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CLT: sample of 16, P(x̄<160)

Short answer

CLT: for large n, sample mean is approximately N(μ, σ²/n). With normal population, x̄ ~ N(150, 4²) exactly; (a) z=2.5 gives P=0.9938, (b) z=-2 gives P=0.9772.

Define Central Limit Theorem. In a sample of 16 observations from a normal distribution with mean of 150 and a variance of 256, what is (a) P(x̄ < 160) (b) P(x̄ > 142)

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

Population normal with μ = 150, σ² = 256 so σ = 16; n = 16.

SE = σ/√n = 16/4 = 4, so x̄ ~ N(150, 4²) (exactly normal since the population is normal).

(a) P(x̄ < 160)

z = (160 − 150)/4 = 2.50

P(x̄ < 160) = P(Z < 2.50) = 0.5 + 0.4938 = 0.9938

(b) P(x̄ > 142)

z = (142 − 150)/4 = -2.00

P(x̄ > 142) = P(Z > −2.00) = 0.5 + 0.4772 = 0.9772

Answer: (a) 0.9938 (b) 0.9772

✓ Answer checked by Claude Sonnet 5.5

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