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- 1. The Normal Distribution James H. Steiger
- 2. Types of ProbabilityDistributionsThere are two fundamental types ofprobability distributions Discrete Continuous
- 3. Discrete Probability DistributionsThese are used to model a situation where The number of events (things that can happen) is countable (i.e., can be placed in correspondence with integers) Probabilities sum to 1 Each event has a probability
- 4. A Discrete Probability Function Discrete Uniform (1,6) Distribution0.1670.000 0 1 2 3 4 5 6 7
- 5. Continuous ProbabilityDistributionsThese are used to model situations where thenumber of things that can happen is not countableProbability of a particular outcome cannot bedefinedInterval probabilities can be defined: Probabilityof an interval is the area under the probabilitydensity curve between the endpoints of the interval
- 6. Probability Density Function Uniform(0,6) Distribution0.1670.000 -2 -1 0 1 2 3 4 5 6 7 8
- 7. The Normal Distribution Family Standard Normal Distribution0.4 34.13%0.3 σ0.20.10.0 µ µ+σ
- 8. The Standard NormalDistribution Standard Normal Distribution0.4 34.13%0.3 σ0.20.10.0 -3 -2 -1 0 1 2 3
- 9. The Normal Curve TableZ F(Z) Z F(Z)0.25 59.87 1.75 95.990.50 69.15 1.96 97.50.75 77.34 2.00 97.721.00 84.13 2.25 98.781.25 89.44 2.326 99.001.50 93.32 2.50 99.381.645 95.00 2.576 99.50
- 10. Some Typical “Routine”Problems Distribution of IQ Scores Normal(100,15)0.0280.0240.0200.0160.0120.0080.0040.000 55 70 85 100 115 130 145
- 11. Some Typical “Routine”Problems1.What percentage of people have IQ scoresbetween 100 and 115?2.What percentage of people have IQ scoresgreater than 130?3.What percentage of people have IQ scoresbetween 70 and 85?4. What is the percentile rank of an IQ score of122.5?5. What IQ score is at the 99th percentile?
- 12. General Strategy for NormalCurve ProblemsALWAYS draw the pictureEstimate the answer from the picture,remembering: Symmetry Total area is 1.0 Area to the left or right of center is .50Convert to Z-score formCompute interval probability by subtraction
- 13. Problem #1 Distribution of IQ Scores Normal(100,15)0.0280.0240.0200.0160.0120.0080.0040.000 55 70 85 100 115 130 145
- 14. Problem #2 Distribution of IQ Scores Normal(100,15)0.0280.0240.0200.0160.0120.0080.0040.000 55 70 85 100 115 130 145
- 15. Problem #2 (ctd) 130 − 100 Z130 = = +2.00 15
- 16. The Normal Curve TableZ F(Z) Z F(Z)0.25 59.87 1.75 95.990.50 69.15 1.96 97.50.75 77.34 2.00 97.721.00 84.13 2.25 98.781.25 89.44 2.326 99.001.50 93.32 2.50 99.381.645 95.00 2.576 99.50
- 17. Problem #3 Distribution of IQ Scores Normal(100,15)0.0280.0240.0200.0160.0120.0080.0040.000 55 70 85 100 115 130 145
- 18. Problem #3 70 − 100 Z 70 = = −2.00 15 85 − 100 Z 85 = = −1.00 15
- 19. The Normal Curve TableZ F(Z) Z F(Z)0.25 59.87 1.75 95.990.50 69.15 1.96 97.50.75 77.34 2.00 97.721.00 84.13 2.25 98.781.25 89.44 2.326 99.001.50 93.32 2.50 99.381.645 95.00 2.576 99.50
- 20. Problem #4 Distribution of IQ Scores Normal(100,15)0.0280.0240.0200.0160.0120.0080.0040.000 55 70 85 100 115 130 145
- 21. Problem #4 122.5 − 100Z122.5 = = 1.50 15
- 22. The Normal Curve TableZ F(Z) Z F(Z)0.25 59.87 1.75 95.990.50 69.15 1.96 97.50.75 77.34 2.00 97.721.00 84.13 2.25 98.781.25 89.44 2.326 99.001.50 93.32 2.50 99.381.645 95.00 2.576 99.50
- 23. Problem #5 Distribution of IQ Scores Normal(100,15)0.0280.0240.0200.0160.0120.0080.0040.000 55 70 85 100 115 130 145
- 24. Problem #5Find Z-score corresponding to 99thpercentile.Convert to a raw score.
- 25. Relative Tail ProbabilityProblemsThese problems involve analyzing relativeprobabilities of extreme events in twopopulations. Such problems: Are seldom found in textbooks Are often relevant to real world problems Sometimes produce results that are counterintuitive
- 26. Relative Tail Probability #1Group A has a mean of 100 and a standarddeviation of 15. Group B has a mean of 100and a standard deviation of 17. What is therelative likelihood of finding a person witha score above 145 in Group B, relative toGroup A?
- 27. The Normal Curve TableZ Area Above Z2.6 .00472.65 .004052.7 .00353.0 .00135Answer: About 3 to 1.
- 28. Relative Tail Probability #2Group A has a mean of 100 and a standarddeviation of 15. Group B has a mean of 100and a standard deviation of 17. What is therelative likelihood of finding a person witha score above 160 in Group B, relative toGroup A?
- 29. The Normal Curve TableZ Area Above Z3.529 .0002084.00 .0000317Answer: About 6.57 to 1.

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