Now that we’ve examined the CLT from a high level, let’s get into the details.

The CLT not only establishes that the sampling distribution will be normally distributed, but it also allows us to describe that normal distribution quantitatively. Normal distributions are described by their mean `μ`

(mu) and standard deviation `σ`

(sigma).

Let’s break this up:

- We take samples of size
*n*from a population (that has a true population mean`μ`

and standard deviation of`σ`

) and calculate the sample mean`x`

. - Given that
*n*is sufficiently large (n > 30), the sampling distribution of the means will be normally distributed with:- mean
`x`

approximately equal to the population mean`μ`

- standard deviation equal to the population standard deviation divided by the square root of the sample size. We can write this out as:$Sampling\ Distribution\ St.Dev = \frac{\sigma}{\sqrt{n}}$

- mean

We’ll focus on the first point in this exercise and the second point in the next exercise.

As an example of this, let’s look again at our salmon fish population. Last exercise, we saw that the sampling distribution of the mean was normally distributed. In the plot below, we can see that the mean of the simulated sampling distribution is approximately equal to the population mean.

In the workspace, we’ve simulated a sampling distribution of the mean using a sample size of 50.

### Instructions

**1.**

In the workspace, we’ve set up a simulation of a population that has a mean of 10 and a standard deviation of 10. We’ve set a sample size of 50.

According to the CLT, we should have a sampling distribution of the mean that is normally distributed and has a mean that is close to the population mean.

Run the code once. Does what you see align with the CLT?

*Make sure to scroll down in the web browser to see the sampling distribution of the mean.*

**2.**

Set variable `samp_size`

equal to 6 and run the code.

Why do you think the CLT applies here, even with a smaller sample size?