4B: Probability part B Normal Distributions

1. 4B: Probability part BNormal Distributions 4B: Probability Part B

2. How’s my hair? Looks good. The Normal distributions • Last lecture covered the most popular type of discrete random variable: binomial variables • This lecture covers the most popular continuous random variable: Normal variables • History of the Normal function • Recognized by de Moivre (1667–1754) • Extended by Laplace (1749–1827) 4B: Probability Part B

3. Probability density function (curve) • Illustrative example: vocabulary scores of 947 seventh graders • Smooth curve drawn over histogram is a density function model of the actual distribution • This is the Normal probability density function (pdf) 4B: Probability Part B

4. Areas under curve (cont.) • Last week we introduced the idea of the area under the curve (AUC); the same principals applies here • The darker bars in the figure represent scores ≤ 6.0, • About 30% of the scores were less than or equal to 6 • Therefore, selecting a score at random will have probability Pr(X ≤ 6) ≈ 0.30 4B: Probability Part B

5. Areas under curve (cont.) • Now translate this to a Normal curve • As before, the area under the curve (AUC) = probability • The scale of the Y-axis is adjusted so the total AUC = 1 • The AUC to the left of 6.0 in the figure to the right (shaded) = 0.30 • Therefore, Pr(X ≤ 6) ≈ 0.30 • In practice, the Normal density curve helps us work with Normal probabilities 4B: Probability Part B

6. Density Curves 4B: Probability Part B

7. Normal distributions • Normal distributions = a family of distributions with common characteristics • Normal distributions have two parameters • Mean µ locates center of the curve • Standard deviation quantifies spread (at points of inflection) Arrows indicate points of inflection 4B: Probability Part B

8. 68-95-99.7 rule for Normal RVs • 68% of AUC falls within 1 standard deviation of the mean (µ) • 95% fall within 2 (µ2) • 99.7% fall within 3 (µ  3) 4B: Probability Part B

9. Illustrative example: WAIS Wechsler adult intelligence scores (WAIS) vary according to a Normal distribution with μ = 100 and σ = 15 4B: Probability Part B

10. Illustrative example: male height • Adult male height is approximately Normal with µ = 70.0 inches and  = 2.8 inches (NHANES, 1980) • Shorthand: X ~ N(70, 2.8) • Therefore: • 68% of heights = µ = 70.0  2.8 = 67.2 to 72.8 • 95% of heights = µ 2 = 70.0  2(2.8) = 64.4 to 75.6 • 99.7% of heights = µ 3 = 70.0  3(2.8) = 61.6 to 78.4 4B: Probability Part B

11. 68% (by 68-95-99.7 Rule) ? 16% 16% -1 +1 70 72.8 (height) 84% Illustrative example: male height What proportion of men are less than 72.8 inches tall? (Note: 72.8 is one σ above μ) 4B: Probability Part B

12. ? 68 70 (height) Male Height Example What proportion of men are less than 68 inches tall? 68 does not fall on a ±σ marker. To determine the AUC, we must first standardize the value. 4B: Probability Part B

13. Standardized value = z score To standardize a value, simply subtract μ and divide by σ This is now a z-score The z-score tells you the number of standard deviations the value falls from μ 4B: Probability Part B

14. Example: Standardize a male height of 68” Recall X ~ N(70,2.8) Therefore, the value 68 is 0.71 standard deviations below the mean of the distribution 4B: Probability Part B

15. ? 68 70 (height values) Men’s Height (NHANES, 1980) What proportion of men are less than 68 inches tall? = What proportion of a Standard z curve is less than –0.71? -0.71 0 (standardized values) You can now look up the AUC in a Standard Normal “Z” table. 4B: Probability Part B

16. Using the Standard Normal table Pr(Z≤ −0.71) = .2389 4B: Probability Part B

17. .2389 68 70 (height values) -0.71 0 (standardized values) Summary (finding Normal probabilities) • Draw curve w/ landmarks • Shade area • Standardize value(s) • Use Z table to find appropriate AUC 4B: Probability Part B

18. 68 70 (height values) -0.71 0 (standardized values) Right tail • What proportion of men are greater than 68” tall? • Greater than  look at right “tail” • Area in right tail = 1 – (area in left tail) .2389 1- .2389 = .7611 Therefore, 76.11% of men are greater than 68 inches tall. 4B: Probability Part B

19. Z percentiles • zp the z score with cumulative probability p • What is the 50th percentile on Z? ANS: z.5 = 0 • What is the 2.5th percentile on Z? ANS: z.025 = 2 • What is the 97.5th percentile on Z? ANS: z.975 = 2 4B: Probability Part B

20. Finding Z percentile in the table • Look up the closest entry in the table • Find corresponding z score • e.g., What is the 1st percentile on Z? • z.01 = -2.33 • closest cumulative proportion is .0099 4B: Probability Part B

21. .10 ? 70 (height values) Unstandardizing a value How tall must a man be to place in the lower 10% for men aged 18 to 24? 4B: Probability Part B

22. Table A:Standard Normal Table • Use Table A • Look up the closest proportion in the table • Find corresponding standardized score • Solve for X (“un-standardize score”) 4B: Probability Part B

23. Table A:Standard Normal Proportion .08 1.2 .1003 Pr(Z < -1.28) = .1003 4B: Probability Part B

24. .10 ? 70 (height values) Men’s Height Example (NHANES, 1980) • How tall must a man be to place in the lower 10% for men aged 18 to 24? -1.28 0 (standardized values) 4B: Probability Part B

25. Observed Value for a Standardized Score • “Unstandardize” z-score to find associated x : 4B: Probability Part B

26. Observed Value for a Standardized Score • x = μ + zσ = 70 + (-1.28 )(2.8) = 70 + (3.58) = 66.42 • A man would have to be approximately 66.42 inches tall or less to place in the lower 10% of the population 4B: Probability Part B