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CLT: bulbs average life below 775 h

Short answer

For a normal population, sample mean is normal with mean 800 h and SE=4/√16=1 h. z=(775-800)/1=-25, so P(x̄<775) is practically zero.

State central limit theorem. An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with mean equal to 800 hours and standard deviation of 4 hours. Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours. [Printed as "12775" hours in the scan.]

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

μ = 800 hours, σ = 4 hours, n = 16 (the population is normal, so x̄ is normal).

SE = σ/√n = 4/√16 = 1 hour

z = (775 − 800)/1 = -25

P(x̄ < 775) = P(Z < −25) ≈ 0

Answer: the probability is practically zero (less than 0.0001).

Note: the scanned question shows "12775" hours, read here as 775 hours. If σ were 40 hours (as in the similar question), SE = 10, z = −2.50 and P = 0.0062.

✓ Answer checked by Claude Sonnet 5.5

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