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Chapter 3 Chapter Chapter Outline 5 Normal Probability Distributions Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 1 2 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Section 5.4 Objectives • How to find sampling distributions and verify their properties • How to interpret the Central Limit Theorem • How to apply the Central Limit Theorem to find the probability of a sample mean Section 5.4 Sampling Distributions and the Central Limit Theorem Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 3 Sampling Distributions 4 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Sampling Distribution of Sample Means Sampling distribution • The probability distribution of a sample statistic. • Formed when samples of size n are repeatedly taken from a population. • e.g. Sampling distribution of sample means Population with μ, σ Sample 5 Sample 3 x Sample 1 x1 Sample 2 x 2 3 Sample 4 x5 x4 The sampling distribution consists of the values of the sample means, x 1 , x 2 , x 3 , x 4 , x 5 , . . . Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Larson/Farber 6th ed 5 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 6 1 Chapter 3 Properties of Sampling Distributions of Sample Means Example: Sampling Distribution of Sample Means 1. The mean of the sample means, m x , is equal to the population mean μ. mx = m The population values {1, 3, 5, 7} are written on slips of paper and put in a box. Two slips of paper are randomly selected, with replacement. a. Find the mean, variance, and standard deviation of the population. 2. The standard deviation of the sample means, s x , is equal to the population standard deviation, σ divided by the square root of the sample size, n. sx = Solution: s Mean: n S( x - m )2 =5 N Standard Deviation: s = 5 » 2.236 7 Example: Sampling Distribution of Sample Means c. List all the possible samples of size n = 2 and calculate the mean of each sample. Solution: Solution: Sample 1, 1 1, 3 1, 5 1, 7 3, 1 3, 3 3, 5 3, 7 Probability Histogram of Population of x 0.25 Probability All values have the same probability of being selected (uniform distribution) x 1 3 5 7 Population values Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 9 Example: Sampling Distribution of Sample Means 1 0.0625 2 3 4 5 2 3 4 3 0.1250 0.1875 0.2500 0.1875 6 7 2 1 0.1250 0.0625 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Larson/Farber 6th ed x 1 2 3 4 2 3 4 5 Sample 5, 1 5, 3 5, 5 5, 7 7, 1 7, 3 7, 5 7, 7 x 3 4 5 6 4 5 6 7 These means form the sampling distribution of sample means. 10 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Example: Sampling Distribution of Sample Means d. Construct the probability distribution of the sample means. Solution: f Probability x x f Probability 1 8 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Example: Sampling Distribution of Sample Means b. Graph the probability histogram for the population values. P(x) Sx =4 N Variance: s 2 = • Called the standard error of the mean. Copyright © 2015, 2012, and 2009 Pearson Education, Inc. m= e. Find the mean, variance, and standard deviation of the sampling distribution of the sample means. Solution: The mean, variance, and standard deviation of the 16 sample means are: mx = 4 s x2 = 5 = 2.5 2 s x = 2.5 » 1.581 These results satisfy the properties of sampling distributions of sample means. mx = m = 4 11 s x = s = 5 » 2.236 » 1.581 n 2 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 2 12 2 Chapter 3 The Central Limit Theorem Example: Sampling Distribution of Sample Means 1. If samples of size n ³ 30, are drawn from any population with mean = m and standard deviation = s, f. Graph the probability histogram for the sampling distribution of the sample means. Solution: P(x) Probability 0.25 Probability Histogram of Sampling Distribution of x 0.20 0.15 0.10 0.05 x 2 3 4 5 6 x m then the sampling distribution of the sample means approximates a normal distribution. The greater the sample size, the better the approximation. The shape of the graph is symmetric and bell shaped. It approximates a normal distribution. xx x x x x x x x x x x 7 m Sample mean 13 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. The Central Limit Theorem x Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 14 The Central Limit Theorem 2. If the population itself is normally distributed, • In either case, the sampling distribution of sample means has a mean equal to the population mean. mx = m • The sampling distribution of sample means has a variance equal to 1/n times the variance of the population and a standard deviation equal to the population standard deviation divided by the square root of n. 2 x m the sampling distribution of the sample means is normally distribution for any sample size n. xx x x x x x x x x x x s x2 = s n x m Copyright © 2015, 2012, and 2009 Pearson Education, Inc. sx = 15 Any Population Distribution Distribution of Sample Means, n ≥ 30 2. Normal Population Distribution Larson/Farber 6th ed Standard deviation (standard error of the mean) Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 16 Cellular phone bills for residents of a city have a mean of $63 and a standard deviation of $11. Random samples of 100 cellular phone bills are drawn from this population and the mean of each sample is determined. Find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution