Sampling Distributions and the CLT: Lesson, Examples and Practice

1Learning goal: what you will be able to do

A single sample mean is one draw from a distribution of possible sample means. By the end of this lesson you will be able to describe that distribution, calculate probabilities about a sample mean, and explain why the Central Limit Theorem makes the normal curve so useful in statistics.

Time55 to 60 minutes

You needPen, paper, a calculator and a normal table or calculator function

AssumedMean and standard deviation, the normal distribution and z-scores

Learning outcomes

FoundationDistinguish a population from a sample, and a parameter from a statistic. Define a sampling distribution and the standard error.
CoreState the mean, standard deviation and shape of the sampling distribution of the sample mean, and calculate probabilities about a sample mean using z-scores.
CoreState the Central Limit Theorem, check the conditions for using it, and choose a sample size to reach a target standard error.
ExtensionApply the same reasoning to a sample proportion, and explain why the standard error shrinks with the square root of n and not with n.

Where this lesson fits

This is the first lesson of Module 4 in Introductory Statistics. It is the bridge between probability and inference: everything after it, from confidence intervals to hypothesis tests, rests on the idea you meet here. The other lessons shown are planned and are not yet published.

Module 3, Probability distributions and the normal modelPlanned
Module 4, Lesson 1: Sampling distributions and the Central Limit TheoremThis lesson
Module 4, Lesson 2: Confidence intervals for a meanOpen
Module 5, Lesson 1: Hypothesis tests for a meanOpen
Module 5, Lesson 2: Inference for proportionsPlanned

Where this topic appears in courses

Sampling distributions and the Central Limit Theorem are standard content in a first university statistics course, in AP Statistics (the unit on sampling distributions) and in the IB Mathematics higher level options. The treatment here follows the usual introductory-course scope: the sample mean and sample proportion, with the normal approximation. A proof of the theorem, using moment generating functions or characteristic functions, belongs to a later probability course. This lesson supports, and does not replace, your own syllabus and instructor, so check them for the exact requirements.

2A familiar example: your commute

Your commute takes 30 minutes on average, but individual days vary a lot: the standard deviation is 10 minutes. Yesterday took 41 minutes. Is that surprising? Now suppose you record 25 mornings in a row and find an average of 41 minutes. Is that surprising?

Predict first: which would be more unusual, one 41-minute day or a 25-day average of 41 minutes?

Think about how an average behaves. One awful day is normal. A whole month of awful days is not.

The 25-day average of 41 minutes is far more unusual. A single 41-minute day is only 1.1 standard deviations above the mean (41 − 30 = 11, and 11/10 = 1.1), which happens quite often. A 25-day average varies much less, with a spread of only 10/√25 = 2 minutes, so 41 is 5.5 of those spreads above 30. You will learn to calculate exactly this.

Look at what happened in your thinking. You treated the average of 25 days as a thing that has its own spread, smaller than the spread of single days, and you knew it before doing any calculation. Good days cancel bad days. That cancelling is the whole subject of this lesson.

Statisticians say the sample mean has a sampling distribution: the pattern you would see if you could repeat the 25-day recording thousands of times and plot each average. We cannot repeat real mornings thousands of times, but a computer can. Use the simulator below to watch the pattern appear.

Try it: the sampling simulator

Each press of a draw button takes random samples of size n from the population on the left and records each sample mean on the right. Start with the skewed population and n = 1, then increase n and draw 1,000 samples each time. Watch the shape and the width.

Population

Sample size n =1

Overlay Normal curve with mean μ and SD σ/√n

Population (individual values). Dashed line = μ.

Sample means (one per sample). Green curve = prediction.

Things to notice: with n = 1 the right-hand picture copies the population. As n grows the picture narrows, centres on μ, and turns bell-shaped even though the population stays skewed. The observed spread of the means tracks the predicted SE.

3Clear explanation

Populations, samples and the two kinds of number

A population is the whole group you want to know about, such as every morning you will ever commute. A sample is the part of it you actually observe. A number that describes the population is a parameter, written with Greek letters: the population mean μ and standard deviation σ. A number computed from a sample is a statistic, written with Latin letters: the sample mean x̄. Parameters are fixed but usually unknown. Statistics are known but change from sample to sample.

The sampling distribution

Take a random sample of size n and compute x̄. Take another sample and you get a different x̄. If you could take every possible sample of size n, the collection of all those means would form a distribution. That is the sampling distribution of the sample mean. Its standard deviation has a special name: the standard error (SE). Do not confuse the three distributions in play: the population (individual values), one sample (the data you hold), and the sampling distribution (all possible sample means).

