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.

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The sample average is:

$\huge s_n:=\frac{\sum_i X_i}{N}$

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.

Standard error of the mean

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 $\bar x$?

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.

Check out that link.

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