Section 6.5 The Central Limit Theorem
Idea Will allow us to combine the theory from 6.4 (sampling distribution idea) with our central limit theorem and that will allow us the do hypothesis testing in the next few chapters. We are going to use samples to represent populations.
Central Limit Theorem From a population with a mean and a standard deviation. The Central Limit Theorem will be broken up into three situations. Sample size n > 30 Sample size n 30 and normal distribution Sample size n 30
Sample size n > 30 The sampling distribution of sample means is normally distributed. Does not matter what the population looked like. Mean μ x=μ Standard Deviation σ x= σ n
Spoken Mean μ x=μ average of the sample means is the average Standard Deviation σ x= σ n standard deviation of sample means is the quotient of the standard deviation and the square root of the sample size
Meaning If you take a sample size greater than 30 from any type of population the sampling distribution of sample means is normally distributed. The ages in this class may or may not make a normal distribution, but if we let the whole college be our population and we take samples of size 31 or more and we found the average age of each sample and put those averages in order, the distribution of averages will be normally distributed. We are not looking at the ages being normally distributed each possible sample creates a normal bell shape curve.
Sample size n > 30 and normal distribution The sampling distribution of sample means is normally distributed. Does not matter what the population looked like. Mean μ x=μ Standard Deviation σ x= σ n
Sample size n > 30 You can not solve this problem YET :)
Why Before we learned how to find the z score for an individual data value z= x μ σ Now we are going to alter our z score formula so we can find the z score for an average group of values, a sample. z= x μ x σ x
Z score Now we need to use our definitions to have a final z score definition for an average group : Recall μ x=μ σ x= σ n Substitute them into z= x μ x σ x New Z score z= x μ σ n
Examples A population of men have a mean of 172 pounds and a standard deviation of 10 pounds. Find the probability that a random selected man will weight more than 175 pounds. A population mean of 172 pounds with a standard deviation of 10 pounds. Find the probability that a group of 20 men will have an average weight of more than 175 pounds. Assume weight is normally distributed.
Question of Examples Which one is is collecting information on a single data value and which one is collecting information for a group of values?
Single Data Problem A population of men have a mean of 172 pounds and a standard deviation of 10 pounds. Find the probability that a random selected man will weight more than 175 pounds. Let us solve the problem
Single Data Problem A population of men have a mean of 172 pounds and a standard deviation of 10 pounds. Find the probability that a random selected man will weight more than 175 pounds. Let us solve the problem Z = (175 172) / 10 Z = 3 / 10 = 0.33 P(Z> 0.33) = 1-0.6293 = 0.3707 The probability of randomly selecting a man whose weight is more than 175 pounds is 0.3707
Group of Data Problem A population mean of 172 pounds with a standard deviation of 10 pounds. Find the probability that a group of 20 men will have an average weight of more than 175 pounds. Assume weight is normally distributed. Now let us solve the problem
Group of Data Problem A population mean of 172 pounds with a standard deviation of 10 pounds. Find the probability that a group of 20 men will have an average weight of more than 175 pounds. Assume weight is normally distributed. Now let us solve the problem Z = (175-172) / (10 / 20) Z = 3 / (10/4.472) = 3 / 2.236 = 1.34 P(Z > 1.34) = 1 0.9099 = 0.0901 The probability of randomly selecting a group of 20 men whose average weight is more than 175 pounds is 0.901
Why did it drop so much? To find a whole group of men whose average weight is more than 175 pounds is more rare than finding one man whose weight is more than 175 pounds.
Example When women were allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ACES II ejection seats were designed for men weighing between 140 pounds and 211 pounds. The average weight of women is 165 pounds with a standard deviation of 45.6 pounds. If one women is randomly selected find the probability that her weight is between 140 pounds and 211 pounds.
Example P(a < z < b) = z = (x µ) / σ a = (140 165) / 45.6 z1 = -0.55 b = (211 165) / 45.6 z2 = 1.01 P(-0.55<z<1.01) = P(z<1.01) P(z<-0.55) P(-0.55<z<1.01) =.8430 -.2912 P(-0.55<z<1.01) =.5518
Example If 26 different women are randomly selected, find the probability that their mean weight is between 140 pounds and 211 pounds.
Example If 36 different women are randomly selected, find the probability that their mean weight is between 140 pounds and 211 pounds.
Example P(a < z < b) = z = (x µ) / (σ / n) A = (140 165) / (45.6/ 36) z1 = -3.29 b = (211 165) / (45.6/ 36) z2 = 6.05 P(-3.29<z<6.05) = P(z<6.05) P(z<-3.29) P(-3.29<z<6.05) =.9999 -.0005 P(-3.29<z<6.05) =.9994
Example When redesigning the fighter jet which probability is better to work with the single women or 36 women? The single women because ejection seats will be occupied by single women not groups of 36.
Example When redesigning the fighter jet which probability is better to work with the single women or 36 women?