Key takeaway: The central limit theorem tells us that averages of large random samples form a bell-shaped pattern around the population mean, however the original data are shaped, with spread equal to σ over the square root of n.
Grade 12 – AP StatisticsAP Statistics Unit 2Statistics and Data
Definition of Central Limit Theorem
The central limit theorem (CLT) is the reason normal-curve methods work so widely. Even if individual measurements are skewed or lumpy, the averages of many random samples pile up in a bell shape around the population mean.
The bigger the samples, the closer the bell and the narrower it becomes. A common rule of thumb treats \(n\ge 30\) as large enough. For populations that are already normal, any sample size works. The theorem lets us compute probabilities about sample means with z-scores.
Formula and notation
Let a population have center \(\mu\) and spread \(\sigma\). Draw many random samples, each of a large size \(n\). Their averages follow
\[ \,\overline{x}\ \approx\ N\!(\mu,\ \frac{\sigma}{\sqrt{n}}), \qquad z=\frac{\,\overline{x}-\mu}{\sigma/\sqrt{n}}. \]
The second formula standardizes a sample mean.
Key facts
- The central limit theorem ignores the shape of the population once the samples are large.
- The center of the distribution of means is the population mean.
- Quadrupling the sample size cuts the spread of the means in half.
- For strongly skewed populations, a sample size well above 30 may be needed.
Central Limit Theorem example
A population of delivery times has mean 40 minutes and standard deviation 12 minutes and is right-skewed. For random samples of 36 deliveries, what is the probability that the sample mean exceeds 43 minutes?
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With \(n=36\), the sample means are roughly normal with center 40 and spread \(\frac{12}{\sqrt{36}}=2\).
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Standardize: \(z=\frac{43-40}{2}=1.5\).
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Use the normal table: \(P(Z\gt 1.5)=1-0.9332=0.0668\).
The chance that a sample of 36 averages more than 43 minutes is about 0.0668.
Common mistake
Watch out
A student standardizes with the population spread, \(z=\frac{43-40}{12}=0.25\), and finds a probability of about 0.40.
Why it is wrong
That z-score would suit a single delivery, and only if the population were normal. For a sample mean the spread is \(\frac{\sigma}{\sqrt{n}}=2\), so \(z=1.5\) and the probability drops to 0.0668. Averages vary less than individual values, which is the message of the central limit theorem.
Related terms and lessons
Browse the glossary: A–Z list · AP Statistics vocabulary · Statistics and Data