation. Choice C uses the population standard deviation instead of the standard error. Choice D is incorrect because the CLT applies regardless of the population shape when n≥30.
A sociologist surveys household income, which follows a log-normal distribution with μ = $45,000 and σ = $15,000. She takes samples of size 36. According to CLT, the sampling distribution of sample means will be approximately normal. What assumption is most critical for this conclusion?
Answer: The population mean and variance must be finite, which they are When you encounter questions about the Central Limit Theorem (CLT), focus on its fundamental requirements rather than getting distracted by the specific distribution mentioned. The CLT states that sample means will be approximately normally distributed regardless of the population's shape, provided certain conditions are met. The most critical assumption for CLT to work is that the population must have finite mean and variance. Since we're told the log-normal distribution has μ = $45,000 and σ = $15,000, both parameters are clearly finite, satisfying this essential requirement. This is why answer C is correct. Now let's examine why the other options miss the mark. Option A suggests you must convert the log-normal distribution to normal first, but this misunderstands CLT entirely—the theorem's power is that it works regardless of the original distribution's shape. Option B claims that n = 36 is automatically sufficient for any continuous distribution, but sample size requirements actually depend on how skewed the population is; some heavily skewed distributions might need larger samples. Option D implies log-normal distributions are "close enough" to normal, but this proximity isn't what makes CLT work—CLT applies to any distribution with finite parameters. Study tip: When facing CLT questions, always check first whether the population has finite mean and variance. Don't get sidetracked by the specific distribution type or assume certain sample sizes are universally adequate. The finite parameters condition is your foundation for applying CLT.
A medical researcher studies blood glucose levels that follow a bimodal distribution with overall μ = 95 mg/dL and σ = 25 mg/dL. She plans to use sample means from samples of size 49 to make probability statements. Which concern about applying CLT is most valid?
Answer: The bimodal shape may require larger samples than usual for good normal approximation When you encounter questions about the Central Limit Theorem (CLT) with non-normal populations, focus on how population shape affects the sample size needed for good normal approximation of sample means. The CLT guarantees that sample means approach normality as sample size increases, regardless of population shape. However, the key insight is that different population shapes require different sample sizes to achieve adequate approximation. While $n = 30$ is often cited as a general rule, this assumes reasonably symmetric populations. Bimodal distributions, with their two peaks and potential valley between them, represent a significant departure from normality that typically requires larger samples than usual for the sampling distribution to become approximately normal. With $n = 49$ , you're in a gray area where the approximation might not be as reliable as desired. Option A is incorrect because there's no universal rule requiring $n ≥ 100$ for bimodal distributions—the required sample size depends on the specific shape and how pronounced the bimodality is. Option B is wrong because CLT applies to all distributions with finite variance; bimodal distributions are not exempt. Option D misses the point entirely—the relationship between mean and standard deviation doesn't affect CLT applicability. The correct answer is C because it accurately reflects that bimodal shapes may indeed require larger samples than the typical $n = 30$ guideline suggests for adequate normal approximation. Study tip: Remember that CLT works for any distribution, but heavily skewed or multimodal populations need larger sample sizes than symmetric ones to achieve good normal approximation of sample means.

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