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Central Limit Theorem-CLT. MM4D1. Using simulation, students will develop the idea of the central limit theorem. Central Limit Theorem - CLT.

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Central limit theorem clt

Central Limit Theorem-CLT

MM4D1. Using simulation, students will develop the idea of the central limit

theorem.


Central limit theorem clt1
Central Limit Theorem - CLT

  • The central limit theorem states that the sampling distribution of any statistic will be normal or nearly normal, if the sample size is large enough. Generally, a sample size is considered "large enough" if any of the following conditions apply:

    • The population distribution is normal.

    • The sample distribution is roughly symmetric, unimodal, without outliers, and the sample size is 15 or less.

    • The sample distribution is moderately skewed, unimodal, without outliers, and sample size is between 16 and 40.

    • The sample size is greater than 40, without outliers



Clt formula
CLT Formula

Use to gain information about a sample mean

Use to gain information about an individual data value


To calculate the clt
To calculate the CLT


Examples
Examples

1. A bottling company uses a filling machine to fill plastic bottles with a popular cola. The bottles are supposed to contain 300 ml. In fact, the contents vary according to a normal distribution with mean µ = 303 ml and standard deviation σ = 3 ml.

  • What is the probability that an individual bottle contains less than 300 ml?

  • What is the probability that an individual bottle contains greater than 300 ml?

  • What is the probability that an individual bottle contains between 300 ml and 350 ml?

  • Now take a random sample of 10 bottles. What are the mean and standard deviation of the sample mean contents x-bar of these 10 bottles?

  • What is the probability that the sample mean contents of the 10 bottles is less than 300 ml?


Solution
Solution

a) use z=(x-µ)/σ) & Table of negative Z-score

z=(300-303)/3 = -1

P(x<300) = 0.1587 or 15.87%,

b) P(x>300) = 1 – 0.1587 = 0.8413 or 84.13%

c) z=(x-µ)/σ)

z=(300-303)/3 = -1 → P(x=300) = 0.1587

z=(310-303)/3 = 2.33 → P(x=310) = 0.9893

P(300<x<310) = 0.9893-0.1587 = 0.8306 or 83.06%

d) mean: 303, stdev: 3/sqrt(10) = 0.94868

e) z=(x-µ)/(σ/sqrt(10) & Table of negative Z-score

z=(300-303)/0.94868 = -3.16

p=0.0008 or 0.08%


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