--- title: "CLT: battery cell average life" subject: "Probability and Statistics" chapter: 4 key: "semester_5/probability_and_statistics:battery-cells-clt-more-15-less" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/battery-cells-clt-more-15-less site: BCT Engineering --- # CLT: battery cell average life Probability and Statistics · Chapter 4: Sampling Distribution - Asked in: 2072 Chaitra, 5 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/battery-cells-clt-more-15-less - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4 ## Question Define the Central Limit Theorem. A sample of 100 mobile battery cells tested to find the length of life produced the following results as mean 13 months and standard deviation of 3 months. Assuming the data to be normally distributed by using Central Limit Theorem what percentage of battery cells expected to have Average life? i) More than 15 months ii) Less than 9 months ## Short answer By CLT, sample mean of 100 cells is normal with mean 13 months and SE=3/√100=0.3 month. For 15 months z=6.67 and for 9 months z=-13.33, so both probabilities are practically 0%. ## 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. #### Numerical n = 100, mean = 13 months, standard deviation = 3 months. The sample is large, so by CLT the sample mean x̄ ~ N(13, SE²). SE = σ/√n = 3/√100 = 0.3 month #### i) Average life more than 15 months z = (15 − 13)/0.3 = 6.67 P(x̄ > 15) = P(Z > 6.67) ≈ 0 (less than 0.0001) #### ii) Average life less than 9 months z = (9 − 13)/0.3 = -13.33 P(x̄ < 9) = P(Z < −13.33) ≈ 0 **Answer: practically 0% in both cases. A sample average of more than 15 or less than 9 months is almost impossible.** *Note:* if the percentage of individual cells (not sample averages) were wanted, z = (15 − 13)/3 = 0.67 gives 25.14% above 15 months, and z = (9 − 13)/3 = −1.33 gives 9.18% below 9 months. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. According to the Central Limit Theorem, for a large sample size n the sampling distribution of the sample mean is approximately: A. Normal with mean μ and standard deviation σ B. Normal with mean μ and standard deviation σ/√n C. Normal with mean μ/√n and standard deviation σ D. The same shape as the parent population with standard deviation σ/n Answer: B. For large n, x̄ is approximately N(μ, σ²/n), so its standard deviation is σ/√n, whatever the shape of the population. (Key point: Statement of CLT) 2. With n = 100 cells and σ = 3 months, what is the standard error of the sample mean? A. 3 months B. 0.03 months C. 0.3 months D. 30 months Answer: C. SE = σ/√n = 3/√100 = 3/10 = 0.3 month. (Key point: Standard error) 3. Which z value is used for P(average life > 15 months)? A. (15 − 13)/3 = 0.67 B. (15 − 13)/0.3 = 6.67 C. (15 − 13)/100 = 0.02 D. (13 − 15)/0.3 = 6.67 and the area is taken as 0.5 Answer: B. For an average we divide by the standard error 0.3, not by σ, giving z = 6.67. (Key point: Standardising the sample mean) 4. Why is the percentage of cells with average life above 15 months practically 0%? A. 15 months is far out in the tail of the distribution of x̄, 6.67 standard errors above the mean B. Because the sample is smaller than 30 C. Because the population mean is 15 months D. Because averages cannot exceed the mean Answer: A. The sample mean is tightly concentrated around 13 (SE 0.3), so 15 is z = 6.67 away and the tail area is essentially zero. 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