> For the complete documentation index, see [llms.txt](https://stephanosterburg.gitbook.io/scrapbook/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://stephanosterburg.gitbook.io/scrapbook/math/statistics-and-probability/central-limit-theorem.md).

# Central Limit Theorem

Sample size of $$n=4$$

$$S\_1 = \[1, 1, 3, 6] \Rightarrow \bar{x\_1}= 2.75 \newline S\_2 = \[3, 4, 3, 1] \Rightarrow \bar{x\_2} = 2.75 \newline S\_3 = \[1, 1, 6, 6] \Rightarrow \bar{x\_3} = 3.50 \newline .\newline.\newline.\newline. \newline S\_n$$

Add your samples together and divide by the number of sample size.

{% embed url="<http://onlinestatbook.com/2/index.html>" %}

The sample average is:

$$\huge s\_n:=\frac{\sum\_i X\_i}{N}$$<br>

For large ​$$n$$, the distribution of sample sums $$​S\_n$$​ is close to normal distribution ​$$N(\mu^\prime,\sigma^\prime)$$where:

* $$​​\mu^\prime=n \times \mu$$
* $$​​\sigma^\prime=\sqrt{n} \times \sigma$$

The **marginOfError** formula can be found [here](https://www.thoughtco.com/margin-of-error-formula-3126275).

#### Standard error of the mean

$$\huge \sigma\_\bar{x}^2 = \frac{\sigma^2}{n} \Longrightarrow \sigma\_\bar{x}=\frac{\sigma}{\sqrt{n}}$$

## Example: Means in quality control

An auto-maker does quality control tests on the paint thickness at different points on its car parts since there is some variability in the painting process. A certain part has a target thickness of $$2\text{ mm}$$. The distribution of thicknesses on this part is skewed to the right with a mean of $$2\text{ mm}$$ and a standard deviation of $$0.5\text{ mm}$$.

A quality control check on this part involves taking a random sample of $$100$$ points and calculating the mean thickness of those points.

**Assuming the stated mean and standard deviation of the thicknesses are correct, what is the probability that the mean thickness in the sample of** $$100$$**points is within** $$0.1\text{ mm}$$ **of the target value?**

### **Part 1: Establish normality**

Approximately normal. Since $$n=100 \geq 30$$, the central limit theorem applies. Even though the population of thicknesses is skewed to the right, the sample means will be normally distributed since the sample size is large.

### Part 2: Find the mean and standard deviation of the sampling distribution

The sampling distribution of a sample mean $$\bar x$$ has:

$$\begin{aligned} \mu\_{\bar x}&=\mu \\\ \sigma\_{\bar x}&=\dfrac{\sigma}{\sqrt n} \end{aligned}$$

Note: For this standard deviation formula to be accurate, our sample size needs to be $$10%$$ or less of the population so we can assume independence.

**What is the mean of the sampling distribution of** $$\bar x$$**?**&#x20;

On average, the sample means will equal the population mean.

$$\mu\_{\bar x}=\mu=2\text{ mm}$$

**What is the standard deviation of the sampling distribution of** $$\bar x$$**?**

It's reasonable to assume that there are more than $$1000$$ points on the part, so we can safely use this formula:

$$\sigma\_{\bar x}=\dfrac{\sigma}{\sqrt n}=\dfrac{0.5}{\sqrt {100}}=\dfrac{0.5}{10}=0.05\text{ mm}$$

### Part 3: Use normal calculations to find the probability in question

**Assuming the stated mean and standard deviation of the thicknesses are correct, what is the approximate probability that the mean thickness in the sample of** $$100$$ **points is within** $$0.1\text{ mm}$$ **of the target value?**

In any normal distribution, we know that approximately $$68%$$ of the data falls within one standard deviation of the mean, $$95%$$9 of data falls within two standard deviations of the mean, and $$99.7%$$ of data falls within three standard deviations of the mean.

We already established that the sample mean thickness $$\bar x$$ is normally distributed with $$\mu\_{\bar x}=2\text{ mm}$$ and $$\sigma\_{\bar x}=0.05\text{ mm}$$. "Within $$0.1\text{ mm}$$ of the target value" is exactly two standard deviations above and below the mean.

<div align="left"><img src="/files/-LNFxGFHjAW3p26zXClO" alt=""></div>

Check out that [link](https://osterburg.gitbook.io/scrapbook/~/edit/drafts/-LMvY91E6c4kssbA4IaZ/statistics-and-probability/modeling-data-distribution).

**There is about a** $$95%$$ **probability that sample mean is within** $$0.1\text{ mm}$$ **of the target value.**