1. Understanding Basic Statistics Chapter Seven Normal Distributions

2. The Normal Distribution

3. Properties of The Normal Distribution The curve is bell-shaped with the highest point over the mean, . 

4. Properties of The Normal Distribution The curve is symmetrical about a vertical line through . 

5. Properties of The Normal Distribution The curve approaches the horizontal axis but never touches or crosses it. 

6. Properties of The Normal Distribution The transition points between cupping upward and downward occur above  +  and  – .  –     

7. The Empirical Rule Approximately 68.2% of the data values lie is within one standard deviation of the mean. One standard deviation from the mean. 68.2%

8. The Empirical Rule Approximately 95.4% of the data values lie within two standard deviations of the mean. Two standard deviations from the mean. 95.4%

9. 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%

10. Application of the Empirical Rule The life of a particular type of lightbulb is normally distributed with a mean of 1100 hours and a standard deviation of 100 hours. What is the probability that a lightbulb of this type will last between 1000 and 1200 hours? Approximately 68.2%

11. Control Chart a statistical tool to track data over a period of equally spaced time intervals or in some sequential order

12. 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.

13. To Construct a Control Chart Draw a center horizontal line at . Draw dashed lines (control limits) at     and   . The values of  and s 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.

14. Control Chart      12345671234567

15. Control Chart      12345671234567

16. Out-Of-Control Warning Signals IOne point beyond the 3  level IIA run of nine consecutive points on one side of the center line IIIAt least two of three consecutive points beyond the 2  level on the same side of the center line.

17. Is the Process in Control?      12345671234567

18. Is the Process in Control?      1 2 3 4 5 6 7 8 9 10 11 12 13

19. Is the Process in Control?      12345671234567

20. Is the Process in Control?      12345671234567

21. 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.

22. Formula for z score

23. 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.

24. Calculating z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Convert 29.7 minutes to a z score.

25. 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.

26. Standard Normal Distribution:  = 0  = 1 1 Values are converted to z scores where z = 0

27. Importance of the Standard Normal Distribution: 1 0 11  Areas will be equal. Any Normal Distribution : Standard Normal Distribution:

28. Use of the Normal Probability Table (Table 4) - Appendix I Entries give the probability that a standard normally distributed random variable will assume a value between the mean (zero) and a given z-score.

29. To find the area between z = 0 and z = 1.34 _____________________________________z 0.020.030.04 _____________________________________ 1.2.3888.3907..3925 1.3.4066.4082.4099 1.4.4222.4236.4251

30. Patterns for Finding Areas Under the Standard Normal Curve To find the area between a given z value and mean: Use Table 4 (Appendix I) directly. z

31. Patterns for Finding Areas Under the Standard Normal Curve To find the area between z values on either side of zero: Add area from z 1 to mean to area from mean to z 2. z2z2 z1z1

32. Patterns for Finding Areas Under the Standard Normal Curve To find the area between z values on the same side of mean: Subtract area from mean to z 1 from area from mean to z 2. z2z2 0 z1z1

33. Patterns for Finding Areas Under the Standard Normal Curve To find the area to the right of a positive z value or to the left of a negative z value: Subtract the area from mean to z from 0.5000. z 0 0.500 0 table

34. Patterns for Finding Areas Under the Standard Normal Curve To find the area to the left of a positive z value or to the right of a negative z value: Add 0.5000 to the area from mean to z. z 0 0.500 0 table

35. Use of the Normal Probability Table a.P(0 < z < 1.24) = ______ b. P(0 < z < 1.60) = _______ c.P( - 2.37 < z < 0) = ______.3925.4452.4911

36. Normal Probability d.P(  3 < z < 3 ) = ________ e. P(  2.34 < z < 1.57 ) = _____ f.P( 1.24 < z < 1.88 ) = _______.9974.9322.0774

37. Normal Probability g. P(  2.44 < z <  0.73 ) = _______ h.P( z < 1.64 ) = __________ i. P( z > 2.39 ) = _________.9495.0084.2254

38. Normal Probability j.P ( z >  1.43 ) = __________ k. P( z <  2.71 ) = __________.9236.0034

39. 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

40. Finding Z Scores When Probabilities (Areas) Are Given 1.Find the indicated z score:.3907 0 z =1.23

41. Find the indicated z score:.1331 z 0 z = – 0.34

42. Find the indicated z score:.3560 0 z z = 1.06

43. Find the indicated z score:.4792 z 0 z = – 2.04

44. Find the indicated z score: z = 2.33.01 0 z.4900

45. Find the indicated z score: z = – 2.575 z 0.005.4950

46. Find the indicated z score: If area A + area B =.01, z = __________ A B – z 0 z 2.575 or 2.58 =.005.4950

47. Find the indicated z score: If area A + area B =.05, z = __________ A B – z 0 z 1.96 =.025.4750

48. 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?

49. Application of Determining z Scores Mean = 500, standard deviation = 100 =.04.4600 z = 1.75 The cut-off score is 1.75 standard deviations above the mean.

50. Application of Determining z Scores Mean = 500, standard deviation = 100 =.04.4600 z = 1.75 The cut-off score is 500 + 1.75(100) = 675.

51. Normal Approximation Of The Binomial Distribution: If n (the number of trials) is sufficiently large, a binomial random variable has a distribution that is approximately normal.

52. Define “sufficiently large” The sample size, n, is considered to be "sufficiently large" if np and nq are both greater than 5.

53. Mean and Standard Deviation: Binomial Distribution

54. 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

55. Using the Binomial Probability Formula n = x = p = q = 1  p = P(10) = 0.176197052 20 10 0.5

56. Normal Approximation of the Binomial Distribution First calculate the mean and standard deviation:  = np = 20 (.5) = 10

57. 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.

58. The Continuity Correction Continuity Correction is needed because we are approximating a discrete probability distribution with a continuous distribution.

59. The Continuity Correction We are using the area under the curve to approximate the area of the rectangle. 9.5 - 10.5

60. Using the Normal Distribution P(9.5 < x < 10.5 ) = ? for x = 9.5: z =  0.22 P(  0.22 < z < 0 ) =.0871

61. Using the Normal Distribution for x = 10.5: z = = 0.22 P( 0 < z <.22) =.0871 P(9.5 < x < 10.5 ) =.0871 +.0871 =.1742

62. 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)

63. Using the Normal Approximation to the Binomial Distribution Is n sufficiently large? Check: n p = n q =

64. Using the Normal Approximation to the Binomial Distribution Is n sufficiently large? n p = 22 n q = 78 Both are greater than five.

65. Find the mean and standard deviation  = 100(.22) = 22 and  =

66. 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

67. Applying the Normal Distribution z = 29.5 – 22 = 1.81 4.14 Find P( z  1.81) 0 1.81 =.0351.4649 The probability that at least 30 of the patients will have side effects is 0.0351.

68. 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.