1. 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.

2. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 2 Formula for z score

3. 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.

4. 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.

5. 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.

6. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 6 Standard Normal Distribution:  = 0  = 1 1 Values are converted to z scores where 0

7. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 7 Importance of the Standard Normal Distribution: 1 0 11  Areas will be equal. Any Normal Distribution: Standard Normal Distribution:

8. 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.

9. 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.

10. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 10 To find the area to the left of z = 1.34 _____________________________________ z…0.030.040.05..… _____________________________________. 1.2….8907.8925.8944 …. 1.3 ….9082.9099.9115 …. 1.4….9236.9251.9265 …..

11. 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

12. 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

13. 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

14. 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

15. 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 1.0000 the area to the left of the given z. z 0 Area under entire curve = 1.000.

16. 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( - 2.37 < z < 0) = ______.8925.4452.4911

17. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 17 Normal Probability d.P( - 3 < z < 3 ) = ________ e. P( - 2.34 < z < 1.57 ) = _____ f.P( 1.24 < z < 1.88 ) = _______.9974.9322.0774

18. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 18 Normal Probability g. P( - 2.44 < z < - 0.73 ) = _______ h.P( z < 1.64 ) = __________ i. P( z > 2.39 ) = _________.9495.0084.2254

19. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 19 Normal Probability j.P ( z > - 1.43 ) = __________ k. P( z < - 2.71 ) = __________.9236.0034

20. 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. __________.3413.9938.1251

21. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 21 Inverse Normal Distribution Finding z scores when probabilities (areas) are given

22. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 22 Find the indicated z score:.8907 0 z = 1.23

23. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 23 Find the indicated z score:.6331 z 0 z = – 0.34.3669

24. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 24 Find the indicated z score:.3560 0 z = 1.06.8560

25. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 25 Find the indicated z score:.4792 z = 0 – 2.04.0208

26. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 26 Find the indicated z score: 0 z =.4900 2.33

27. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 27 Find the indicated z score: z = 0.005 – 2.575

28. 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  2.575 or  2.58 =.005

29. 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?

30. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 30...students whose Verbal SAT scores are in the top 4%. Mean = 500, standard deviation = 100 =.04.9600 z = 1.75 The cut-off score is 1.75 standard deviations above the mean.

31. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 31 Application of Determining z Scores Mean = 500, standard deviation = 100 =.04.9600 z = 1.75 The cut-off score is 500 + 1.75(100) = 675.