Three facts about the sample mean

For a random sample of size n from a population with mean μ and standard deviation σ:

Centre. The mean of the sampling distribution equals the population mean: E(x̄) = μ. The sample mean is an unbiased estimator: it does not lean high or low in the long run.
Spread. The standard error is SE = σ/√n, provided the observations are independent.
Shape. If the population is normal, the sampling distribution is exactly normal for every n. If it is not normal, the Central Limit Theorem says that the sampling distribution gets closer to normal as n grows.

Central Limit Theorem (CLT). Let X₁, …, X_n be independent observations from any population with finite mean μ and finite standard deviation σ. Then for large n the distribution of x̄ is approximately normal with mean μ and standard deviation σ/√n. The shape of the population does not matter, provided the standard deviation is finite.

Why the square root?

Variances of independent quantities add. The sum of n independent observations has variance nσ². The mean divides that sum by n, and dividing a quantity by n divides its variance by n², so Var(x̄) = nσ²/n² = σ²/n. Taking the square root gives SE = σ/√n. The consequence is diminishing returns:

Sample size n	SE as a multiple of σ	Meaning
1	1	No averaging at all
4	1/2	Four times the data halves the error
16	1/4	Sixteen times the data quarters it
100	1/10	A hundred times the data cuts it tenfold

From a sample mean to a probability

Once the sampling distribution is approximately normal, a probability about x̄ is a normal-table lookup. Standardise with the standard error, not with σ:

z = (x̄ − μ) / (σ/√n)

Conditions that must hold

Random sample. The data come from random sampling or random assignment. The CLT cannot repair a biased sample.
Independence. Observations do not influence one another. When sampling without replacement from a finite population, a common check is that n is no more than 10% of the population.
Large enough n, or a normal population. A common rule of thumb is n ≥ 30. A population that is only mildly skewed may need less. A heavily skewed one, or one with extreme outliers, may need much more. Treat 30 as a guide, not a law.

Extension: the sample proportion

A proportion is a mean of 0-and-1 data, so the same reasoning applies. For a population proportion p and sample size n, the sample proportion p̂ has mean p and standard error √(p(1 − p)/n). It is approximately normal when both np and n(1 − p) are at least 10.

Common mistakes

Mistake	Why it is wrong	Fix
Dividing by σ instead of σ/√n when finding a probability for x̄	That gives the probability for a single observation	Ask: is the question about one value or about a mean? Use SE for means
"The CLT makes my data normal"	The data keep the population's shape. It is the distribution of the means that approaches normal	Say which distribution you mean
"A bigger sample removes bias"	Larger n shrinks random variation only. A biased sampling method stays biased	Check how the sample was chosen first
Applying the CLT with n = 5 to a very skewed population	The approximation needs a larger n when the population is far from normal	Check the shape and the sample size together
Doubling n to halve the error	SE falls by a factor of √2, not 2. Halving needs four times the data	Remember the square root

4Worked examples

Read each example with a pencil in hand. Cover the steps, attempt the first line yourself, then compare. Every example states what is being asked, what is known, and what the answer means in context.

Example 1 Foundation 2 marks: the standard error

A population has standard deviation σ = 12. Find the standard error of the sample mean for samples of size 36. Then say what happens to it if the sample size is increased to 144.

SE = σ/√n = 12/√36 = 12/6 = 2.
With n = 144, √144 = 12, so SE = 12/12 = 1.
The sample size rose four times (36 to 144), so the standard error halved (2 to 1).

Answer: SE = 2 for n = 36. Quadrupling the sample size halves it, to 1.

Example 2 Core 4 marks: one day versus a 25-day average

Your commute has mean μ = 30 minutes and standard deviation σ = 10 minutes. Assume the sampling distribution of the mean is approximately normal. (a) Find the probability that a single day takes more than 32 minutes, treating single days as roughly normal. (b) Find the probability that the average of 25 days is more than 32 minutes.

(a) For one day use σ itself: z = (32 − 30)/10 = 0.2. Then P(X > 32) = 1 − Φ(0.2) = 1 − 0.5793 = 0.4207.
(b) For a mean of 25 days use the standard error: SE = 10/√25 = 2. Then z = (32 − 30)/2 = 1.
P(x̄ > 32) = 1 − Φ(1) = 1 − 0.8413 = 0.1587.
Sense check: averages cancel good days against bad days, so a high average is rarer than a high single day. 0.1587 < 0.4207 agrees.

Answer: (a) about 0.42; (b) about 0.16. The same 32-minute threshold is 0.2 standard deviations away for one day but 1 standard error away for a 25-day mean.

