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1325 LearnersLast updated on November 25, 2025

The central limit theorem is a fundamental concept in statistics. It states that the sampling distribution of the mean will be normal regardless of the shape of the population distribution, as the sample size is large. In this topic, we will learn about it in detail.
The central limit theorem states that if we consider numerous random samples from a population, the sample mean distribution will form a normal distribution, which is a bell curve. The distribution of the original population can be skewed, Poisson, or binomial, but the mean distribution of the sample will be normal when the sample size is greater than or equal to 30. So, if the sample size is larger, it allows for better estimation of population characteristics.
The Central Limit Theorem diagram clarifies a counterintuitive concept: how random chaos becomes predictable order. It visually demonstrates that no matter how weird or skewed the original "parent" population looks, the distribution of its sample averages will always shape-shift into a perfect Bell Curve.
So far, we have discussed what the central limit theorem states. Let us now learn about the key components of the theorem. The key concepts are:
Here are the four key assumptions of the Central Limit Theorem (CLT) in point form:


Now let’s look at the formula for the central limit theorem. Let X be a random variable with a known or unknown probability distribution. The standard deviation is σ, and the mean of X is μ. According to the central limit theorem, if the large number sample are drawn of size n, then the new random variable is x̄, then,
\(\bar{x} \sim N\!\left(\mu, \frac{\sigma}{\sqrt{n}}\right)\), where σ / √n is the standard error. The z score of the random variable x̄ is \( z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \)
Below is the step-by-step proof of the Central Limit Theorem.
Let \(X_1, X_2, ..., X_n\) be a sequence of independent and identically distributed (i.i.d) random variables with:
We want to prove that the standardized sum converges to a Standard Normal Distribution \((Z \sim N(0,1))\) as \(n \to \infty.\)
First, we shift and scale the variables to make the math cleaner. Let \(Y_i \) be the standardized version of \(X_i\):
\(Y_i = \frac{X_i - \mu}{\sigma}\)
Consequently, the properties of \(Y_i\) are:
\(E[Y_i] = 0\)
\(Var(Y_i) = E[Y_i^2] - (E[Y_i])^2 = 1\)
We define the statistic \(Z_n\) (the standardized sample mean) as:
\(Z_n = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = \frac{1}{\sqrt{n}} \sum_{i=1}^{n} Y_i\)
We need to find the MGF of \(Z_n\), denoted as \(M_{Z_n}(t)\).
\(M_{Z_n}(t) = E\left[ e^{t Z_n} \right] = E\left[ e^{t \frac{1}{\sqrt{n}} \sum Y_i} \right]\)
Using the property of exponentials \((e^{a+b} = e^a e^b)\) and the independence of \(Y_i\):
\(M_{Z_n}(t) = E\left[ \prod_{i=1}^{n} e^{\frac{t}{\sqrt{n}} Y_i} \right] = \prod_{i=1}^{n}E\left[ e^{\frac{t}{\sqrt{n}} Y_i} \right]\)
Since all \(Y_i\) are identical, we can write this in terms of the MGF of a single Y:
\(M_{Z_n}(t) = \left( M_Y\left( \frac{t}{\sqrt{n}} \right) \right)^n\)
We expand \(M_Y(s)\) using a Taylor series around 0.
\(M_Y(s) = M_Y(0) + M_Y'(0)s + \frac{M_Y''(0)s^2}{2} + o(s^2)\)
Recall the moments of Y:
\(M_Y(0) = 1\)
\(M_Y'(0) = E[Y] = 0\)
\(M_Y''(0) = E[Y^2] = 1\)
Now, substitute \(s = \frac{t}{\sqrt{n}}\) into the expansion:
\(M_Y\left( \frac{t}{\sqrt{n}} \right) = 1 + 0 \cdot \frac{t}{\sqrt{n}} + \frac{1}{2} \left( \frac{t}{\sqrt{n}} \right)^2 + o\left( \frac{t^2}{n} \right)\)
\(M_Y\left( \frac{t}{\sqrt{n}} \right) = 1 + \frac{t^2}{2n} + o\left( \frac{1}{n} \right)\)
Now substitute this back into our equation for \(M_{Z_n}(t)\) and take the limit as \(n \to \infty\):
\(\lim_{n \to \infty} M_{Z_n}(t) = \lim_{n \to \infty} \left( 1 + \frac{t^2/2}{n} \right)^n\)
Recall the standard calculus limit definition of \(e^x\): \(\lim_{n \to \infty} (1 + \frac{x}{n})^n = e^x\).
Here, \(x = t^2/2\).
