Central Limit Theorem: Formula, Solved Examples & Practice

5 fully solved probability examples and 3 practice problems on sample means

What is the Central Limit Theorem?

The Central Limit Theorem (CLT) states that when sufficiently large random samples are drawn from a population with mean \( \mu \) and standard deviation \( \sigma \), the sampling distribution of the sample mean \( \bar X \) is approximately normal, with mean \( \mu \) and standard deviation \( \sigma/\sqrt{n} \).

Key Facts

Central Limit Theorem Formulas

Quantity Formula
Mean of the sampling distribution \( \mu_{\bar X} = \mu \)
Standard error of the mean \( \sigma_{\bar X} = \dfrac{\sigma}{\sqrt{n}} \)
Z-score for a sample mean \( Z = \dfrac{\bar X - \mu_{\bar X}}{\sigma_{\bar X}} = \dfrac{\bar X - \mu}{\sigma/\sqrt{n}} \)

Why \( \sigma/\sqrt{n} \)? If \( n \) independent observations each have variance \( \sigma^2 \), then \( \bar X \) has variance \( \sigma^2/n \). Taking square roots gives \( \sigma_{\bar X} = \sigma/\sqrt{n} \).

Sampling Distributions in Action — A Fair Die

Consider the discrete uniform population on \( \{1, 2, 3, 4, 5, 6\} \), like rolling a fair die.

Uniform probability distribution of a die: six bars of equal height 1/6 for values 1 through 6
Figure 1 — Uniform population distribution (fair die).

1. Population parameters

\[ \mu = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5 \] \[ \sigma^2 = \frac{6^2 - 1}{12} = \frac{35}{12} \approx 2.9167 \] \[ \sigma = \sqrt{\frac{35}{12}} \approx 1.7078 \]

2. Sampling distribution of \( \bar X \) for \( n = 2 \)

Enumerate all \( 6^2 = 36 \) equally likely samples of size 2 (with replacement). The 36 possible sample means are:

1.01.52.02.53.03.5
1.52.02.53.03.54.0
2.02.53.03.54.04.5
2.53.03.54.04.55.0
3.03.54.04.55.05.5
3.54.04.55.05.56.0

Tabulating their frequencies:

\( \bar X \) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
\( P(\bar X) \) 1/362/363/364/365/366/36 5/364/363/362/361/36
Triangular distribution of sample means for n = 2 from a uniform die, peak at 3.5
Figure 2 — Sampling distribution of \( \bar X \) for \( n = 2 \) (triangular shape).

Mean of the sampling distribution:

\[ \mu_{\bar X} = \sum_i \bar X_i\,P(\bar X_i) = 3.5 = \mu \]

Standard error via the formula:

\[ \sigma_{\bar X} = \frac{\sigma}{\sqrt{n}} = \frac{1.7078}{\sqrt{2}} \approx 1.2076 \]

Standard error via direct computation (from the table of probabilities):

\[ \sigma_{\bar X} = \sqrt{\sum_i (\bar X_i - \mu_{\bar X})^2\,P(\bar X_i)} = \sqrt{\tfrac{35}{24}} \approx 1.2076 \]

They match exactly. The standard error formula \( \sigma/\sqrt{n} \) is not an approximation — it holds exactly for samples of independent observations. What the CLT adds is that the shape of the sampling distribution becomes approximately normal as \( n \) grows. For \( n = 2 \) from a uniform die, the shape is a symmetric triangle (peak at \( 3.5 \)), not yet a bell.

Worked Examples

Example 1 — Sample mean below 19

A population has mean \( \mu = 20 \) and standard deviation \( \sigma = 4 \). A random sample of \( n = 64 \) observations is selected. What is the probability that \( \bar X < 19 \)?

View solution

\( n = 64 \ge 30 \), so the CLT applies.

\[ \mu_{\bar X} = 20, \qquad \sigma_{\bar X} = \frac{4}{\sqrt{64}} = \frac{4}{8} = 0.5 \] \[ Z = \frac{19 - 20}{0.5} = -2.0 \] \[ P(\bar X < 19) = P(Z < -2.0) \approx 0.0228 \]

Interpretation: about a 2.28% chance that a sample of 64 observations averages below 19.

Example 2 — Investment returns between 4% and 5%

251 investing companies had an average return of 4.5% with standard deviation 1.5%. For a random sample of 40 companies, what is the probability that the average return is between 4% and 5%?

View solution

\( n = 40 \ge 30 \), so the CLT applies.

\[ \mu_{\bar X} = 4.5\%, \qquad \sigma_{\bar X} = \frac{1.5\%}{\sqrt{40}} \approx 0.2372\% \]

Z-scores for the two boundaries:

\[ Z_1 = \frac{4 - 4.5}{0.2372} \approx -2.108, \qquad Z_2 = \frac{5 - 4.5}{0.2372} \approx 2.108 \] \[ P(4\% < \bar X < 5\%) = P(-2.108 < Z < 2.108) \approx 0.965 \]

Interpretation: about a 96.5% chance.

