1. HS 1674B: Probability Part B1 4B: Probability part B Normal Distributions

2. HS 1674B: Probability Part B2 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) How’s my hair? Looks good.

3. HS 1674B: Probability Part B3 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)

4. HS 1674B: Probability Part B4 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

5. HS 1674B: Probability Part B5 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

6. HS 1674B: Probability Part B6 Density Curves

7. HS 1674B: Probability Part B7 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

8. HS 1674B: Probability Part B8 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  )

9. HS 1674B: Probability Part B9 Illustrative example: WAIS Wechsler adult intelligence scores (WAIS) vary according to a Normal distribution with μ = 100 and σ = 15

10. HS 1674B: Probability Part B10 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

11. HS 1674B: Probability Part B11 Illustrative example: male height What proportion of men are less than 72.8 inches tall? (Note: 72.8 is one σ above μ) ? 70 72.8 (height) +1 84% 68% (by 68-95-99.7 Rule) 16% 16%

12. HS 1674B: Probability Part B12 Male Height Example What proportion of men are less than 68 inches tall? ? 68 70 (height) 68 does not fall on a ±σ marker. To determine the AUC, we must first standardize the value.

13. HS 1674B: Probability Part B13 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

14. HS 1674B: Probability Part B14 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

15. HS 1674B: Probability Part B15 Men’s Height (NHANES, 1980) -0.71 0 (standardized values) 68 70 (height values) ? What proportion of men are less than 68 inches tall? = What proportion of a Standard z curve is less than –0.71? You can now look up the AUC in a Standard Normal “Z” table.

16. HS 1674B: Probability Part B16 Using the Standard Normal table z.00.01.02  0.8.2119.2090.2061  0.7.2420.2389.2358  0.6.2743.2709.2676 Pr(Z ≤ −0.71) =.2389

17. HS 1674B: Probability Part B17 Summary (finding Normal probabilities) Draw curve w/ landmarks Shade area Standardize value(s) Use Z table to find appropriate AUC -0.71 0 (standardized values) 68 70 (height values).2389

18. HS 1674B: Probability Part B18 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) -0.71 0 (standardized values) 68 70 (height values).23891-.2389 =.7611 Therefore, 76.11% of men are greater than 68 inches tall.

19. HS 1674B: Probability Part B19 Z percentiles z p  the z score with cumulative probability p What is the 50 th percentile on Z? ANS: z.5 = 0 What is the 2.5 th percentile on Z? ANS: z.025 = 2 What is the 97.5 th percentile on Z? ANS: z.975 = 2

20. HS 1674B: Probability Part B20 Finding Z percentile in the table Look up the closest entry in the table Find corresponding z score e.g., What is the 1 st percentile on Z? z.01 = -2.33 closest cumulative proportion is.0099 z.02.03.04  2.3.0102.0099.0096

21. HS 1674B: Probability Part B21 Unstandardizing a value.10 ? 70 (height values) How tall must a man be to place in the lower 10% for men aged 18 to 24?

22. HS 1674B: Probability Part B22 Use Table A Look up the closest proportion in the table Find corresponding standardized score Solve for X (“un-standardize score”) Table A: Standard Normal Table

23. HS 1674B: Probability Part B23 Table A: Standard Normal Proportion z.07.09  1.3.0853.0838.0823.1020.0985  1.1.1210.1190.1170  1.2.08.1003 Pr(Z < -1.28) =.1003

24. HS 1674B: Probability Part B24 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).10 ? 70 (height values)

25. HS 1674B: Probability Part B25 Observed Value for a Standardized Score “Unstandardize” z-score to find associated x :

26. HS 1674B: Probability Part B26 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