--- title: "Central limit theorem and applications" subject: "Probability and Statistics" chapter: 4 key: "semester_5/probability_and_statistics:clt-applications" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-applications site: BCT Engineering --- # Central limit theorem and applications Probability and Statistics · Chapter 4: Sampling Distribution - Asked in: 2073 Chaitra, 4 marks; 2070 Chaitra, 4 marks; 2073 Shrawan, 5 marks (asked 3 times) - Video: [Real-world application of the Central Limit Theorem (CLT)](https://www.youtube.com/watch?v=N7wW1dlmMaE) — 365 Data Science - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/clt-applications - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4 ## Question What do you mean by central limit theorem and discuss its applications. ## Short answer CLT states that for large n (≥30), sample mean x̄ from any population with mean μ, variance σ² is approx. normal N(μ, σ²/n). Applications: computing probabilities of sample means, confidence intervals, hypothesis testing, quality control charts, and normal approximations for binomial/Poisson. ## Answer **Central Limit Theorem (CLT):** if X₁, X₂, …, Xₙ is a random sample of size *n* from any population with mean μ and finite variance σ², then for large *n* (n ≥ 30) the sampling distribution of the sample mean x̄ is approximately normal with mean μ and standard deviation σ/√n, whatever the shape of the population: x̄ ~ N(μ, σ²/n), and z = (x̄ − μ)/(σ/√n) ~ N(0, 1). If the population itself is normal, x̄ is exactly normal for every n. ```text Any population Sampling distribution of x̄ (skewed, etc.) for large n ___ .-'''-. / \___ / | \ / \___ .' | '. --------------- --'-----+-----'-- μ μ spread σ spread σ/√n (narrower) ``` **Key points** - Mean of x̄ = μ; standard error σ/√n (σ/√n × √((N−n)/(N−1)) for a finite population). - The larger the sample, the closer to normal and the smaller the spread. - If σ is unknown, the sample standard deviation *s* is used for large n. #### Applications 1. **Probability of sample means:** the probability that x̄ lies in a given range is found with the normal table (z = (x̄ − μ)/(σ/√n)), even when the population is not normal. 2. **Estimation:** confidence intervals for the mean, x̄ ± z·σ/√n, are based on it. 3. **Testing of hypothesis:** the z-test for a mean (and the difference of two means) uses the normal distribution of x̄ for large samples. 4. **Quality control:** control charts for the mean (x̄ charts) use limits μ ± 3σ/√n. 5. **Approximations:** the Binomial and Poisson distributions (sums of independent trials or events) are approximated by the normal distribution for large n or λ. 6. **Sampling and survey design:** it explains why sample averages are stable and helps decide the sample size needed for a required precision. **Example:** bulb life has mean 800 h and σ = 40 h. For n = 64 bulbs, x̄ ~ N(800, (5)²), so P(x̄ > 810) = P(Z > 2) = 0.0228. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. According to the CLT, for large n the sample mean x̄ is approximately: A. N(μ, σ^2/n) B. N(μ, σ^2) C. N(0, 1) always D. N(μ/n, σ^2) Answer: A. x̄ has mean μ and standard deviation σ/√n, whatever the population shape. (Key point: CLT statement) 2. As n increases, the spread of x̄: A. increases B. stays the same C. decreases as 1/√n D. becomes zero for n = 30 Answer: C. The standard error σ/√n shrinks as n grows. (Key point: Effect of sample size) 3. Which is an application of the CLT? A. Confidence intervals and z-tests for a mean B. Computing exact Poisson means C. Making populations normal D. Removing sampling error Answer: A. Normality of x̄ underlies confidence intervals, hypothesis tests and x̄ control charts. (Key point: Applications) 4. Bulb life mean 800, σ = 40, n = 64: the standard error and P(x̄ > 810) are: A. SE = 5, P = 0.0228 B. SE = 40, P = 0.4013 C. SE = 5, P = 0.4772 D. SE = 0.625, P = 0.0228 Answer: A. SE = 40/8 = 5, z = 10/5 = 2, so P(Z > 2) = 0.0228. 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