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On this page
  • Example: Means in quality control
  • Part 1: Establish normality
  • Part 2: Find the mean and standard deviation of the sampling distribution
  • Part 3: Use normal calculations to find the probability in question
  1. math
  2. Statistics and probability (khan academy)

Central Limit Theorem

PreviousGeometric random variableNextSignificance Tests (hypothesis testing)

Last updated 6 years ago

Sample size of n=4n=4n=4

S1=[1,1,3,6]⇒x1ˉ=2.75S2=[3,4,3,1]⇒x2ˉ=2.75S3=[1,1,6,6]⇒x3ˉ=3.50....SnS_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 S1​=[1,1,3,6]⇒x1​ˉ​=2.75S2​=[3,4,3,1]⇒x2​ˉ​=2.75S3​=[1,1,6,6]⇒x3​ˉ​=3.50....Sn​

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

The sample average is:

sn:=∑iXiN\huge s_n:=\frac{\sum_i X_i}{N}sn​:=N∑i​Xi​​

For large ​nnn, the distribution of sample sums ​Sn​S_n​Sn​​ is close to normal distribution ​N(μ′,σ′)N(\mu^\prime,\sigma^\prime)N(μ′,σ′)where:

  • ​​μ′=n×μ​​\mu^\prime=n \times \mu​​μ′=n×μ

  • ​​σ′=n×σ​​\sigma^\prime=\sqrt{n} \times \sigma​​σ′=n​×σ

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 mm2\text{ mm}2 mm. The distribution of thicknesses on this part is skewed to the right with a mean of 2 mm2\text{ mm}2 mm and a standard deviation of 0.5 mm0.5\text{ mm}0.5 mm.

A quality control check on this part involves taking a random sample of 100100100 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 100100100points is within 0.1 mm0.1\text{ mm}0.1 mm of the target value?

Part 1: Establish normality

Approximately normal. Since n=100≥30n=100 \geq 30n=100≥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 xˉ\bar xxˉ has:

μxˉ=μσxˉ=σn\begin{aligned} \mu_{\bar x}&=\mu \\\\ \sigma_{\bar x}&=\dfrac{\sigma}{\sqrt n} \end{aligned}μxˉ​σxˉ​​=μ=n​σ​​

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

What is the mean of the sampling distribution of xˉ\bar xxˉ?

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

μxˉ=μ=2 mm\mu_{\bar x}=\mu=2\text{ mm}μxˉ​=μ=2 mm

What is the standard deviation of the sampling distribution of xˉ\bar xxˉ?

It's reasonable to assume that there are more than 100010001000 points on the part, so we can safely use this formula:

σxˉ=σn=0.5100=0.510=0.05 mm\sigma_{\bar x}=\dfrac{\sigma}{\sqrt n}=\dfrac{0.5}{\sqrt {100}}=\dfrac{0.5}{10}=0.05\text{ mm}σxˉ​=n​σ​=100​0.5​=100.5​=0.05 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 100100100 points is within 0.1 mm0.1\text{ mm}0.1 mm of the target value?

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

We already established that the sample mean thickness xˉ\bar xxˉ is normally distributed with μxˉ=2 mm\mu_{\bar x}=2\text{ mm}μxˉ​=2 mm and σxˉ=0.05 mm\sigma_{\bar x}=0.05\text{ mm}σxˉ​=0.05 mm. "Within 0.1 mm0.1\text{ mm}0.1 mm of the target value" is exactly two standard deviations above and below the mean.

There is about a 95%95\%95% probability that sample mean is within 0.1 mm0.1\text{ mm}0.1 mm of the target value.

The marginOfError formula can be found .

Check out that .

🪶
here
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Online Statistics Education: A Free Resource for Introductory Statistics