Central Limit Theorem Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Central Limit Theorem.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

For sufficiently large sample size (n \geq 30 as a rule of thumb), the sampling distribution of the sample mean is approximately normal with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}, regardless of the shape of the population distribution.

Roll a single die and the outcomes are flat (uniform). But average the rolls of 30 dice and the result looks like a bell curve every time. No matter how weird the original data looksโ€”skewed, bimodal, flatโ€”the averages of large enough samples always settle into a normal shape. It's one of the most surprising facts in all of mathematics.

Read the full concept explanation โ†’

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: The CLT is why the normal distribution dominates statistics: it guarantees that sample means are approximately normal for large n, giving us a universal framework for inference.

Common stuck point: The CLT applies to sample means (and sums), not to individual observations. A single data point from a skewed population is still skewed.

Worked Examples

Example 1

medium
A highly skewed population (times between bus arrivals) has \mu=15 min and \sigma=8 min. For samples of n=64, describe the shape, mean, and SD of the sampling distribution of \bar{X}, and find P(\bar{X} < 14).

Solution

  1. 1
    CLT: despite skewed population, with n=64 \geq 30, \bar{X} is approximately normally distributed
  2. 2
    Mean: \mu_{\bar{X}} = 15 min; SE: \sigma_{\bar{X}} = \frac{8}{\sqrt{64}} = \frac{8}{8} = 1 min
  3. 3
    \bar{X} \sim N(15, 1)
  4. 4
    P(\bar{X} < 14) = P(Z < \frac{14-15}{1}) = P(Z < -1) = 0.1587

Answer

\bar{X} \sim N(15, 1); P(\bar{X} < 14) \approx 0.159 despite non-normal population.
The CLT's power: even with a skewed population, sample means become normally distributed with large enough n. This allows us to use normal-distribution methods (z-scores, standard tables) for any population shape, which is why CLT is central to statistical inference.

Example 2

hard
A fair die (ฮผ=3.5, ฯƒ=1.71) is rolled n=100 times. By CLT, find the approximate probability that the sample mean is between 3.3 and 3.7.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
State the Central Limit Theorem in your own words, including what conditions must be met and what it tells us about the shape of the sampling distribution.

Example 2

hard
Customers arrive at a store with mean \mu=2 per minute, \sigma=1.5 per minute (Poisson-like). For 36-minute observation windows, find P(\text{total arrivals} > 80) using CLT.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

sampling distributionnormal distribution