AP Statistics Topic 2.12: Sampling Distributions and the Central Limit Theorem

01 · Start here

The concept in 60 seconds

Suppose you randomly choose $30$ café customers and calculate their mean wait. A different random sample of $30$ customers will probably give a different mean. That is sampling variability: a statistic changes from sample to sample, even when the population stays the same.

A sampling distribution describes the possible values of a statistic and how likely they are under a specified sampling method and fixed sample size. Think of a graph where each dot represents $1$ sample’s answer.

The key change:
A dot showing $1$ customer’s wait belongs to a distribution of individual values.
A dot showing the mean of $30$ customers’ waits belongs to a distribution of sample means.

The Central Limit Theorem (CLT) helps explain why, with suitable independent random sampling and a sufficiently large sample size, a distribution of sample means can be approximately normal. The original wait times may still be strongly right-skewed.

All numerical examples here are original fictional teaching examples. The simulated café model is chosen to make sampling variability visible; it does not describe collected customer records.

Quick check: does

1

sample of

30

give me

30

sample means?

No. It gives $30$ individual observations and $1$ sample mean. To obtain $30$ sample means, take $30$ separate samples and calculate a mean for each.

02 · Ask what each dot represents

Three different distributions

Visual guide 1: what does

1

plotted value represent?

Population distribution $1$ person

Values: waits for everyone in the population, or the probability model for an individual wait.

Each value: $1$ wait, such as $7$ minutes.

Sample distribution $1$ selected person

Values: the $n$ individual waits in $1$ sample.

Each value: $1$ sampled wait, such as $7$ minutes.

Sampling distribution $1$ sample’s answer

Values: a statistic from each possible sample, approximated by repeated sampling.

Each value: $1$ mean of $n$ waits, such as $9.8$ minutes.

The horizontal axis can have the same units in all three graphs. Read what a plotted value summarizes before deciding which distribution you are looking at.

Parameter

A number describing the population or its probability model. It is fixed for that population.

$\mu$ is the population mean; $\sigma$ is the population SD.

Statistic

A number calculated from a sample. Before sampling, its value can vary.

$\bar{x}$ , read “x-bar,” is a sample mean; $\hat{p}$ , read “p-hat,” is a sample proportion.

Sample size $n$

The number of observations in $1$ sample. Keep $n$ fixed when building one sampling distribution.

$n = 30$ means $30$ waits are averaged each time.

Repetitions $B$

The number of samples generated in a simulation. $B$ is a convenient simulation label used in these notes.

$B = 5{,}000$ means $5{,}000$ sample means are recorded.

You can build a sampling distribution for a mean, proportion, median, difference between $2$ means, or another statistic. The CLT statement about means should not be applied automatically to every statistic. A sample minimum, for example, need not become normal as $n$ grows.

Quick check: a histogram contains the

30

waits from

1

sample. Which distribution is it?

A sample distribution of individual wait times. It becomes a simulated sampling distribution of means only when its plotted values are means from repeated samples.

03 · A complete example you can count

All possible samples: an exact sampling distribution

A tiny population has $4$ people’s recorded waits: $2, 4, 6,\text{ and }8$ minutes. Choose a simple random sample of $2$ different people, without replacement. Each unordered pair is equally likely. There are $6$ pairs.

$6$ simple random samples of $2$ people without replacement.
Selected waits (minutes)	Sample mean (minutes)	Probability of this sample
$2, 4$	$3$	$\frac{1}{6}$
$2, 6$	$4$	$\frac{1}{6}$
$2, 8$	$5$	$\frac{1}{6}$
$4, 6$	$5$	$\frac{1}{6}$
$4, 8$	$6$	$\frac{1}{6}$
$6, 8$	$7$	$\frac{1}{6}$

Visual guide 2: individuals above; sample means below

Each upper dot is a person. Each lower dot is $1$ of the $6$ equally likely samples. The stacked dots at $5$ retain $2$ distinct samples; both panels use the same horizontal scale.

The population mean is $\frac{2 + 4 + 6 + 8}{4} = 5$ minutes. The $6$ equally likely sample means are $3, 4, 5, 5, 6,\text{ and }7$ minutes. Their average is also $5$ minutes.

Do not remove the second $5$ . $2$ distinct samples produce it, so $P(\bar{x} = 5) = \frac{2}{6} = \frac{1}{3}$ . Each of the other $4$ possible means has probability $\frac{1}{6}$ . The probabilities add to $1$ .

This is an exact sampling distribution because every possible sample under the stated design has been included. In a large population, listing them all is usually impractical, so we use a model or simulation.

The sampling method matters: With replacement, pairs such as $(2,2)$ become possible. The probabilities and possible means would change. State whether you sample with or without replacement before counting outcomes.

