1. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1 Understandable Statistics S eventh Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Chapter Six Normal Distributions

2. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 2 The Normal Distribution

3. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 3 Properties of The Normal Distribution The curve is bell-shaped with the highest point over the mean, . 

4. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 4 Properties of The Normal Distribution The curve is symmetrical about a vertical line through . 

5. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 5 Properties of The Normal Distribution The curve approaches the horizontal axis but never touches or crosses it. 

6. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 6 Properties of The Normal Distribution The transition points between cupping upward and downward occur above  +  and  – .  –     

7. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 7 The Normal Density Function This formula generates the density curve which gives the shape of the normal distribution.

8. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 8 The Empirical Rule Approximately 68% of the data values lie is within one standard deviation of the mean. One standard deviation from the mean. 68%

9. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 9 The Empirical Rule Approximately 95% of the data values lie within two standard deviations of the mean. Two standard deviations from the mean. 95%

10. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 10 The Empirical Rule Almost all (approximately 99.7%) of the data values will be within three standard deviations of the mean. Three standard deviations from the mean. 99.7%

11. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 11 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%

12. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 12 Control Chart a statistical tool to track data over a period of equally spaced time intervals or in some sequential order

13. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 13 Statistical Control A random variable is in statistical control if it can be described by the same probability distribution when it is observed at successive points in time.

14. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 14 To Construct a Control Chart Draw a center horizontal line at . Draw dashed lines (control limits) at     and   . The values of  and  may be target values or may be computed from past data when the process was in control. Plot the variable being measured using time on the horizontal axis.

15. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 15 Control Chart      12345671234567

16. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 16 Control Chart      12345671234567

17. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 17 Out-Of-Control Warning Signals IOne point beyond the 3  level IIA run of nine consecutive points on one side of the center line at target  IIIAt least two of three consecutive points beyond the 2  level on the same side of the center line.

18. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 18 Probability of a False Alarm

19. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 19 Is the Process in Control?      12345671234567

20. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 20 Is the Process in Control?      1 2 3 4 5 6 7 8 9 10 11 12 13

21. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 21 Is the Process in Control?      12345671234567

22. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 22 Is the Process in Control?      12345671234567

23. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 23 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.

24. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 24 Formula for z score

25. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 25 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.

26. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 26 Calculating z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Convert 29.7 minutes to a z score.

27. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 27 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.

28. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 28 Standard Normal Distribution:  = 0  = 1 1 Values are converted to z scores where z = 0

29. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 29 Importance of the Standard Normal Distribution: 1 0 11  Areas will be equal. Any Normal Distribution: Standard Normal Distribution:

30. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 30 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.

31. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 31 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.

32. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 32 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 …..

33. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 33 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

34. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 34 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

35. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 35 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 z 1 from area to left of z 2. z2z2 0 z1z1

36. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 36 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 z 1 from area to left of z 2. z2z2 0 z1z1

37. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 37 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.

38. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 38 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

39. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 39 Normal Probability d.P( - 3 < z < 3 ) = ________ e. P( - 2.34 < z < 1.57 ) = _____ f.P( 1.24 < z < 1.88 ) = _______.9974.9322.0774

40. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 40 Normal Probability g. P( - 2.44 < z < - 0.73 ) = _______ h.P( z < 1.64 ) = __________ i. P( z > 2.39 ) = _________.9495.0084.2254

41. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 41 Normal Probability j.P ( z > - 1.43 ) = __________ k. P( z < - 2.71 ) = __________.9236.0034

42. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 42 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

43. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 43 Inverse Normal Distribution Finding z scores when probabilities (areas) are given

44. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 44 Find the indicated z score:.8907 0 z = 1.23

45. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 45 Find the indicated z score:.6331 z 0 z = – 0.34.3669

46. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 46 Find the indicated z score:.3560 0 z = 1.06.8560

47. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 47 Find the indicated z score:.4792 z = 0 – 2.04.0208

48. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 48 Find the indicated z score: 0 z =.4900 2.33

49. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 49 Find the indicated z score: z = 0.005 – 2.575

50. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 50 Find the indicated z score: If area A + area B =.01, z = __________ A B – z 0 z  2.575 or  2.58 =.005

51. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 51 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?

52. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 52...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.

53. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 53 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.

54. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 54 Normal Approximation Of The Binomial Distribution: Under certain conditions, a binomial random variable has a distribution that is approximately normal.

55. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 55 Using the normal distribution to approximate the binomial distribution If n, p, and q are such that: np and nq are both greater than 5.

56. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 56 Mean and Standard Deviation: Binomial Distribution

57. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 57 Experiment: tossing a coin 20 times Problem: Find the probability of getting exactly 10 heads. Distribution of the number of heads appearing should look like : 10 20 0

58. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 58 Using the Binomial Probability Formula n = x = p = q = 1  p = P(10) = 0.176197052 20 10 0.5

59. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 59 Normal Approximation of the Binomial Distribution First calculate the mean and standard deviation:  = np = 20 (.5) = 10

60. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 60 The Continuity Correction Continuity Correction: to compute the probability of getting exactly 10 heads, find the probability of getting between 9.5 and 10.5 heads.

61. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 61 The Continuity Correction Continuity Correction is needed because we are approximating a discrete probability distribution with a continuous distribution.

62. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 62 The Continuity Correction We are using the area under the curve to approximate the area of the rectangle. 9.5 - 10.5

63. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 63 Using the Normal Distribution P(9.5 < x < 10.5 ) = ? for x = 9.5: z =  0.22 P(z <  0.22 ) =.4129

64. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 64 Using the Normal Distribution for x = 10.5: z = = 0.22 P( z <.22) =.5871 P(9.5 < x < 10.5 ) =.5871 -.4129 =.1742

65. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 65 Application of Normal Distribution If 22% of all patients with high blood pressure have side effects from a certain medication, and 100 patients are treated, find the probability that at least 30 of them will have side effects. Using the Binomial Probability Formula we would need to compute: P(30) + P(31) +... + P(100) or 1  P( x < 29)

66. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 66 Using the Normal Approximation to the Binomial Distribution Is it appropriate to use the normal distribution? Check: n p = n q =

67. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 67 Using the Normal Approximation to the Binomial Distribution n p = 22 n q = 78 Both are greater than five.

68. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 68 Find the mean and standard deviation  = 100(.22) = 22 and  =

69. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 69 Applying the Normal Distribution To find the probability that at least 30 of them will have side effects, find P( x  29.5) 22 29.5 Find this area

70. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 70 Applying the Normal Distribution z = 29.5 – 22 = 1.81 4.14 Find P( z  1.81) 0 1.81 =.0351.9649 The probability that at least 30 of the patients will have side effects is 0.0351.

71. Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 71 Reminders: Use the normal distribution to approximate the binomial only if both np and nq are greater than 5. Always use the continuity correction when approximating the binomial distribution.