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Chapter 11

Chapter 11. The Normal Distribution s. Note: Normal distributions are the most popular probability models in statistics. Other popular models include binomial distributions, t distributions, etc. Recall: Rules About Density Curves. Recall the relationship between AUCs and probabilities.

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Chapter 11

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  1. Chapter 11 The Normal Distributions Note: Normal distributions are the most popular probability models in statistics. Other popular models include binomial distributions, t distributions, etc.

  2. Recall: Rules About Density Curves

  3. Recall the relationship between AUCs and probabilities Area Under Curve = probability for that range 30% of students had scores ≤ 6 30% of area under the curve (AUC) is shaded 30%

  4. The Normal pdf is: • A familyof related bell-shaped density • Each family member has a different μ(mean) and σ(standard deviation) • μ  location • σ spread • There are an infinite number of Normal density curves (curve changes every time µ and/or σchanges) • Notation: X~N(µ,σ) is read “X is distributed as …”

  5. Draw Normal curves accurately! • Symmetrical around μ • Infection points at μ ± σ • Horizontal asymptotes Drawing a half circle is not acceptable

  6. 68-95-99.7 Rule • μ± 1σ contains 68% of AUC • μ± 2σ contains 95% of AUC • μ± 3σ contains 99.7% of AUC The 68-95-99.7 rule applies only to Normal densities

  7. 68-95-99.7 Rule- Example • Assume male height (X) has Normal distribution with μ= 70.0 inches and σ= 2.8 inches • Notation:X~ N(70, 2.8) • µ = 70.0  2.8 = 67.2 to 72.8  68% of men between 67.2 and 72.8 inches tall • µ 2 = 70.0  2(2.8) = 64.4 to 75.6 (inches) 95% in this range • µ 3 = 70.0  3(2.8) = 61.6 to 78.4 (inches) 99.7% in this range

  8. 68-95-99.7 Ruleexample (cont.) 61.6 64.4 67.2 70 72.8 75.6 78.4 Height in inches 9/5/2014 9

  9. Example 2 (Male Height) 16% + 68% = 84% 16% + 68% = 84% 16% + 68% = 84% What proportion of men are less than 72.8" tall? X~ N(70, 2.8) By 68-95-99.7 rule 68% Total AUC =100%  32% split evenly in tails 16% 16% 16% 16% 84% of AUC to the left of 72.8 9/5/2014 10

  10. Draw an reasonably accurate Normal curve with landmarks at µ and µ ±σ Place values on the curve & shade relevant area Standardize values (i.e., z scores) Use Table A to determine the AUC Finding Normal Probabilities: Method

  11. Finding Normal Probabilities: Draw & Shade What proportion of men are less than 68” tall? ? 68 70

  12. Finding Normal Probabilities: Standardize In X~N(70, 2.8), the value 68 has: This indicates that 68 is 0.71 standard deviations below the mean

  13. Finding Normal Probabilities: Standardize Proportion of men less than68” tall = Pr(Z < -0.71): 68 70 (height) -0.71 0 (z score)

  14. Z is the Normal random variable with µ = 0 and σ= 1 Z~N(0,1) There is only one Standard Normal Z variable Z table in text (pp. 677-8) and in “Formulas” handout Standard Normal “Z” Table Notation: zp where p≡ cumulative probability = AUC to left p

  15. Finding Normal Probabilities: Table A .01 0.7 .2389 z score of −0.71→ AUCto left is .2389

  16. Pr(Z < −0.7) as an Image Conclude: .2389 (23.89%) of men are less than 68 inches in height From Table A .2389 68 70 (height) -0.71 0 (z score) 9/5/2014 17

  17. Pr(Z > z)  AUC to the right Ask: What proportion of men greater than68” tall? “Greater than” = “area to right of”= 1 – (AUC to left) 1 – .2389 = .7611 .2389 68 70 (height) -0.71 0 (z score) 9/5/2014 18

  18. AUC Between Two Points Total AUC = 1. Therefore: Pr(a < Z < b) = Pr(Z < b) – Pr(Z < a) .6915 .1003 -1.28 0.5 0.5 -1.28 Pr(−1.28 < Z < 0.5) = Pr(Z < 0.5) – Pr(Z < −1.28) = .6915 – .1003 = .5912

  19. Finding z percentiles and associated values At times, we must find the z value or X value that is associated with a given probability. To do this: 1. State the problem 2. Sketch the curve 3. Look up the related z-score is Z table 4. If you need to know X, “unstandardize” z with this formula:

  20. State & Sketch How tall must a man be to be taller than 10% of men (i.e., what is the 10th percentile of heights)? Given: X~N(70, 2.8) .10 .10 70 (height) 70 (height) ? 0 (z equivalent) z.10

  21. Find z score in Table A .08 .1003 1.2  z.10 = -1.28

  22. Visual Relationship Notation: z.1003 = −1.28 z-score AUC to left X~N(70, 2.8) .10 70 (height) Height (inches) 70 ? Height (z score) 0 −1.28 Chapter 3

  23. Unstandardize & Conclude • x = μ + z∙σ = 70 + (-1.28 )(2.8) = 70 + (3.58) = 66.42 • Conclude: A man who is 66.42 inches in height is taller than 10% of men • We can also conclude that he is shorter than (1 – 10%) or 90% of men Chapter 3

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