Skip to main content

CLT: battery cell average life

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%.

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

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.

✓ Answer checked by Claude Sonnet 5.5

All Probability and Statistics chapter 4 questions →