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Unformatted text preview: The Central Limit Theorem Revisited Two more views Here again is the Central Limit Theorem : Central Limit Theorem If X 1 , X 2 , . . . X n are a sample of n independent and identically distributed trials from any distribution with mean μ and standard deviation σ , then for n large enough, ) , ( ~ σ μ n n N S n , or ) , ( ~ n N X n σ μ (these are equivalent statements) That is, the sampling distribution of the sample mean is approximately normal. What does this mean?”!?!! STAT 101a106a Introduction to Statistics 169 Central Limit Theorem Interpretation 1 Take sample of size n from any distribution and calculate the sample mean. Repeat this process many times. As long as n is ‘large enough’ : The histogram of all the sample means will be approximately normal. 2. The mean of the sample means will be the same as the mean of the original distribution. 3. The standard deviation of the sample means will be the standard deviation of the original distribution divided by n . Example (discrete) : Poisson. Suppose we take a sample of size 10 from a Poisson distribution with mean 2 (so standard deviation equals 2 ). Repeat this process 1000 times. The original distribution does not look normal (because it’s Poisson!!!). However, the histogram of the sample means is approximately normal. The mean of the sample means is 2.0003 (about 2, the mean of the original distribution), and the standard deviation of the sample means is 0.46, about equal to the standard deviation predicted by the Central Limit Theorem : 45 . 10 2 = STAT 101a106a Introduction to Statistics 170 Here are two normal quantile plots : the one of the left is of data randomly selected from Poisson with mean 2 (NOT normal). On the right, a quantile plot of the 1000 sample means (each of size 10) – approximately normal. STAT 101a106a Introduction to Statistics 171 Events Probability 12 11 10 9 8 7 6 5 4 3 2 1 0.30 0.25 0.20 0.15 0.10 0.05 0.00 Poisson Distribution with Mean= 2 sample means Density 3.2 2.8 2.4 2.0 1.6 1.2 0.8 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 Histogram of sample means, n=10 Mean of Sample Means = 2.0003 St.Dev of Sample Means = 0.4624 Percent 9 8 7 6 5 4 3 2 1 99.99 99 95 80 50 20 5 1 0.01 Quantile Plot of Poisson, Mean 2 Percent 3.5 3.0 2.5 2.0 1.5 1.0 99.99 99 95 80 50 20 5 1 0.01 Quantile plot of Sample Means, n= 10 Example (continuous) : Camel. Below is a bimodal distribution with mean 11 and standard deviation 3.16.Take sample of size 10 from this distribution, calculate sample mean. Repeat 1000 times. Make a histogram of these 1000 sample means. Distribution of sample means is approximately normal, mean=10.97 (about 11), and std....
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 Fall '05
 JonathanReuningSchererDonaldGreen
 Central Limit Theorem, Normal Distribution, Standard Deviation, Camel Distribution

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