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The Central Limit Theorem questions
Average enough independent observations from almost any distribution and the average behaves normally, with a spread that shrinks as the square root of the sample size.
7 original questions · 27 marks · the the central limit theorem notes · Further Statistics 1
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State the Central Limit Theorem, including what it says about the shape of the parent distribution.
Worked answer
For a population with mean μ and variance σ², the mean of a random sample of size n is approximately N(μ, σ²/n) when n is large, whatever the shape of the parent distribution. The approximation improves as n grows. B1 for the mean, B1 for the variance σ²/n, B1 for it holding whatever the shape of the parent. State that the sample is random and that n is large, since both conditions carry credit.A population has mean 75 and standard deviation 16. Write down the approximate distribution of the mean of a sample of 64, and state the standard error.
Worked answer
Approximately N(75, 4), since 256/64 = 4. The standard error is √4 = 2, which is also 16/√64. M1 for 256/64, A1 for the distribution, B1 for the standard error 2. Divide the variance by n rather than the standard deviation, and write the distribution with a variance in the bracket, not a standard deviation.A population has mean 75 and standard deviation 16, and a random sample of 64 is taken. Find the probability that the sample mean exceeds 78.
Worked answer
z = (78 − 75)/2 = 1.5, so P(X̄ > 78) = P(Z > 1.5) = 1 − 0.9332 = 0.0668. M1 for standardising, A1 for z = 1.5, M1 for 1 − 0.9332, A1 for 0.0668. Dividing by the standard error of the mean rather than by 16 is the whole point, and the parent's shape never entered the working.Counts follow Po(9). Find the approximate distribution of the mean of 36 such counts, and the probability that this mean is below 8.5.
Worked answer
A Poisson has mean and variance both 9, so the sample mean is approximately N(9, 9/36) = N(9, 0.25), standard error 0.5. Then z = (8.5 − 9)/0.5 = −1, giving P ≈ 0.1587. B1 for the mean and variance both being 9, M1 for N(9, 0.25), M1 for z = −1, A1 for 0.1587. A discrete, skewed parent is no obstacle at all.A population has standard deviation 6. Find the smallest sample size for which the standard error of the mean is at most 0.5.
Worked answer
6/√n ≤ 0.5 gives √n ≥ 12, so n ≥ 144. M1 for 6/√n ≤ 0.5, A1 for √n ≥ 12, A1 for n = 144. Halving that requirement to 0.25 would need n ≥ 576: four times the data for half the error, because the root is what does the work.A student claims that as the sample size grows, the sample values themselves become normally distributed. Explain what is wrong with this, and state what actually becomes normal.
Worked answer
Individual observations keep the parent's distribution no matter how many are collected, since taking more of them changes nothing about any single one. What becomes approximately normal is the distribution of the sample mean, taken over repeated samples of size n. The theorem is a statement about a statistic, not about the raw data. B1 for individual observations keeping the parent distribution, B1 for the sample mean being what becomes normal, B1 for the theorem being a statement about a statistic.A population has standard deviation 5 and unknown mean μ. A random sample of size n is taken. Find the smallest n for which the sample mean lies within 1 of μ with probability at least 0.95, and state what the Central Limit Theorem contributes to the argument.
Worked answer
By the Central Limit Theorem X̄ is approximately N(μ, 25/n) for large n, so Z = (X̄ − μ)/(5/√n) is approximately standard normal.
P(|X̄ − μ| < 1) ≥ 0.95 requires 1 ≥ 1.96 × 5/√n, so √n ≥ 9.8 and n ≥ 96.04. The normal approximation therefore estimates the least such n as n = 97.
Check the boundary: with n = 96 the margin is 1.96 × 5/√96 = 1.0002, just too wide, and with n = 97 it is 0.9950. Always round a sample size up, whatever the decimal, since rounding down breaks the requirement.
Now say what the theorem does and does not give you. It makes X̄ approximately normal for large n without any assumption about the parent, which is why no shape was ever named, but it is an approximation rather than a guarantee at any fixed n. How large n must be before that approximation can be trusted depends on the parent, since a heavily skewed or long-tailed population settles down more slowly. So 97 is the least n the normal approximation estimates, not a smallest n that is certain to work for every population. An exact minimum would need the parent distribution itself, or a result that bounds the error in the normal approximation. The variance also has to be known, and here it is 25. Halving the margin to 0.5 would need n ≥ 385, roughly four times as many observations. M1 for N(μ, 25/n), A1 for the standardised form, M1 for 1 ≥ 1.96 × 5/√n, A1 for √n ≥ 9.8, A1 for n = 97, B1 for the theorem needing no assumption about the parent, B1 for it being an approximation rather than a guarantee.
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