Last updated on 8 September 2025
The central limit theorem is a fundamental concept of statistics, which 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 that 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.
As we have discussed, the central limit theorem. Let us now learn about the key components of the central limit theorem. The key concepts are:
As we know, the central limit theorem is a fundamental concept in statistics and probability, as it helps to understand how the population estimates the behavior under repeated sampling. Now we will discuss how it works and the formulas.
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. The z score of the random variable, x̄ is \(
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
\)
The central limit theorem is used in the real world to analyze the population characteristics. Let’s discuss the real-world application of the central limit theorem.
Mistakes are common among students when working on the central limit theorem, as they tend to get confused. So to master the central limit theorem, we can learn a few common mistakes and ways to avoid them.
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 $45000
The standard deviation of the sample mean is $1000
The mean of the sample mean is equal to the population mean
So the sample mean is $45000
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!