Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1 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) 2002 Houghton Mifflin Company. All rights reserved. 2 Formula for z score
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 3 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.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 4 Calculating z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Convert 29.7 minutes to a z score.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 5 Interpreting z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Interpret a z score of 1.6. The delivery time is 28.2 minutes.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 6 Standard Normal Distribution: = 0 = 1 1 Values are converted to z scores where 0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 7 Importance of the Standard Normal Distribution: 1 0 11 Areas will be equal. Any Normal Distribution: Standard Normal Distribution:
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 8 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) 2002 Houghton Mifflin Company. All rights reserved. 9 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) 2002 Houghton Mifflin Company. All rights reserved. 10 To find the area to the left of z = 1.34 _____________________________________ z… … _____________________________________. 1.2… …. 1.3 … …. 1.4… …..
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 11 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 0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 12 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. z 0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 13 Patterns for Finding Areas Under the Standard Normal Curve To find the area between z values Subtract area to left of z 1 from area to left of z 2. z2z2 0 z1z1
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 14 Patterns for Finding Areas Under the Standard Normal Curve To find the area between z values, subtract area to left of z 1 from area to left of z 2. z2z2 0 z1z1
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 15 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 the area to the left of the given z. z 0 Area under entire curve =
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 16 Use of the Normal Probability Table a.P(z < 1.24) = ______ b. P(0 < z < 1.60) = _______ c.P( < z < 0) = ______
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 17 Normal Probability d.P( - 3 < z < 3 ) = ________ e. P( < z < 1.57 ) = _____ f.P( 1.24 < z < 1.88 ) = _______
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 18 Normal Probability g. P( < z < ) = _______ h.P( z < 1.64 ) = __________ i. P( z > 2.39 ) = _________
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 19 Normal Probability j.P ( z > ) = __________ k. P( z < ) = __________
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 20 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.a. ___________ b.less than 30 minutes.b. __________ c.less than 22.7 minutes.c. __________
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 21 Inverse Normal Distribution Finding z scores when probabilities (areas) are given
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 22 Find the indicated z score: z = 1.23
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 23 Find the indicated z score:.6331 z 0 z = –
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 24 Find the indicated z score: z =
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 25 Find the indicated z score:.4792 z = 0 –
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 26 Find the indicated z score: 0 z =
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 27 Find the indicated z score: z = – 2.575
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 28 Find the indicated z score: If area A + area B =.01, z = __________ A B – z 0 z or 2.58 =.005
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 29 Application of Determining z Scores The Verbal SAT test has a mean score of 500 and a standard deviation of 100. Scores are normally distributed. A major university determines that it will accept only students whose Verbal SAT scores are in the top 4%. What is the minimum score that a student must earn to be accepted?
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved students whose Verbal SAT scores are in the top 4%. Mean = 500, standard deviation = 100 = z = 1.75 The cut-off score is 1.75 standard deviations above the mean.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 31 Application of Determining z Scores Mean = 500, standard deviation = 100 = z = 1.75 The cut-off score is (100) = 675.