Quick check: what is

P(\bar{x} \ge 6)

in this example?

The qualifying means are $6$ and $7$ . They arise from $2\text{ of the }6$ equally likely pairs, so $P(\bar{x} \ge 6) = \frac{2}{6} = \frac{1}{3}$ . The denominator is the number of possible samples, not the $4$ people.

04 · Keep the process fixed; let chance vary the answer

Build and read a sampling simulation

For our café example, assume $1$ customer’s wait $W$ follows a right-skewed exponential model with mean $10$ minutes and SD $10$ minutes. The model has many short waits and some long waits. Draw observations independently from that same model.

You do not need to learn exponential probability formulas for this lesson. It is simply a specified population model that lets us generate positive, skewed data.

Visual guide 3:

1

repetition produces

1

statistic

Step 1Draw $1$ sample

Generate $n = 30$ independent waits from the same specified café model.

Step 2Calculate $1$ mean

Add the $30$ waits and divide by $30$ . Record that $1$ mean in minutes.

Step 3Start again

Generate a fresh sample of $30$ . Repeat until you have $B = 5{,}000$ means.

Step 4Graph and count

Plot the $5{,}000$ means. Count those satisfying your event and divide by $5{,}000$ .

For this plan, $150{,}000$ raw waits lead to $5{,}000$ plotted means. Keep the model, sample size and statistic fixed throughout the run.

A complete plan for $n = 30$

Define $1$ repetition: generate $30$ independent waits from the exponential model with mean $10$ minutes.
Calculate the statistic: average those $30$ waits.
Record $1$ result: store the sample mean, measured in minutes.
Repeat independently: start a fresh sample of $30$ , keeping the population model and sampling rule unchanged.
Build the distribution: graph all recorded means and describe its shape, center and spread.
Estimate an event: count the means that satisfy the requested condition and divide by the total repetitions.

What this lesson’s actual simulation produced

We generated $5{,}000$ independent samples for each $n = 1$ , $5$ and $30$ . Each setting used fresh draws from the same population model. The random seed was 2,122,026, making this run reproducible.

One reproducible café-model simulation: $5{,}000$ repetitions per sample size.
Sample size $n$	Observed mean of the means (min)	Observed SD of the means (min)	Theoretical SD $\frac{\sigma }{\sqrt{n}}$ (min)
$1$	$9.82$	$9.87$	$10.00$
$5$	$9.97$	$4.51$	$4.47$
$30$	$9.99$	$1.82$	$1.83$

For $n = 30$ , $682$ of $5{,}000$ sample means were at least $12$ minutes. The simulation estimate is $\frac{682}{5{,}000} = 0.1364$ , or $13.64\%$ . This describes groups of $30$ customers whose average wait is at least $12$ minutes; it does not describe the percentage of individual customers waiting that long.

Another seed will usually produce a slightly different fraction. Simulated proportions approximate model probabilities; they are not exact just because the software displays many decimal places.

How to reproduce the model with technology

Use a random-number tool’s exponential generator with $\text{mean or scale} = 10$ . Some tools instead request a $\text{rate} = 0.1$ ; check how the tool labels its input. If only uniform random numbers are available, an equivalent wait is $W=-10\ln(U)$ , where $U$ is uniform on $0 < U < 1$ . Regenerate $U$ if the tool returns either endpoint, $0$ or $1$ .

Arrange generated waits into $B$ rows and $n$ columns. Calculate the mean of each row, then plot the $B$ row means. Do not plot all $B \times n$ raw waits and call that a sampling distribution of means.

Quick check:

2{,}000

repetitions with

n = 25

give how many raw waits and how many plotted means?

Raw waits: $2{,}000 \times 25 = 50{,}000$ . Recorded sample means: $2{,}000$ . Each group of $25$ contributes $1$ mean.

05 · Averages can become nearly bell-shaped

The Central Limit Theorem

For independent, identically distributed observations from a population with a finite mean and finite positive variance, the standardized sample mean approaches a standard normal distribution as $n$ increases. The standardized mean $Z_n=\frac{\bar X-\mu}{\sigma/\sqrt n}$ uses the SD of the sample mean; $\bar X$ represents the mean before the sample is drawn. In practice, this lets us approximate the shape of a sample mean’s sampling distribution by a normal curve when $n$ is sufficiently large.

“Identically distributed” means each observation comes from the same population model. “Finite variance” rules out unusual mathematical models whose variability is infinite.

Visual guide 4: average more waits; compare the resulting distributions

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Sampling Distributions and the Central Limit Theorem

By the end of this lesson, you should be able to:

The concept in 60 seconds