of sample means. Distribution of Sample Means, (any n) Copyright © 2015, 2012, and 2009 Pearson Education, Inc. n Example: Interpreting the Central Limit Theorem The Central Limit Theorem 1. s Variance 17 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 18 3 Chapter 3 Solution: Interpreting the Central Limit Theorem Solution: Interpreting the Central Limit Theorem • The mean of the sampling distribution is equal to the population mean m x = m = 63 • Since the sample size is greater than 30, the sampling distribution can be approximated by a normal distribution with s x = $1.10 m x = $63 • The standard error of the mean is equal to the population standard deviation divided by the square root of n. s x = s = 11 = 1.1 n 100 19 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Example: Interpreting the Central Limit Theorem Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Solution: Interpreting the Central Limit Theorem The training heart rates of all 20-years old athletes are normally distributed, with a mean of 135 beats per minute and standard deviation of 18 beats per minute. Random samples of size 4 are drawn from this population, and the mean of each sample is determined. Find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution of sample means. • The mean of the sampling distribution is equal to the population mean m x = m = 135 • The standard error of the mean is equal to the population standard deviation divided by the square root of n. s x = s = 18 = 9 n Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 21 Solution: Interpreting the Central Limit Theorem Larson/Farber 6th ed 4 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 22 Probability and the Central Limit Theorem • Since the population is normally distributed, the sampling distribution of the sample means is also normally distributed. sx = 9 m x = 135 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 20 • To transform x to a z-score z= 23 of 104 23 x - mx x - m Value-Mean = = s Standard Error sx n Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 24 4 Chapter 3 Example: Probabilities for Sampling Distributions Solution: Probabilities for Sampling Distributions The graph shows the length of time people spend driving each day. You randomly select 50 drivers age 15 to 19. What is the probability that the mean time they spend driving each day is between 24.7 and 25.5 minutes? Assume that σ = 1.5 minutes. From the Central Limit Theorem (sample size is greater than 30), the sampling distribution of sample means is approximately normal with m x = m = 25 25 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Solution: Probabilities for Sampling Distributions Normal Distribution μ = 25 σ = 0.21213 24.7 - 25 z1 = = = -1.41 s 1 .5 n 50 P(24.7 < x < 25.5) z2 = x-m s n = 25 25.5 z 0 2.36 P(24 < x < 54) = P(-1.41 < z < 2.36) = 0.9909 – 0.0793 = 0.9116 27 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Solution: Probabilities for x and x Normal Distribution μ = 3173 σ = 1120 z= P(x < 2700) s = 2700 - 3173 » -0.42 1120 P(z < -0.42) 0.3372 x 2500 2870 z -0.42 0 Larson/Farber 6th ed An education finance corporation claims that the average credit card debts carried by undergraduates are normally distributed, with a mean of $3173 and a standard deviation of $1120. (Adapted from Sallie Mae) Solution: You are asked to find the probability associated with a certain value of the random variable x. 28 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 2. You randomly select 25 undergraduates who are credit card holders. What is the probability that their mean credit card balance is less than $2700? Solution: You are asked to find the probability associated with a sample mean x. m x = m = 3173 P( x < 2700) = P(z < 0.42) = 0.3372 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 26 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Example: Probabilities for x and x Standard Normal Distribution μ=0 σ=1 x-m 1.5 » 0.21213 50 1. What is the probability that a randomly selected undergraduate, who is a credit card holder, has a credit card balance less than $2700? 0.9909 0.0793 -1.41 = P(-1.41 < z < 2.36) 25.5 - 25 = 2.36 1.5 50 x 24.7 n Example: Probabilities for x and x Standard Normal Distribution μ=0 σ=1 x-m s sx = 29 s x = s = 1120 = 224 n Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 25 30 5 Chapter 3 Solution: Probabilities for x and x Normal Distribution μ = 3173 σ = 1120 z= Standard Normal Distribution μ=0 σ=1 x-m s Solution: Probabilities for x and x n = 2700 - 3173 -473 = » -2.11 1120 224 25 P(z < -2.11) P(x < 2700) 0.0174 x 2500 2870 z -2.11 0 • There is a 34% chance that an undergraduate will have a balance less than $2700. • There is only a 2% chance that the mean of a sample of 25 will have a balance less than $2700 (unusual event). • It is possible that the sample is unusual or it is possible that the corporation’s claim that the mean is $3173 is incorrect. P( x < 2700) = P(z < -2.11) = 0.0174 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 31 Copyright © 2015, 2012, and 2009 Pearson Education, Inc. 32 Section 5.4 Summary • Found sampling distributions and verify their properties • Interpreted the Central Limit Theorem • Applied the Central Limit Theorem to find the probability of a sample mean Copyright © 2015, 2012, and 2009 Pearson Education, Inc. Larson/Farber 6th ed 33 6