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Central Limit Theorem
1. The random variable
x^ has a distribution (which
may or may not be normal) with mean
μ^ and
standard deviation
σ.
2. Samples all of the same size
n^ are randomly
selected from the population of
x^ values.
Given:
1. The distribution of sample
x^ will, as the
sample size increases, approach a
normal
distribution.2. The mean of the sample means will be thepopulation mean
3. The standard deviation of the sample meanswill approach
σ/^
. n
Conclusions:
Central Limit Theorem
Practical RulesCommonly Used:
- For samples of size
n^ larger than 30, the distribution of the sample means can be approximated reasonably wellby a normal distribution. The approximation gets betteras the sample size
n^ becomes larger.
- If the original population is itself normally distributed,then the sample means will be normally distributed forany sample size
n^ (not just the values of
n^ larger than 30).
the mean of the sample means
μ = x^
μ Notation
Notation
the mean of the sample meansthe standard deviation of sample mean
μ = x^
σ= x^
σ n
Notation
the mean of the sample meansthe standard deviation of sample mean (often called standard error of the mean)
μ = x^
σ= x^
σ n
Distribution of 200 digits fromSocial Security Numbers(Last 4 digits from 50 students)
Figure 5-
Example:
Given the population of men has normally distributed weights with a mean of 172 lb and a standarddeviation of 29 lb,a) if one man is randomly selected, the probability that hisweight is greater than 167 lb. is 0.5675.
Example:
Given the population of men has normally distributed weights with a mean of 172 lb and a standarddeviation of 29 lb,b) if 12 different men are randomly selected, find theprobability that their mean weight is greater than 167 lb.
Example:
Given the population of men has normally distributed weights with a mean of 172 lb and a standarddeviation of 29 lb,b) if 12 different men are randomly selected, find theprobability that their mean weight is greater than 167 lb.
Example:
Given the population of men has normally distributed weights with a mean of 172 lband a standard deviation of 29 lb,b) if 12 different men are randomly selected, find z = 167 – 172 = –0.60the probability that their mean weight is greater^29 than 167 lb.^36
Example:^ z^ = 167 – 172 = –0.60^2936
Given the population of men has normally distributed weights with a mean of 143 lb and a standarddeviation of 29 lb,b.) if 12 different men are randomly selected, theprobability that their mean weight is greater than 167 lb is0.7257.
Example:
Given the population of men has normally distributed weights with a mean of 172 lb and a standarddeviation of 29 lb,b) if 12 different men are randomly selected, their meanweight is greater than 167 lb. a) if one man is randomly selected, find the probabilitythat his weight is greater than 167 lb.P( x^ > 167) = 0.5675P( x^ > 167) = 0.7257It is much easier for an individual to deviate from themean than it is for a group of 12 to deviate from the mean.
