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Chapter 37 The Normal Probability Distribution 2010 reserved Pearson Prentice Hall. All rights reserved © 2010 Pearson Prentice Hall. All©rights Section 7.2 The Standard Normal Distribution © 2010 Pearson Prentice Hall. All rights reserved 7-2 © 2010 Pearson Prentice Hall. All rights reserved 7-3 © 2010 Pearson Prentice Hall. All rights reserved 7-4 © 2010 Pearson Prentice Hall. All rights reserved 7-5 © 2010 Pearson Prentice Hall. All rights reserved 7-6 The table gives the area under the standard normal curve for values to the left of a specified Z-score, zo, as shown in the figure. © 2010 Pearson Prentice Hall. All rights reserved 7-7 EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve to the left of z = -0.38. Area left of z = -0.38 is 0.3520. © 2010 Pearson Prentice Hall. All rights reserved 7-8 Area under the normal curve to the right of zo = 1 – Area to the left of zo © 2010 Pearson Prentice Hall. All rights reserved 7-9 EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve to the right of Z = 1.25. Area right of 1.25 = 1 – area left of 1.25 = 1 – 0.8944 = 0.1056 © 2010 Pearson Prentice Hall. All rights reserved 7-10 EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve between z = -1.02 and z = 2.94. Area between -1.02 and 2.94 = (Area left of z = 2.94) – (area left of z = -1.02) = 0.9984 – 0.1539 = 0.8445 © 2010 Pearson Prentice Hall. All rights reserved 7-11 © 2010 Pearson Prentice Hall. All rights reserved 7-12 © 2010 Pearson Prentice Hall. All rights reserved 7-13 EXAMPLE Finding a z-score from a Specified Area to the Left Find the z-score such that the area to the left of the z-score is 0.7157. The z-score such that the area to the left of the z-score is 0.7157 is z = 0.57. © 2010 Pearson Prentice Hall. All rights reserved 7-14 EXAMPLE Finding a z-score from a Specified Area to the Right Find the z-score such that the area to the right of the z-score is 0.3021. The area left of the z-score is 1 – 0.3021 = 0.6979. The approximate z-score that corresponds to an area of 0.6979 to the left (0.3021 to the right) is 0.52. Therefore, z = 0.52. © 2010 Pearson Prentice Hall. All rights reserved 7-15 EXAMPLE Finding a z-score from a Specified Area Find the z-scores that separate the middle 80% of the area under the normal curve from the 20% in the tails. Area = 0.8 Area = 0.1 Area = 0.1 z1 is the z-score such that the area left is 0.1, so z1 = -1.28. z2 is the z-score such that the area left is 0.9, so z2 = 1.28. © 2010 Pearson Prentice Hall. All rights reserved 7-16 The notation zα (prounounced “z sub alpha”) is the z-score such that the area under the standard normal curve to the right of zα is α. © 2010 Pearson Prentice Hall. All rights reserved 7-17 EXAMPLE Finding the Value of z Find the value of z0.25 We are looking for the z-value such that the area to the right of the z-value is 0.25. This means that the area left of the z-value is 0.75. z0.25 = 0.67 © 2010 Pearson Prentice Hall. All rights reserved 7-18 © 2010 Pearson Prentice Hall. All rights reserved 7-19 Notation for the Probability of a Standard Normal Random Variable P(a < Z < b) represents the probability a standard normal random variable is between a and b P(Z > a) represents the probability a standard normal random variable is greater than a. P(Z < a) represents the probability a standard normal random variable is less than a. © 2010 Pearson Prentice Hall. All rights reserved 7-20 EXAMPLE Finding Probabilities of Standard Normal Random Variables Find each of the following probabilities: (a) P(Z < -0.23) (b) P(Z > 1.93) (c) P(0.65 < Z < 2.10) (a) P(Z < -0.23) = 0.4090 (b) P(Z > 1.93) = 0.0268 (c) P(0.65 < Z < 2.10) = 0.2399 © 2010 Pearson Prentice Hall. All rights reserved 7-21 For any continuous random variable, the probability of observing a specific value of the random variable is 0. For example, for a standard normal random variable, P(a) = 0 for any value of a. This is because there is no area under the standard normal curve associated with a single value, so the probability must be 0. Therefore, the following probabilities are equivalent: P(a < Z < b) = P(a < Z < b) = P(a < Z < b) = P(a < Z < b) © 2010 Pearson Prentice Hall. All rights reserved 7-22