Difficulty: Intermediate | Prerequisites: Part 1 of these notes (sampling distributions, standard error, z-scores for x̄).
Big picture: Part 1 established that when the population is normal, the sampling distribution of x̄ is also normal. But what if the population is not normal? The Central Limit Theorem (CLT) answers that question and is one of the most important results in all of statistics. It is the reason we can use normal-distribution methods on data from populations that are skewed, uniform, or otherwise non-normal, provided the sample size is large enough. Nearly every inferential method you will learn going forward relies on the CLT.
The Central Limit Theorem says that for a sufficiently large sample size, the sampling distribution of the sample mean is approximately normal, regardless of the shape of the original population. This means you can use z-scores and normal probability tables for sample means even when the underlying data are not normally distributed. The usual rule of thumb is n ≥ 30.
Central Limit Theorem (CLT)
If you draw sufficiently large random samples from any population with mean μ and standard deviation σ, the sampling distribution of x̄ will be approximately normal: x̄ ~ N(μ, σ²/n). This holds regardless of the shape of the original population distribution. Think of it as: "Average enough random things together and the result starts looking normal."
Sufficiently large sample size
The sample size at which the CLT approximation becomes reasonable. The common rule of thumb is n ≥ 30, though the more skewed the population, the larger n needs to be. For populations that are already close to symmetric, even n = 15 or 20 can be adequate.
z-score for a sample mean
z = (x̄ − μ) / (σ / √n). Converts a sample mean into standard normal units so you can look up probabilities in a z-table.
If you have a sufficiently large sample size from any population, the sampling distribution of x̄ will be approximately normally distributed, regardless of the shape of the original population distribution.
Formally: x̄ ~ N(μ, σ²/n) approximately, for large n.
The two properties from Part 1 still hold exactly:
μ_x̄ = μ
σ_x̄ = σ / √n
The sample size must be "sufficiently large."
The standard threshold taught in most intro courses is n ≥ 30.
For highly skewed populations (e.g., exponential), you may need a larger n.
For populations that are already roughly symmetric, smaller n may suffice.
If the population is exactly normal, you do not need the CLT at all, as the sampling distribution is exactly normal for any n.
The lecture included a comparison of sampling distributions at n = 2, 5, 10, and 20 for three different population shapes:
Normal population: Sampling distribution is normal at every n.
Uniform population: At n = 2, the sampling distribution is triangular. By n = 5 it is bell-shaped. By n = 20 it is virtually indistinguishable from a normal curve.
Exponential population: Heavily right-skewed. At n = 2, still very skewed. By n = 10, roughly symmetric. By n = 20, close to normal.
The takeaway: the more non-normal the population, the larger the sample size you need before the CLT kicks in.
Identify μ and σ for the population.
Determine the sample size n.
Compute the standard error: σ_x̄ = σ / √n.
Set up the probability statement in terms of x̄.
Convert to a z-score: z = (x̄ − μ) / σ_x̄.
Use the standard normal table (z-table) to find the probability.
Quantity | Formula |
|---|---|
Standard error | σ_x̄ = σ / √n |
z-score for x̄ | z = (x̄ − μ) / (σ / √n) |
CLT approximation | x̄ ~ N(μ, σ²/n) for large n |
For a uniform distribution on [a, b]:
μ = (a + b) / 2
σ = (b − a) / √12
Given: μ = 100, σ = 10.
(a) P(individual value < 95):
z = (95 − 100) / 10 = −0.50
P(z < −0.50) = 0.3085
(b) P(x̄ < 95) for a random sample of n = 25:
σ_x̄ = 10 / √25 = 2
z = (95 − 100) / 2 = −2.5
P(z < −2.5) = 0.0062
Notice how the probability drops dramatically when you are asking about a sample mean rather than an individual value. The sample mean is much less likely to be far from μ.
Given: Arrivals are uniform on [0, 10] minutes.
(a) P(single arrival < 6):
For a uniform distribution: P(X < 6) = (6 − 0) / (10 − 0) = 0.6
(b) P(x̄ < 6) for n = 40 students:
μ = (0 + 10) / 2 = 5
σ = (10 − 0) / √12 = 10/√12 ≈ 2.887
σ_x̄ = 2.887 / √40 ≈ 0.4564
z = (6 − 5) / 0.4564 = 2.19
P(z < 2.19) = 0.9857
Even though the population is uniform (not normal), the CLT lets us treat the sampling distribution as approximately normal because n = 40 is large enough.