Example 3 Core 4 marks: a probability between two values

Scores on a standardised test have mean 50 and standard deviation 10. A random sample of 25 students is taken. Find the probability that the sample mean lies between 48 and 53.

SE = 10/√25 = 2.
Convert both ends: z₁ = (48 − 50)/2 = −1 and z₂ = (53 − 50)/2 = 1.5.
P(48 < x̄ < 53) = Φ(1.5) − Φ(−1) = 0.9332 − 0.1587 = 0.7745.
Sketch the curve and shade between the two z-values to confirm that the area is a little over three quarters.

Answer: about 0.77. Roughly three samples in four of this size will have a mean between 48 and 53.

Example 4 Core 5 marks: a skewed population and the CLT

Waiting times at a clinic are strongly right-skewed with mean 20 minutes and standard deviation 20 minutes. A random sample of 40 patients is taken. Find the approximate probability that their mean wait exceeds 25 minutes. Why is the answer different from the probability that one patient waits more than 25 minutes?

Check conditions. The sample is random, 40 patients are far below 10% of all patients, and n = 40 is at least 30. The CLT applies, although the skew means this is an approximation.
SE = 20/√40 = 3.162.
z = (25 − 20)/3.162 = 1.581.
P(x̄ > 25) = 1 − Φ(1.58) ≈ 0.057.
For one patient the normal model would be wrong, because the population is skewed. If the waits follow an exponential distribution, the exact probability is e^−25/20 = 0.2865.

Answer: about 0.057 for the mean of 40, against about 0.29 for one patient. The CLT lets us use the normal model for the mean even though it would be wrong for individual waits.

Example 5 Core 4 marks: choosing a sample size

A population has σ = 12. A researcher wants the margin 1.96 × SE to be at most 1.5 units, which gives a 95% margin of error for the mean. What is the smallest sample size that achieves this?

Require 1.96 × 12/√n ≤ 1.5.
Rearrange: √n ≥ 1.96 × 12/1.5 = 15.68.
Square: n ≥ 245.86.
Sample sizes are whole numbers and we must not fall short, so round up: n = 246.
Check: with n = 246 the margin is 1.96 × 12/√246 = 1.4996, which is within 1.5. With n = 245 it is 1.5026, which is not.

Answer: n = 246. Always round up, never to the nearest.

Example 6 Extension 5 marks: a sample proportion

40% of a large population supports a proposal. A random sample of 100 people is surveyed. Find the probability that at least half the sample supports it, that is, p̂ > 0.5 approximately.

Check conditions. np = 40 and n(1 − p) = 60 are both at least 10, so the normal approximation is reasonable.
SE = √(0.4 × 0.6/100) = √0.0024 = 0.0490.
z = (0.5 − 0.4)/0.0490 = 2.04.
P(p̂ > 0.5) = 1 − Φ(2.04) ≈ 0.021.

Answer: about 0.02, or roughly 1 sample in 50. A poll of 100 would rarely put support at half or more if the true figure is 40%.

Example 7 Core 4 marks: judging whether the conditions hold

For each situation, decide whether it is reasonable to treat the sampling distribution of the mean as approximately normal. (a) n = 8 from a normally distributed population. (b) n = 8 from a population with a long right tail. (c) n = 200 customers who all entered a shop during a free-gift promotion, used to estimate the average spend of all customers on a normal day.

(a) Yes. When the population is normal, the sampling distribution is exactly normal for every n, even 1.
(b) Not safely. A sample of 8 is too small for the CLT to smooth a long tail. The sampling distribution probably keeps some skew.
(c) The shape is fine because n = 200 is large, but the sample is not random for the population of interest. The sampling distribution is approximately normal around the wrong centre, so conclusions about a normal day are biased. Size cannot repair bias.

Answer: (a) yes; (b) not safely; (c) approximately normal in shape but biased, so not valid for the stated purpose.

6Answer explanations

Attempt every question before you open these. A right answer with a wrong method is luck, so read each explanation even when you were correct. Questions 3, 5 and 7 show their own worked solution as soon as you answer.

Question 1: the sampling distribution for n = 50

Answer: approximately normal, with mean μ and standard deviation σ/√50. The Central Limit Theorem says the distribution of the sample mean approaches a normal shape as n grows, whatever the population shape. Its centre is μ and its spread is the standard error σ/√n.

Why the others fail: the population keeps its skew, but the means do not (option a confuses the two distributions). "Exactly normal" is only true for a normal population, and the spread σ ignores the averaging effect (option c). Dividing the mean by √50 is meaningless, since averaging does not move the centre (option d).

Question 2: SE for σ = 20, n = 64

SE = 20/√64 = 20/8 = 2.5.

