Central Limit Theorem

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Transcript Central Limit Theorem

Central Limit Theorem-CLT
MM4D1. Using simulation, students will
develop the idea of the central limit
theorem.
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-continue
CLT Formula
Use to gain information about
a sample mean
Use to gain information about
an individual data value
To calculate the CLT
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.
a. What is the probability that an individual bottle contains
less than 300 ml?
b. What is the probability that an individual bottle contains
greater than 300 ml?
c. What is the probability that an individual bottle contains
between 300 ml and 350 ml?
d. 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?
e. What is the probability that the sample mean contents of
the 10 bottles is less than 300 ml?
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%