Example 3 — Average 5-year return above 90%

A group of mutual funds had an average 5-year return of 80% with standard deviation 30%. For a random sample of 50 funds, what is the probability that their average return exceeds 90%?

View solution

\( n = 50 \ge 30 \), so the CLT applies.

\[ \mu_{\bar X} = 80\%, \qquad \sigma_{\bar X} = \frac{30\%}{\sqrt{50}} \approx 4.2426\% \] \[ Z = \frac{90 - 80}{4.2426} \approx 2.357 \] \[ P(\bar X > 90\%) = 1 - P(Z < 2.357) \approx 1 - 0.9908 = 0.0092 \]

Interpretation: only about 0.92% chance.

Example 4 — Tools within 0.05 cm of the target

A factory produces tools with average length 10 cm and standard deviation 0.3 cm. For a random sample of 200 tools, what is the probability the sample mean is within 0.05 cm of the true mean?

View solution

\( n = 200 \ge 30 \), so the CLT applies.

\[ \mu_{\bar X} = 10, \qquad \sigma_{\bar X} = \frac{0.3}{\sqrt{200}} \approx 0.0212 \]

The interval \( 10 \pm 0.05 \) corresponds to \( 9.95 \) to \( 10.05 \):

\[ Z_1 = \frac{9.95 - 10}{0.0212} \approx -2.358, \qquad Z_2 = \frac{10.05 - 10}{0.0212} \approx 2.358 \] \[ P(9.95 < \bar X < 10.05) \approx 0.9816 \]

Interpretation: about a 98.2% chance.

Example 5 — Airline baggage weight limit

An airplane with 200 seats has a total baggage limit of 6,000 kg. Individual passenger baggage weights have mean 28 kg and standard deviation 15 kg (unknown distribution). What is the probability that 200 passengers' total baggage stays under the limit?

View solution

Convert to a sample-mean constraint. The total must satisfy \( 200 \cdot \bar X \le 6000 \), i.e. \( \bar X \le 30 \) kg.

\( n = 200 \ge 30 \), so the CLT applies even for a skewed distribution of individual weights.

\[ \mu_{\bar X} = 28, \qquad \sigma_{\bar X} = \frac{15}{\sqrt{200}} \approx 1.0607 \] \[ Z = \frac{30 - 28}{1.0607} \approx 1.886 \] \[ P(\bar X \le 30) = P(Z \le 1.886) \approx 0.9703 \]

Interpretation: about a 97.0% chance the total stays under 6,000 kg.

Practice Problems

Practice 1 — Sample mean above 52

A population has mean \( \mu = 50 \) and standard deviation \( \sigma = 8 \). For a random sample of \( n = 100 \), what is the probability that \( \bar X > 52 \)?

Answer
\[ \sigma_{\bar X} = \frac{8}{\sqrt{100}} = 0.8, \qquad Z = \frac{52 - 50}{0.8} = 2.5 \] \[ P(\bar X > 52) = P(Z > 2.5) = 1 - 0.9938 = 0.0062 \]

Practice 2 — Interval probability

A population has mean \( \mu = 75 \) and standard deviation \( \sigma = 12 \). For a random sample of \( n = 36 \), what is the probability that \( 72 < \bar X < 78 \)?

Answer
\[ \sigma_{\bar X} = \frac{12}{6} = 2, \qquad Z_1 = \frac{72 - 75}{2} = -1.5, \qquad Z_2 = \frac{78 - 75}{2} = 1.5 \] \[ P(-1.5 < Z < 1.5) \approx 0.8664 \]

Practice 3 — Finding a required sample size

A population has standard deviation \( \sigma = 20 \). How large a sample is needed so that the standard error of the mean is at most 2?

Answer

Solve \( \sigma/\sqrt{n} \le 2 \):

\[ \frac{20}{\sqrt{n}} \le 2 \implies \sqrt{n} \ge 10 \implies n \ge 100 \]

So \( n = 100 \) is the minimum sample size.

Frequently Asked Questions

What is the Central Limit Theorem?

For sufficiently large random samples from any population with mean \(\mu\) and standard deviation \(\sigma\), the sampling distribution of the sample mean \(\bar X\) is approximately normal with mean \(\mu\) and standard deviation \(\sigma/\sqrt{n}\). The approximation improves with larger \(n\).

How large does the sample need to be for the CLT?

A common rule of thumb is \(n \ge 30\), but this depends on the population's shape: symmetric populations converge quickly (sometimes \(n = 10\) suffices); heavily skewed populations may need \(n \ge 50\) or more. If the population is already normal, the sample mean is normal for any \(n\).

What is the standard error of the mean?

It's \(\sigma_{\bar X} = \sigma/\sqrt{n}\), the standard deviation of the sampling distribution of \(\bar X\). It shrinks by a factor of \(\sqrt{n}\) compared to the population standard deviation.

Do I need a continuity correction when using the CLT?

Only if you're approximating a discrete count or proportion with a continuous normal distribution — then a ±0.5 continuity correction improves accuracy. For the sample mean of a continuous variable, no continuity correction is needed.

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Last updated: October 20, 2026