A common slip is dividing by n (20/64 = 0.3125) instead of by √n. Remember that the variance is divided by n, so the standard deviation is divided by its square root.

Question 3: generated standard error

Divide the population standard deviation by the square root of the sample size. When √n is a whole number the work is mental arithmetic. Use New numbers to repeat the question until the method is automatic.

Question 4: P( sample mean > 103 ) with μ = 100, σ = 15, n = 36

SE = 15/√36 = 15/6 = 2.5. Then z = (103 − 100)/2.5 = 1.2.

P(x̄ > 103) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.

A common wrong answer is 0.4207, which comes from using σ = 15 instead of the standard error: z = 3/15 = 0.2. That is the probability for one person, not for a mean of 36.

Question 5: generated probability for a sample mean

The method never changes: (1) standard error = σ/√n; (2) z = (x̄ − μ)/SE; (3) look up Φ(z); (4) for a "greater than" question subtract the table value from 1, and for a "less than" question use it directly. Sketching the curve and shading the required tail catches most direction errors.

Question 6: doubling the sample size

Answer: multiplied by about 0.71. The new SE divided by the old SE is (σ/√100)/(σ/√50) = √50/√100 = √0.5 = 0.707. Doubling n cuts the standard error by about 29%, not 50%.

To halve the standard error you need four times the sample size, as the table in the explanation shows.

Question 7: generated sample size

Set σ/√n equal to the target, solve for n = (σ/target)², and round up. Rounding to the nearest whole number can leave you slightly short of the target, as Example 5 showed.

Question 8: sample proportion defective

Check conditions: np = 20 and n(1 − p) = 80 are both at least 10.

SE = √(0.2 × 0.8/100) = √0.0016 = 0.04. Then z = (0.15 − 0.20)/0.04 = −1.25.

P(p̂ < 0.15) = Φ(−1.25) = 0.1056. A common slip is using the population standard deviation formula √(p(1 − p)) = 0.4 without dividing by √n.

Question 9: when the normal model is least reasonable

Answer: a random sample of 6 from a population with a very long right tail. The sample is far too small for the CLT to remove a heavy skew.

Why the others are acceptable: a normal population needs no CLT (option a); 100 observations are ample for moderate skew (option c); 40 is above the usual guide of 30 for a roughly symmetric population (option d).

7Short recap

A parameter describes a population; a statistic describes a sample. The sample mean x̄ changes from sample to sample.
The sampling distribution is the distribution of a statistic over all possible samples of size n.
The sample mean has mean μ and standard error σ/√n (for independent observations).
If the population is normal, x̄ is exactly normal. Otherwise the Central Limit Theorem makes it approximately normal for large n.
For a probability about a mean, use z = (x̄ − μ)/(σ/√n). For one observation, use σ.
Check the conditions: random sample, independence (about 10% rule), and large n unless the population is normal.
Halving the standard error needs four times the data. Round sample sizes up.
For a proportion, SE = √(p(1 − p)/n), with np and n(1 − p) both at least 10.

Tick what you can now do

I can distinguish a population from a sample and define a sampling distribution and the standard error.
I can state the centre, spread and shape of the sampling distribution of the mean and calculate probabilities with z-scores.
I can state the CLT, check its conditions and choose a sample size for a target standard error.
I can apply the same reasoning to a proportion and explain the square-root law.

If you ticked fewer than four, return to the matching worked example, redo it without looking, then try the generated questions again. This idea carries straight into confidence intervals and then hypothesis tests, the next lessons in this course.

Related Encyclopedia entries

Confidence Intervals for a Mean: the next lesson, which uses the standard error to build interval estimates.
Statistical significance, p-value and margin of error: where sampling distributions lead next.
Nonresponse bias: why a large sample cannot fix a biased one.
Ecological Sampling virtual lab: practise random sampling in a hands-on setting.

References and further reading

Titles and publishers only. Search each site for the title to reach the exact page.

OpenStax, Introductory Statistics 2e, the chapter on the Central Limit Theorem. Free and open textbook. https://openstax.org/
Diez, D. M., Barr, C. D. and Çetinkaya-Rundel, M., OpenIntro Statistics, the section on the sampling distribution of the mean. Free and open textbook. https://www.openintro.org/
Freedman, D., Pisani, R. and Purves, R., Statistics, the chapters on the expected value and standard error and on the normal approximation for sample averages. W. W. Norton.
MIT OpenCourseWare, 18.05 Introduction to Probability and Statistics, lecture notes on the Central Limit Theorem. https://ocw.mit.edu/

Sampling Distributions and the Central Limit Theorem