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# Chapter Six Normal Distributions

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1. Understandable StatisticsSeventh EditionBy Brase and BrasePrepared by: Lynn SmithGloucester County College Chapter Six 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 Normal Density Function This formula generates the density curve which gives the shape of the normal distribution.

8. 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.

9. The Empirical Rule Approximately 95% of the data values lie within two standard deviations of the mean. 95% Two standard deviations from the mean.

10. The Empirical Rule Almost all (approximately 99.7%) of the data values will be within three standard deviations of the mean. 99.7% Three standard deviations from the mean.

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

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. 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. Control Chart      1 2 3 4 5 6 7

16. Control Chart      1 2 3 4 5 6 7

17. Out-Of-Control Warning Signals I One point beyond the 3 level II A run of nine consecutive points on one side of the center line at target  III At least two of three consecutive points beyond the 2 level on the same side of the center line.

18. Probability of a False Alarm

19. Is the Process in Control?      1 2 3 4 5 6 7

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

21. Is the Process in Control?      1 2 3 4 5 6 7

22. Is the Process in Control?      1 2 3 4 5 6 7

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. Formula for z score

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

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

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

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. 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. 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 …. .

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. 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. 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 . z2 0 z1

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 z1 from area to left of z2 . z1 z2 0

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. Area under entire curve = 1.000. z 0

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. Normal Probability .9974 d. P( - 3 < z < 3 ) = ________ e. P( - 2.34 < z < 1.57 ) = _____ f. P( 1.24 < z < 1.88 ) = _______ .9322 .0774

40. Normal Probability .2254 g. P( - 2.44 < z < - 0.73 ) = _______ h. P( z < 1.64 ) = __________ i . P( z > 2.39 ) = _________ .9495 .0084

41. Normal Probability j. P ( z > - 1.43 ) = __________ k. P( z < - 2.71 ) = __________ .9236 .0034

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. Inverse Normal Distribution Finding z scores when probabilities (areas) are given

44. Find the indicated z score: Find the indicated z score: .8907 0 z = 1.23

45. z 0 Find the indicated z score: .6331 .3669 z = – 0.34

46. Find the indicated z score: .3560 .8560 0 z = 1.06

47. Find the indicated z score: .4792 .0208 – 2.04 z = 0

48. Find the indicated z score: .4900 0 z = 2.33

49. Find the indicated z score: .005 z = 0 – 2.575

50. Find the indicated z score: = .005 A B – z 0 z  2.575 or  2.58 If area A + area B = .01, z = __________