The CLT is why opinion polls, quality control processes, and medical trials all work. A drug company does not need every patient's response to be normally distributed; it only needs a large enough sample for the average response to follow a normal pattern, which then enables hypothesis testing and confidence intervals. Similarly, a factory measuring the average weight of cereal boxes in batches of 50 can use normal-based methods regardless of how individual box weights are distributed.
"The CLT says individual data points become normally distributed." No. Individual data points keep whatever distribution the population has. The CLT applies only to the distribution of the sample mean (or other sample statistics).
"You always need n ≥ 30." This is a rule of thumb, not a law. If the population is nearly symmetric, smaller samples suffice. If it is heavily skewed, you may need more than 30.
"The CLT makes the population normal." The population does not change. The CLT describes the behaviour of the sampling distribution, not the population itself.
"If n is large, individual observations are normally distributed." This is the same error phrased differently. The CLT never changes the shape of individual observations.
⚠️ You will be asked to recognise when to apply the CLT vs. when the population is already normal (making the CLT unnecessary).
⚠️ Questions comparing P(X < value) for an individual vs. P(x̄ < value) for a sample mean are very common. The denominators in the z-score calculation are different (σ vs. σ/√n).
⚠️ For uniform distributions, remember: μ = (a + b)/2 and σ = (b − a)/√12. These feed into the CLT calculation.
⚠️ When the z-score is negative, be careful with the table lookup. P(z < −k) is in the left tail. If the question asks for P(x̄ > value), compute 1 − P(z < z-score).
True or False: The CLT requires the population to be normally distributed.
Fill in the blank: The common rule of thumb for the CLT is n ≥ ___.
True or False: For a uniform population with n = 40, you can approximate the sampling distribution of x̄ as normal.
True or False: P(x̄ < 95) and P(X < 95) will always be the same value.
Fill in the blank: For a uniform distribution on [0, 10], the population mean is ___ and the population standard deviation is ___.
Answers: 1. False (any population shape). 2. 30. 3. True. 4. False (sample means have less variability). 5. μ = 5, σ ≈ 2.887.
Q: A population has μ = 100 and σ = 10. Compare P(X < 95) for an individual to P(x̄ < 95) for a sample of 25. Why are they different?
A: Individual: z = (95 − 100)/10 = −0.50, P = 0.3085. Sample mean: z = (95 − 100)/2 = −2.50, P = 0.0062. The sample mean has far less variability (σ_x̄ = 2 vs. σ = 10), so it is much less likely to fall far from μ.
Q: Bags of crisps have μ = 1700 in³ and σ = 230 in³. For a random sample of 15 bags, find P(x̄ > 1775).
A: σ_x̄ = 230/√15 ≈ 59.39. z = (1775 − 1700)/59.39 ≈ 1.26. P(z > 1.26) = 1 − 0.8962 = 0.1038.
Q: For the same population of bags, find P(individual bag > 1775). How does this compare to the sample mean probability?
A: z = (1775 − 1700)/230 ≈ 0.326. P(z > 0.326) ≈ 1 − 0.6293 = 0.3707. This is much higher than P(x̄ > 1775) = 0.1038. Individual values have more variability than sample means, so extreme values are more likely for individuals.
Q: Lobster lengths have x̄ = 3.9 in and σ = 2.01 in. Lobsters under 3.35 in must be returned. For a random sample of 60, find P(x̄ < 3.35).
A: σ_x̄ = 2.01/√60 ≈ 0.2594. z = (3.35 − 3.9)/0.2594 ≈ −2.12. P(z < −2.12) = 1 − 0.9830 = 0.0170.
Q: For the same lobster population, if lobsters over 4.06 in cost extra, find P(x̄ > 4.06) for n = 60.
A: z = (4.06 − 3.9)/0.2594 ≈ 0.617. P(z > 0.617) = 1 − 0.7291 = 0.2709.
Q: 1,000 students aged 18–24 are randomly selected. 19.4% are binge drinkers. Is 19.4% a parameter or a statistic?
A: It is a sample statistic, because it describes a sample of 1,000, not the entire population of 18–24 year olds.
The CLT is the direct foundation for confidence intervals (Chapter 8) and hypothesis testing (Chapter 9). When you construct a 95% confidence interval for a population mean, the ±1.96 comes from the normal distribution, which is justified by the CLT. The comparison between individual probabilities P(X < value) and sample mean probabilities P(x̄ < value) also foreshadows the logic of hypothesis testing: if the sample mean lands in an extremely unlikely region, you question whether the assumed μ is correct.
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