Central Limit Theorem

Draw repeated samples and watch the distribution of sample means become normal

Intermediate

Population

Sample size n

Population: Exponential(orange dots: the last sample, red marker: its mean)

Sample means(red curve: predicted normal N(μ, σ/√n))

Population

Mean μ: 2.000
Std. dev. σ: 2.000

Sample means (n = 5)

Samples drawn: 0
Last sample mean: –
Mean of means: –(μ = 2.000)
SD of means: –(σ/√n = 0.894)

As n grows the histogram gets narrower (σ/√n) and more bell-shaped, whatever the population looks like.

The exponential population has no upper limit; the rare values above 10 are not drawn but still count.

What the theorem says

Take a population with mean μ and finite standard deviation σ. Draw a random sample of n independent values and compute its mean x̄. The central limit theorem says that, as n grows, the distribution of x̄ gets closer and closer to a normal distribution with mean μ and standard deviation σ/√n, whatever the shape of the population.

That last part is the surprise. A flat, skewed or two-humped population all give bell-shaped sample means once n is large enough. How large depends on the population: symmetric ones like the uniform look normal by n ≈ 5, while a strongly skewed one like the exponential needs n ≈ 30 or more before the skew disappears.

The theorem is why so many statistical methods (confidence intervals, z-tests, t-tests) use the normal distribution even when the raw data are not normal: they are about averages, and averages are approximately normal.

Population vs sample

The population is the distribution you draw from. A sample is n values drawn from it, and its mean is a random number that changes every time you draw.

Sampling distribution

Draw many samples and record each mean. The histogram of those means is the sampling distribution of the mean.

Center stays at μ

Sample means are unbiased: on average they equal the population mean μ, for any n.

Spread shrinks as σ/√n

Averaging cancels out noise. Quadrupling the sample size halves the spread of the sample means.

Key formulas

Sampling distribution of the mean

E[x̄] = μ, SD(x̄) = σ / √n

x̄: mean of a sample of n independent values
μ, σ: population mean and standard deviation

Central limit theorem

for large n: x̄ ≈ N(μ, σ/√n), so z = (x̄ − μ) / (σ/√n) ≈ N(0, 1)

Standard error

SE = σ / √n (estimated by s / √n)

s: sample standard deviation, used when σ is unknown

Populations in the visualizer

Population	μ	σ	Shape
Uniform on [0, 10]	5	10/√12 ≈ 2.887	Flat, symmetric
Exponential, λ = 0.5	2	2	Strongly right-skewed
Bimodal (U[0, 3] or U[7, 10])	5	√13 ≈ 3.606	Two humps, nothing in the middle
Fair die	3.5	√(35/12) ≈ 1.708	Discrete, flat
Loaded die (0.40, 0.25, 0.15, 0.10, 0.06, 0.04)	2.29	≈ 1.430	Discrete, right-skewed

How to use this visualizer

1
Pick a population. The top chart shows its shape and its mean μ.
2
Pick a sample size n. Changing either setting clears the histogram.
3
Press Draw 1 sample a few times. The orange dots are the n values drawn, and the red marker is their mean, which is added to the histogram below.
4
Press Draw 1000 (or Auto-draw) to build up the histogram of sample means, and compare it with the red predicted normal curve.
5
Compare the observed SD of the means with σ/√n in the side panel. They agree more closely the more samples you draw.
6
Try the exponential population with n = 1, 5 and 30, and watch the skewed shape turn into a bell.

Common mix-up

The CLT is about the distribution of the sample mean, not the data. Drawing a bigger sample from an exponential population does not make the individual values normal; it makes their average normal.

Practice problems

Problem 1: Average of dice rolls

A fair die is rolled 36 times. Using the CLT, estimate the probability that the average roll is greater than 4.

Show solution

For one roll: μ = 3.5, σ = √(35/12) ≈ 1.708

SD of the mean: σ/√n = 1.708 / √36 = 1.708 / 6 ≈ 0.2846

z = (4 − 3.5) / 0.2846 ≈ 1.76

P(x̄ > 4) ≈ 1 − Φ(1.76) ≈ 1 − 0.9608

P ≈ 0.039, about a 4% chance

Problem 2: Waiting times

Waiting times at a help desk are exponential with mean 2 minutes (so σ = 2 minutes). For a random sample of 100 customers, what is the probability that the average wait is between 1.6 and 2.4 minutes? How many customers would you need for the standard error to be 0.1 minutes?

Show solution

SE = σ/√n = 2 / √100 = 0.2 minutes

z for 1.6: (1.6 − 2) / 0.2 = −2; z for 2.4: (2.4 − 2) / 0.2 = 2

P(−2 < Z < 2) = Φ(2) − Φ(−2) ≈ 0.9772 − 0.0228 = 0.9545

For SE = 0.1: 2 / √n = 0.1, so √n = 20 and n = 400

P ≈ 0.954; n = 400 customers (4× the sample halves the SE)

Discussion

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