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The Normal Distribution Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 1 Properties of The Normal Distribution The curve is bell-shaped with the highest point over the mean, . Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 2 Properties of The Normal Distribution The curve is symmetrical about a vertical line through . Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 3 Properties of The Normal Distribution The curve approaches the horizontal axis but never touches or crosses it. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 4 Properties of The Normal Distribution – The transition points between cupping upward and downward occur above + and – . Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 5 The Empirical Rule Approximately 68% of the data values lie is within one standard deviation of the mean. 68% One standard deviation from the mean. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 6 The Empirical Rule Approximately 95% of the data values lie within two standard deviations of the mean. 95% x Two standard deviations from the mean. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 7 The Empirical Rule Almost all (approximately 99.7%) of the data values will be within three standard deviations of the mean. 99.7% x Three standard deviations from the mean. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 8 Application of the Empirical Rule The life of a particular type of light bulb is normally distributed with a mean of 1100 hours and a standard deviation of 100 hours. What is the probability that a light bulb of this type will last between 1000 and 1200 hours? Approximately 68% Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 9 Z Score • The z value or z score tells the number of standard deviations the original measurement is from the mean. • The z value is in standard units. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 10 Formula for z score x z Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 11 Calculating z-scores The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. Convert 21 minutes to a z score. x 21 25 z 2.00 2 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 12 Calculating z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Convert 29.7 minutes to a z score. x 29.7 25 z 2.35 2 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 13 Interpreting z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Interpret a z score of 1.6. x z 1.6( 2 ) 25 28 .2 The delivery time is 28.2 minutes. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 14 Standard Normal Distribution: =0 =1 -1 0 1 Values are converted to z scores wherexz = Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 15 Importance of the Standard Normal Distribution: Standard Normal Distribution: Any Normal Distribution: 0 1 Areas will be equal. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 1 16 Use of the Normal Probability Table (Table 5) - Appendix II Entries give the probability that a standard normally distributed random variable will assume a value to the left of a given negative z-score. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 17 Use of the Normal Probability Table (Table 5a) - Appendix II Entries give the probability that a standard normally distributed random variable will assume a value to the left of a given positive z value. Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 18 To find the area to the left of z = 1.34 _____________________________________ z … 0.03 0.04 0.05 ..… _____________________________________ . . 1.2 … .8907 .8925 .8944 …. 1.3 … .9082 .9099 .9115 …. 1.4 … .9236 .9251 .9265 …. . Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 19 Patterns for Finding Areas Under the Standard Normal Curve To find the area to the left of a given negative z : Use Table 5 (Appendix II) directly. z Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 0 20 Patterns for Finding Areas Under the Standard Normal Curve To find the area to the left of a given positive z : Use Table 5 a (Appendix II) directly. 0 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . z 21 Patterns for Finding Areas Under the Standard Normal Curve To find the area between z values on either side of zero: Subtract area to left of z1 from area to left of z2 . z1 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 0 z2 22 Patterns for Finding Areas Under the Standard Normal Curve To find the area between z values on the same side of zero: Subtract area to left of z1 from area to left of z2 . 0 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . z1 z2 23 Patterns for Finding Areas Under the Standard Normal Curve To find the area to the right of a positive z value or to the right of a negative z value: Subtract from 1.0000 the area to the left of the given z. Area under entire curve = 1.000. 0 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . z 24 Use of the Normal Probability Table a. .8925 P(z < 1.24) = ______ b. .4452 P(0 < z < 1.60) = _______ c. .4911 P( - 2.37 < z < 0) = ______ Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 25 Normal Probability d. .9974 P( - 3 < z < 3 ) = ________ e. .9322 P( - 2.34 < z < 1.57 ) = _____ f. .0774 P( 1.24 < z < 1.88 ) = _______ Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 26 Normal Probability g. .2254 P( - 2.44 < z < - 0.73 ) = _______ h. .9495 P( z < 1.64 ) = __________ i. .0084 P( z > 2.39 ) = _________ Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 27 Normal Probability j. .9236 P ( z > - 1.43 ) = __________ k. .0034 P( z < - 2.71 ) = __________ Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 28 Application of the Normal Curve The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. If you order a pizza, find the probability that the delivery time will be: a. between 25 and 27 minutes. .3413 a. ___________ b. less than 30 minutes. .9938 b. __________ c. less than 22.7 minutes. .1251 c. __________ Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 29 Inverse Normal Distribution Finding z scores when probabilities (areas) are given Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 30 Find the indicated z score: Find the indicated z score: .8907 0 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . z= 1.23 31 Find the indicated z score: .6331 .3669 z = – 0.34 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 32 Find the indicated z score: .3560 .8560 0 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . z= 1.06 33 Find the indicated z score: .4792 .0208 z = – 2.04 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 0 34 Find the indicated z score: .4900 0 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . z= 2.33 35 Find the indicated z score: .005 z = – 2.575 Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 0 36 Find the indicated z score: A = .005 –z B 0 z 2.575 or 2.58 If area A + area B = .01, z = __________ Copyright (C) 2006 Houghton Mifflin Company. All rights reserved . 37