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The Central Limit Theorem April 4, 2009 The Central Limit Theorem in terms of Histograms and Bell Curves I The Central Limit Theorem is about histograms looking like bell curves. The Central Limit Theorem in terms of Histograms and Bell Curves I The Central Limit Theorem is about histograms looking like bell curves. I What types of histograms look like bell curves? The Central Limit Theorem in terms of Histograms and Bell Curves I The Central Limit Theorem is about histograms looking like bell curves. I What types of histograms look like bell curves? I Histograms of averages of large samples. The Central Limit Theorem in terms of Histograms and Bell Curves I The Central Limit Theorem is about histograms looking like bell curves. I What types of histograms look like bell curves? I Histograms of averages of large samples. I Histograms of sums of large samples. The Central Limit Theorem in terms of Distributions I The distribution of sample averages becomes a normal distribution as the size of the sample becomes large. The Central Limit Theorem in terms of Distributions I The distribution of sample averages becomes a normal distribution as the size of the sample becomes large. I The distribution of sample sums becomes a normal distribution as the size of the sample becomes large. The Central Limit Theorem in terms of Distributions I The distribution of sample averages becomes a normal distribution as the size of the sample becomes large. I The distribution of sample sums becomes a normal distribution as the size of the sample becomes large. I How large is large? The Central Limit Theorem in terms of Distributions I The distribution of sample averages becomes a normal distribution as the size of the sample becomes large. I The distribution of sample sums becomes a normal distribution as the size of the sample becomes large. I How large is large? It depends. The Central Limit Theorem in terms of Distributions I The distribution of sample averages becomes a normal distribution as the size of the sample becomes large. I The distribution of sample sums becomes a normal distribution as the size of the sample becomes large. I How large is large? It depends. I If the original population is normally distributed, then a sample of size n = 1 is large enough. The Central Limit Theorem in terms of Distributions I The distribution of sample averages becomes a normal distribution as the size of the sample becomes large. I The distribution of sample sums becomes a normal distribution as the size of the sample becomes large. I How large is large? It depends. I If the original population is normally distributed, then a sample of size n = 1 is large enough. I If the original population is nearly normally distributed, then a sample of size n = 5 might be large enough. The Central Limit Theorem in terms of Distributions I The distribution of sample averages becomes a normal distribution as the size of the sample becomes large. I The distribution of sample sums becomes a normal distribution as the size of the sample becomes large. I How large is large? It depends. I If the original population is normally distributed, then a sample of size n = 1 is large enough. I If the original population is nearly normally distributed, then a sample of size n = 5 might be large enough. I If the original population is not at all normally distributed, then a sample of size n = 30 might be large enough. The Central Limit Theorem in terms of Distributions I No matter what the original distribution is, if we choose a large enough sample size n, then the distributions of sample averages and sample sums will be approximately normal. Where is the Center? Where is the Center? I Another way of saying this is, what is µ in the normal distribution we are approaching? Where is the Center? I Another way of saying this is, what is µ in the normal distribution we are approaching? I That depends on whether we are talking about averages or sums. Where is the Center for Averages? Let X be the random variable that measures the sample average and let µX be the mean of the sample average X . Where is the Center for Averages? Let X be the random variable that measures the sample average and let µX be the mean of the sample average X . I If µX is the population mean, then µX = µX Where is the Center for Averages? Let X be the random variable that measures the sample average and let µX be the mean of the sample average X . I If µX is the population mean, then µX = µX I That is, the mean of the normal distribution we are approaching for averages and the mean of the population are the same. Where is the Center for Sums? P Let X be the random variable that measures Pthe sum of the sample and let µP X be the mean of the sum X. Where is the Center for Sums? P Let X be the random variable that measures Pthe sum of the sample and let µP X be the mean of the sum X. I If µX is the population mean, then µP X = nµX Where is the Center for Sums? P Let X be the random variable that measures Pthe sum of the sample and let µP X be the mean of the sum X. I If µX is the population mean, then µP X = nµX I That is, the mean of the normal distribution we are approaching for sums is the mean of the population multiplied by the size of the sample n. How large is the spread? How large is the spread? I Another way of saying this is, what is σ in the normal distribution we are approaching? How large is the spread? I Another way of saying this is, what is σ in the normal distribution we are approaching? I That depends on whether we are talking about averages or sums. How large is the spread for averages? Let X be the random variable that measures the sample average and let σX be the standard deviation of the sample average X . How large is the spread for averages? Let X be the random variable that measures the sample average and let σX be the standard deviation of the sample average X . I If σX is the standard deviation of the population, then √ σX = σX / n How large is the spread for averages? Let X be the random variable that measures the sample average and let σX be the standard deviation of the sample average X . I If σX is the standard deviation of the population, then √ σX = σX / n I That is, the standard deviation of the normal distribution we are approaching for averages is the standard deviation of the original population divided by the square root of the sample size n. How large is the spread for averages? Let X be the random variable that measures