Therefore:
\(\lim_{n \to \infty} M_{Z_n}(t) = e^{t^2/2}\)
The function \(e^{t^2/2}\) is the Moment Generating Function of the Standard Normal Distribution.
By the Uniqueness Theorem, since the MGF of \(Z_n\) converges to the MGF of the Standard Normal distribution, the distribution of \(Z_n\) must converge to the Standard Normal distribution. Q.E.D.
Here are the key properties of the Central Limit Theorem:
As we know, the central limit theorem is a fundamental concept in statistics and probability; it helps us understand how the population estimates the behavior under repeated sampling. Now we will discuss how it works and the formulas.
Write down the Sample Size (n), Population Mean (\(\mu\)), and Standard Deviation (\(\sigma\)). Identify the inequality in the question:
Sketch a Normal Distribution curve with the Mean (\(\mu\)) in the center.
Apply the CLT formula to convert your value into a Z-score.
\(Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\)
(Note: For “Between” problems, you must calculate two Z-scores: one for the lower limit and one for the upper limit.)
Locate the calculated Z-score in the Z-table to find the corresponding Probability Area (Decimal Value). (Note: Most standard tables give the area to the left of the Z-score).
Use the decimal value from Step 4 based on your problem type:
Look at your graph from Step 2.
Convert the final decimal value into a percentage if required (multiply by 100).
The Central Limit Theorem is often the most confusing concept in introductory statistics because it is abstract. It is counterintuitive—it claims that even if your data is messy and random, the averages of that data will naturally form a perfect, orderly pattern. Here are some tips and tricks to help you with the concept:
Mistakes are common among students when working on the central limit theorem. So to master the central limit theorem, we can learn about a few common mistakes and ways to avoid them.
The central limit theorem is used in the real world to analyze the population characteristics. Let’s discuss the real-world applications of the central limit theorem.
A company reports that the average salary of its employees is $45,000 per year, with a standard deviation of $8,000. If a random sample of 64 employees is taken, what are the mean and standard deviation of the sample mean salaries?
The mean of the sample mean is $45,000
The standard deviation of the sample mean is $1,000
The mean of the sample mean is equal to the population mean
So the sample mean is $45,000
The standard deviation of the sample is calculated by using the formula, \( \sigma \over \sqrt{n}\)
That is, \( {8000 \over \sqrt{64}} \)
\( 8000\over 8\) = 1000.
The study hours of college students follow a distribution with a mean of 15 hours per week and a standard deviation of 5 hours. If a sample of 49 students is taken, what are the mean and standard deviation of the sample mean study hours?
The mean of the sample mean is 15 hours.
The standard deviation of the sample mean is 0.71 hours.
The mean of the sample mean is equal to the population mean
So the sample mean is 15 hours
The standard deviation of the sample is calculated by using the formula, \( \sigma \over \sqrt{n}\)
That is, \( 5\over \sqrt{49}\) = \( 5 \over 7\) = 0.71.
The daily coffee consumption of people in a city follows a distribution with a mean of 2.5 cups and a standard deviation of 0.8 cups. If a random sample of 36 people is selected, what are the mean and standard deviation of the sample mean daily coffee consumption?
The mean of the sample mean is 2.5 cups.
The standard deviation of the sample mean is 0.133 cups.
The mean of the sample mean is equal to the population mean
So the sample mean is 2.5 cups
The standard deviation of the sample is calculated by using the formula, \( \sigma \over \sqrt{n}\)
That is \( 0.8 \over \sqrt{36}\)
= \( 0.8 \over {6}\) = 0.133.
A study on airline flight delays finds that the average delay time is 25 minutes, with a standard deviation of 10 minutes. If a sample of 64 flights is chosen, what are the mean and standard deviation of the sample mean delay times?
The mean of the sample mean is 25 minutes.
The standard deviation of the sample mean is 1.25 minutes.
The mean of the sample mean is equal to the population mean
So the sample mean is 25 minutes
The standard deviation of the sample is calculated by using the formula, \( \sigma \over \sqrt{n}\)
That is, \( 10 \over \sqrt{64}\)
= \( 10\over 8\) = 1.25.
The recorded high temperatures in a city during summer have a mean of 95°F and a standard deviation of 6°F. If a random sample of 81 days is selected, what are the mean and standard deviation of the sample mean temperatures?
The mean of the sample mean is 95°F.
The standard deviation of the sample mean is 0.667°F.
The mean of the sample mean is equal to the population mean
So the sample mean is 95°F
The standard deviation of the sample is calculated by using the formula, \( \sigma \over \sqrt{n}\)
That is, \( 6 \over \sqrt{81}\)
= \( 6\over 9\) = 0.667
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






