Questions on the Central Limit Theorem

Question 1

State the Central Limit Theorem, including what it says about the shape of the parent distribution.

Accepted 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 are required.

Question 2

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.

Accepted 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.

Question 3

State two things that the Central Limit Theorem does not claim.

Accepted Answer

It does not claim that the sample mean is exactly normal for any particular n: the normal distribution is an approximation that improves as n grows. It does not claim that the parent population is normal, nor does it change the parent's mean or variance; μ and σ² are whatever the population has. B1 B1 for any two such statements. Other acceptable answers: it says nothing about how large n must be, since that depends on the parent, and it says nothing about individual observations.

Question 4

A population has mean 20 and variance 9. Write down the approximate distribution of the total of a random sample of 100 observations, and state its standard deviation.

Accepted Answer

The total has mean 100 × 20 = 2000 and variance 100 × 9 = 900, so it is approximately N(2000, 900) with standard deviation √900 = 30. M1 for multiplying the mean and the variance by n, A1 for N(2000, 900), B1 for 30. The variance of a total is nσ², where the variance of a mean is σ²/n: one grows with n and the other shrinks.

Question 5

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.

Accepted 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. Divide by the standard error of the mean, not by 16.

Question 6

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.

Accepted 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.

Question 7

A population has standard deviation 6. Find the smallest sample size for which the standard error of the mean is at most 0.5.

Accepted 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.

Question 8

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.

Accepted 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.

Question 9

Parcels have masses with mean 2 kg and standard deviation 0.5 kg. A van carries a random sample of 50 parcels. Find the probability that their total mass exceeds 105 kg.

Accepted Answer

The total has mean 50 × 2 = 100 and variance 50 × 0.25 = 12.5, so by the Central Limit Theorem it is approximately N(100, 12.5), standard deviation 3.536. Then z = (105 − 100)/3.536 = 1.414 and P(total > 105) = 1 − 0.9214 = 0.0786. M1 for the mean and variance of the total, A1 for N(100, 12.5), M1 for standardising, A1 for 0.0786. Working with the mean instead gives the same answer: a total above 105 is a mean above 2.1, and (2.1 − 2)/(0.5/√50) is the same 1.414.

Question 10

Each of 50 independent counts follows the binomial distribution B(20, 0.3). Find the approximate distribution of the mean of the 50 counts, and the probability that this mean is less than 5.5.

Accepted Answer

A single count has mean 20 × 0.3 = 6 and variance 20 × 0.3 × 0.7 = 4.2, so the mean of 50 is approximately N(6, 4.2/50) = N(6, 0.084), standard error 0.2898. Then z = (5.5 − 6)/0.2898 = −1.725 and P(X̄ < 5.5) = 0.0422. B1 for the mean 6 and variance 4.2 of one count, M1 for dividing the variance by 50, M1 for standardising, A1 for 0.0422. The variance to divide by n is npq of one count, not the variance of anything already averaged.

Question 11

A fair die has mean 3.5 and variance 35/12. It is rolled 60 times. Find the probability that the mean score exceeds 3.8.

Accepted Answer

The mean of 60 rolls is approximately N(3.5, 35/720), and 35/720 = 0.04861, so the standard error is 0.2205. Then z = (3.8 − 3.5)/0.2205 = 1.361 and P(X̄ > 3.8) = 1 − 0.9132 = 0.0868. M1 for the variance 35/12 divided by 60, M1 for standardising, A1 for 0.0868. The parent takes only six equally likely values and is nowhere near normal, but with n = 60 the mean is close enough to normal for this calculation.

Question 12

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.

Accepted 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. 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.

Question 13

A lift has a safe working load of 1200 kg. The masses of the adults who use it have mean 75 kg and standard deviation 12 kg. Using the Central Limit Theorem, find the largest number n of adults for which the probability that their total mass exceeds 1200 kg is less than 0.01. Comment on how far the theorem can be trusted for the value of n you find.

Accepted Answer

The total mass of n adults has mean 75n and variance 144n, so it is approximately N(75n, 144n). The requirement P(total > 1200) < 0.01 needs (1200 − 75n)/(12√n) > 2.326.
The left side falls as n rises, so test whole numbers. n = 14: (1200 − 1050)/(12√14) = 150/44.9 = 3.34, and P(Z > 3.34) = 0.0004 < 0.01. n = 15: (1200 − 1125)/(12√15) = 75/46.5 = 1.61, and P(Z > 1.61) = 0.053, which is too large. So the largest n is 14.
The theorem gives the total an approximately normal distribution for large n, but 14 is not large. It says nothing about how good the approximation is at n = 14: that depends on the shape of the parent distribution of adult masses, which is not given. If masses are themselves roughly normal the answer is sound; if they are skewed, the tail probability at n = 15 could be some way from 0.053, and a safety decision should allow for that. M1 for N(75n, 144n), M1 for standardising 1200, A1 for the condition with 2.326, M1 for testing n = 14 and n = 15, A1 for 14, B1 for the comment that the theorem promises only an approximation for large n and 14 is small.

Question 14

The number of flaws in an item follows Po(0.5), which has mean 0.5 and variance 0.5. Explain why the Central Limit Theorem cannot be used to find the probability that the mean number of flaws in a random sample of 4 items is less than 0.4, and find that probability exactly. Find the corresponding probability for a random sample of 400 items. State what the theorem does and does not claim in each case.

Accepted Answer

With n = 4 the sample is small and the parent is discrete and heavily skewed, so the mean of four counts is nowhere near normal: the theorem describes large n only.
The mean is below 0.4 exactly when the total is below 1.6, that is, when the total is 0 or 1. Independent Poisson counts add, so the total of 4 items follows Po(2), and P(total ≤ 1) = e⁻²(1 + 2) = 0.4060.
With n = 400 the mean is approximately N(0.5, 0.5/400) = N(0.5, 0.00125), standard error 0.03536. Then z = (0.4 − 0.5)/0.03536 = −2.828 and P(X̄ < 0.4) = 0.0023.
The theorem claims that the mean of a large random sample is approximately normal, whatever the parent, and that is used for n = 400. It does not claim anything at n = 4, and it does not say that the normal route is the only one: because the parent is known to be Poisson, the exact distribution of the total was available and is what should be used. B1 for n = 4 being too small for a skewed discrete parent, M1 for the total following Po(2) with the event total ≤ 1, A1 for 0.4060, M1 for N(0.5, 0.00125) and z = −2.83, A1 for 0.0023, B1 for the statement of what the theorem does and does not claim.

Question 15

X₁, X₂, …, X_n are independent observations from a population with mean μ and variance σ², and X̄ is their mean. Prove that E(X̄) = μ and Var(X̄) = σ²/n, stating where independence is used. Explain which part of the Central Limit Theorem this argument does not establish.

Accepted Answer

Expectation is linear, so E(X₁ + … + X_n) = nμ, and dividing by n gives E(X̄) = μ. This step needs no independence.
For independent variables the variance of a sum is the sum of the variances, so Var(X₁ + … + X_n) = nσ². Dividing the variable by n divides its variance by n², so Var(X̄) = nσ²/n² = σ²/n. Independence is used exactly here: without it the variances would not add.
This argument gives the mean and variance of X̄ for every n, but it says nothing about the shape of its distribution. The content of the Central Limit Theorem is that this shape is approximately normal for large n whatever the parent, and that is not established by computing two moments. M1 for E of the sum by linearity, A1 for μ, M1 for the variances adding by independence and the factor 1/n², A1 for σ²/n, B1 for normality being the part not proved.