Why Sample Means Matter: The Central Limit Theorem Explained

Lesson Plan - Get It!

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The Central Limit Theorem explains what happens when you take multiple random samples from a population and calculate their means.

These sample means tend to form a pattern—a normal distribution—regardless of what the original population looks like.

Breaking Down Sample Means and Distributions

Imagine you have a population of students and record their scores on a test. The scores might be all over the place—some students did great, others not so much, and the distribution of scores might not be a neat, symmetrical shape.

Now, say you take a random sample of 30 students and find the mean (average) of their scores. Then, you do it again with another random sample of 30 students. You repeat this process over and over, collecting many sample means.

• What happens when you plot all these sample means on a graph?

Instead of looking like the original test scores, the sample means will form a normal (bell-shaped) curve. This happens regardless of what the original population looked like. This is the Central Limit Theorem at work!

Key observations from this process.

The distribution of sample means looks normal, even if the original data wasn’t.

The mean of the sample means is the same as the mean of the population.

The standard deviation of the sample means is smaller than the standard deviation of the population, meaning the sample means are more consistent.

Understanding the CLT

Under specific conditions, the three observations above will always hold true. Together, they form the basis of the Central Limit Theorem:

If you repeatedly take random samples of size n from a population with mean (μ) and standard deviation (σ), the distribution of sample means will be approximately normal, with mean (μ) and a standard deviation of (σ / √n*).

Here are the conditions for this to work.

Each sample must be the same size.

The samples must be taken randomly.

The sampling process should be repeated an infinite number of times.

The sample size should be large enough (usually 30 or more).

Example Problem

Suppose the number of text messages sent by teens daily has a mean of 60 and a standard deviation of 12.

• If you take random samples of 36 teens, what is the shape, mean, and standard deviation of the distribution of sample means?

Shape: The sample means will form a normal distribution.

Mean: The mean of the distribution will still be 60.

Standard Deviation: The new standard deviation is 12 / √36 = 2.

Confirm your understanding by finding the mean and the standard deviation of the following distributions of sample means.

Now, apply this to probability calculations.

Finding Probabilities With the Normal Curve

Use the same texting example. Suppose you take a sample of 36 teens.

• What is the probability that the average number of texts they send is between 59 and 61 per day?

You already calculated the mean and standard deviation of the sample means.

Mean = 60

Standard Deviation = 2

Now, convert 59 and 61 into z-scores, which measure how many standard deviations a value is from the mean.

Z-score formula:

Z = (X - μ) / (σ/√n)

For X = 59:

Z = (59 - 60) / 2 = -0.5

For X = 61:

Z = (61 - 60) / 2 = 0.5

Now, using a standard normal table (or a calculator), find the area under the normal curve between -0.5 and 0.5. This area represents the probability.

The probability of Z being less than 0.5 is 0.6915.

The probability of Z being less than -0.5 is 0.3085.

So, the probability between 59 and 61 is 0.6915 - 0.3085 = 0.383.

This means there is about a 38.3% chance that a randomly selected sample of 36 teens will have an average number of texts between 59 and 61 per day.

Try solving a similar problem using a different range of values before moving on to the practice section.

Nice! Keep it going in the Got It? section!