the sample average and let σX be the standard deviation of the sample average X . I If σX is the standard deviation of the population, then √ σX = σX / n I That is, the standard deviation of the normal distribution we are approaching for averages is the standard deviation of the original population divided by the square root of the sample size n. I We call the standard deviation of the averages the standard error of the mean. How large is the spread for sums? P Let X be the random variable that measures the sum of P the sample and let σP X be the standard deviation of the sum X. How large is the spread for sums? P Let X be the random variable that measures the sum of P the sample and let σP X be the standard deviation of the sum X. I If σX is the standard deviation of the population, then √ σP X = σX n How large is the spread for sums? P Let X be the random variable that measures the sum of P the sample and let σP X be the standard deviation of the sum X. I If σX is the standard deviation of the population, then √ σP X = σX n I That is, the standard deviation of the normal distribution we are approaching for sums is the standard deviation of the original population multiplied by the square root of the sample size n. The Central Limit Theorem for Averages Suppose X is a random variable with mean µX and standard deviation σX . If we draw a sample of size n, then as n increases the distribution of the random variable X , which finds the sample mean, becomes increasingly normal and σX X ∼ N µX , √ n The Central Limit Theorem for Sums Suppose X is a random variable with mean µX and standard deviation σX . If we draw a sample of P size n, then as n increases the distribution of the random variable X , which finds the sample sum, becomes increasingly normal and X √ X ∼ N nµX , σX n Example 1 Suppose students at Cortland attend 1.2 football games per year on average and with a standard deviation of 0.95. I If we ask 25 students how many games they attended last year, what is the probability that the average number of games attended is less than 1? Example 1 Suppose students at Cortland attend 1.2 football games per year on average and with a standard deviation of 0.95. I If we ask 25 students how many games they attended last year, what is the probability that the average number of games attended is less than 1? sol’n First, we find the standard deviation for X √ σX = 0.95/ 25 = 0.19 Example 1 Suppose students at Cortland attend 1.2 football games per year on average and with a standard deviation of 0.95. I If we ask 25 students how many games they attended last year, what is the probability that the average number of games attended is less than 1? sol’n First, we find the standard deviation for X √ σX = 0.95/ 25 = 0.19 Then, we use our calculator to find P(X < 1) = normalcdf(−100, 1, 1.2, 0.19) = .15 note: -100 could be any number a, so long as P(X < a) = 0. Example 1a Example 1a I If we ask 100 students how many games they attended last year, what is the probability that the average number of games attended is less than 1? Example 1a If we ask 100 students how many games they attended last year, what is the probability that the average number of games attended is less than 1? √ sol’n σX = 0.95/ 100 = 0.095 I P(X < 1) = normalcdf(−100, 1, 1.2, 0.095) = 0.018 note: -100 could be any number a, so long as P(X < a) = 0. Example 2 Suppose the average time waiting in line at a certain grocery store is six minutes and that the standard deviation is 3. Suppose we ask 30 people how long they waited in line. I What is the probability that the the average wait time for these people is between 5 and 7 minutes. Example 2 Suppose the average time waiting in line at a certain grocery store is six minutes and that the standard deviation is 3. Suppose we ask 30 people how long they waited in line. What is the probability that the the average wait time for these people is between 5 and 7 minutes. √ sol’n The standard error of the mean is 3/ 30 = .548. I Example 2 Suppose the average time waiting in line at a certain grocery store is six minutes and that the standard deviation is 3. Suppose we ask 30 people how long they waited in line. What is the probability that the the average wait time for these people is between 5 and 7 minutes. √ sol’n The standard error of the mean is 3/ 30 = .548. I P(5 < X < 7) = normalcdf(5, 7, 6, .548) = .932 Example 2 Suppose the average time waiting in line at a certain grocery store is six minutes and that the standard deviation is 3. Suppose we ask 30 people how long they waited in line. What is the probability that the the average wait time for these people is between 5 and 7 minutes. √ sol’n The standard error of the mean is 3/ 30 = .548. I P(5 < X < 7) = normalcdf(5, 7, 6, .548) = .932 I Find the probability that a person waits between 1 and 10 minutes. Example 2 Suppose the average time waiting in line at a certain grocery store is six minutes and that the standard deviation is 3. Suppose we ask 30 people how long they waited in line. What is the probability that the the average wait time for these people is between 5 and 7 minutes. √ sol’n The standard error of the mean is 3/ 30 = .548. I P(5 < X < 7) = normalcdf(5, 7, 6, .548) = .932 I Find the probability that a person waits between 1 and 10 minutes. We have no way of knowing. The CLT does not apply. Example 3 According to cortland.org, the average yearly precipitation in Cortland is 39.73 inches. Let us suppose that the standard deviation is 1.1 inches. I Find the probability that it rains more than 1000 inches in Cortland over the next two decades. Example 3 According to cortland.org, the average yearly precipitation in Cortland is 39.73 inches. Let us suppose that the standard deviation is 1.1 inches. I Find the probability that it rains more than 1000 inches in Cortland over the next two decades. I The average over 20 years should be 20(39.73) = 794.6. The√standard deviation of the sums should be 1.1 20 = 4.919. The distribution of the sums should be N(794.6, 4.919). Example 3 According to cortland.org, the average yearly precipitation in Cortland is 39.73 inches. Let us suppose that the standard deviation is 1.1 inches. I Find the probability that it rains more than 1000 inches in Cortland over the next two decades. I The average over 20 years should be 20(39.73) = 794.6. The√standard deviation of the sums should be 1.1 20 = 4.919. The distribution of the sums should be N(794.6, 4.919). P( rainfall > 1000) = normalcdf(1000, 10000, 794.6, 4.919) =0