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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 - CLT.

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Central Limit Theorem-CLT

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

  2. 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

  3. CLT-continue

  4. CLT Formula Use to gain information about a sample mean Use to gain information about an individual data value

  5. To calculate the CLT

  6. 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?

  7. 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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