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CLT: average impurity of 50 batches

Short answer

For large samples, the sample mean is approximately normal with mean μ and standard error σ/√n. Here SE=1.5/√50=0.2121 g; z1=-2.36, z2=-0.94, so P(3.5<x̄<3.8)=0.1736-0.0091=0.1645.

Define Central Limit Theorem. The amount of impurity in a batch of a certain chemical product is a random variable with mean value 4.0 gm and standard deviation 1.5 gm. If 50 batches are independently prepared, what is the probability that the sample average amount of impurity is between 3.5 and 3.8 gm?

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 = 4.0 g, σ = 1.5 g, n = 50 (large, so CLT applies).

SE = σ/√n = 1.5/√50 = 1.5/7.0711 = 0.2121

x̄ ~ N(4.0, 0.2121²)

  • z₁ = (3.5 − 4.0)/0.2121 = -2.36
  • z₂ = (3.8 − 4.0)/0.2121 = -0.94

P(3.5 < x̄ < 3.8) = P(−2.36 < Z < −0.94) = P(Z < −0.94) − P(Z < −2.36)

= 0.1736 − 0.0091 = 0.1645

Answer: the probability is about 0.1645.

✓ Answer checked by Claude Sonnet 5.5

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