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Last updated on October 6, 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.
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:
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.
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 \( z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \)
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.
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.
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!