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Understanding Basic Statistics

Understanding Basic Statistics. Chapter Seven Normal Distributions. The Normal Distribution. . Properties of The Normal Distribution. The curve is bell-shaped with the highest point over the mean, . Properties of The Normal Distribution. .

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Understanding Basic Statistics

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  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. 68.2% One standard deviation from the mean.

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

  9. 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.

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

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

  16. 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 III At least two of three consecutive points beyond the 2 level on the same side of the center line.

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

  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?      1 2 3 4 5 6 7

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

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

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

  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 0.02 0.03 0.04 _____________________________________ 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 z1 to mean to area from mean to z2 . z2 z1

  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 z1 from area from mean to z2 . z1 z2 0

  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 . 0.5000 table z 0

  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 . 0.5000 table 0 z

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

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

  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: .4900 .01 0 z z = 2.33

  45. Find the indicated z score: .4950 .005 z 0 z = – 2.575

  46. Find the indicated z score: A B = .005 .4950 – z 0 z 2.575 or 2.58 If area A + area B = .01, z = __________

  47. Find the indicated z score: A = .025 B .4750 – z 0 z 1.96 If area A + area B = .05, z = __________

  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 .4600 = .04 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 .4600 = .04 z = 1.75 The cut-off score is 500 + 1.75(100) = 675.

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