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Central limit theorem and applications

Short answer

CLT states that for large n (≥30), sample mean x̄ from any population with mean μ, variance σ² is approx. normal N(μ, σ²/n). Applications: computing probabilities of sample means, confidence intervals, hypothesis testing, quality control charts, and normal approximations for binomial/Poisson.

What do you mean by central limit theorem and discuss its applications.

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.

 Any population        Sampling distribution of x̄
  (skewed, etc.)       for large n
     ___                    .-'''-.
    /   \___               /   |   \
   /        \___         .'    |    '.
  ---------------      --'-----+-----'--
        μ                      μ
   spread σ              spread σ/√n (narrower)

Key points

  • Mean of x̄ = μ; standard error σ/√n (σ/√n × √((N−n)/(N−1)) for a finite population).
  • The larger the sample, the closer to normal and the smaller the spread.
  • If σ is unknown, the sample standard deviation s is used for large n.

Applications

  1. Probability of sample means: the probability that x̄ lies in a given range is found with the normal table (z = (x̄ − μ)/(σ/√n)), even when the population is not normal.
  2. Estimation: confidence intervals for the mean, x̄ ± z·σ/√n, are based on it.
  3. Testing of hypothesis: the z-test for a mean (and the difference of two means) uses the normal distribution of x̄ for large samples.
  4. Quality control: control charts for the mean (x̄ charts) use limits μ ± 3σ/√n.
  5. Approximations: the Binomial and Poisson distributions (sums of independent trials or events) are approximated by the normal distribution for large n or λ.
  6. Sampling and survey design: it explains why sample averages are stable and helps decide the sample size needed for a required precision.

Example: bulb life has mean 800 h and σ = 40 h. For n = 64 bulbs, x̄ ~ N(800, (5)²), so P(x̄ > 810) = P(Z > 2) = 0.0228.

✓ Answer checked by Claude Sonnet 5.5

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