Gerstman_PP07 for Normal Probability Distributions.ppt
- 1. 7: Normal Probability Distributions 1 Mar 22, 2025 Chapter 7: Chapter 7: Normal Probability Normal Probability Distributions Distributions
- 2. 7: Normal Probability Distributions 2 In Chapter 7: 7.1 Normal Distributions 7.2 Determining Normal Probabilities 7.3 Finding Values That Correspond to Normal Probabilities 7.4 Assessing Departures from Normality
- 3. 7: Normal Probability Distributions 3 §7.1: Normal Distributions • This pdf is the most popular distribution for continuous random variables • First described de Moivre in 1733 • Elaborated in 1812 by Laplace • Describes some natural phenomena • More importantly, describes sampling characteristics of totals and means
- 4. 7: Normal Probability Distributions 4 Normal Probability Density Function • Recall: continuous random variables are described with probability density function (pdfs) curves • Normal pdfs are recognized by their typical bell-shape Figure: Age distribution of a pediatric population with overlying Normal pdf
- 5. 7: Normal Probability Distributions 5 Area Under the Curve • pdfs should be viewed almost like a histogram • Top Figure: The darker bars of the histogram correspond to ages ≤ 9 (~40% of distribution) • Bottom Figure: shaded area under the curve (AUC) corresponds to ages ≤ 9 (~40% of area) 2 2 1 2 1 ) ( x e x f
- 6. 7: Normal Probability Distributions 6 Parameters μ and σ • Normal pdfs have two parameters μ - expected value (mean “mu”) σ - standard deviation (sigma) σ controls spread μ controls location
- 7. 7: Normal Probability Distributions 7 Mean and Standard Deviation of Normal Density μ σ
- 8. 7: Normal Probability Distributions 8 Standard Deviation σ • Points of inflections one σ below and above μ • Practice sketching Normal curves • Feel inflection points (where slopes change) • Label horizontal axis with σ landmarks
- 9. 7: Normal Probability Distributions 9 Two types of means and standard deviations • The mean and standard deviation from the pdf (denoted μ and σ) are parameters • The mean and standard deviation from a sample (“xbar” and s) are statistics • Statistics and parameters are related, but are not the same thing!
- 10. 7: Normal Probability Distributions 10 68-95-99.7 Rule for Normal Distributions • 68% of the AUC within ±1σ of μ • 95% of the AUC within ±2σ of μ • 99.7% of the AUC within ±3σ of μ
- 11. 7: Normal Probability Distributions 11 Example: 68-95-99.7 Rule Wechsler adult intelligence scores: Normally distributed with μ = 100 and σ = 15; X ~ N(100, 15) • 68% of scores within μ ± σ = 100 ± 15 = 85 to 115 • 95% of scores within μ ± 2σ = 100 ± (2)(15) = 70 to 130 • 99.7% of scores in μ ± 3σ = 100 ± (3)(15) = 55 to 145
- 12. 7: Normal Probability Distributions 12 Symmetry in the Tails … we can easily determine the AUC in tails 95% Because the Normal curve is symmetrical and the total AUC is exactly 1…
- 13. 7: Normal Probability Distributions 13 Example: Male Height • Male height: Normal with μ = 70.0˝ and σ = 2.8˝ • 68% within μ ± σ = 70.0 2.8 = 67.2 to 72.8 • 32% in tails (below 67.2˝ and above 72.8˝) • 16% below 67.2˝ and 16% above 72.8˝ (symmetry)
- 14. 7: Normal Probability Distributions 14 Reexpression of Non-Normal Random Variables • Many variables are not Normal but can be reexpressed with a mathematical transformation to be Normal • Example of mathematical transforms used for this purpose: – logarithmic – exponential – square roots • Review logarithmic transformations…
- 15. 7: Normal Probability Distributions 15 Logarithms • Logarithms are exponents of their base • Common log (base 10) – log(100 ) = 0 – log(101 ) = 1 – log(102 ) = 2 • Natural ln (base e) – ln(e0 ) = 0 – ln(e1 ) = 1 Base 10 log function
- 16. 7: Normal Probability Distributions 16 Example: Logarithmic Reexpression • Prostate Specific Antigen (PSA) is used to screen for prostate cancer • In non-diseased populations, it is not Normally distributed, but its logarithm is: • ln(PSA) ~N(−0.3, 0.8) • 95% of ln(PSA) within = μ ± 2σ = −0.3 ± (2)(0.8) = −1.9 to 1.3 Take exponents of “95% range” e−1.9,1.3 = 0.15 and 3.67 Thus, 2.5% of non-diseased population have values greater than 3.67 use 3.67 as screening cutoff
- 17. 7: Normal Probability Distributions 17 §7.2: Determining Normal Probabilities When value do not fall directly on σ landmarks: 1. State the problem 2. Standardize the value(s) (z score) 3. Sketch, label, and shade the curve 4. Use Table B
- 18. 7: Normal Probability Distributions 18 Step 1: State the Problem • What percentage of gestations are less than 40 weeks? • Let X ≡ gestational length • We know from prior research: X ~ N(39, 2) weeks • Pr(X ≤ 40) = ?
- 19. 7: Normal Probability Distributions 19 Step 2: Standardize • Standard Normal variable ≡ “Z” ≡ a Normal random variable with μ = 0 and σ = 1, • Z ~ N(0,1) • Use Table B to look up cumulative probabilities for Z
- 20. 7: Normal Probability Distributions 20 Example: A Z variable of 1.96 has cumulative probability 0.9750.
- 21. 7: Normal Probability Distributions 21 x z Step 2 (cont.) 5 . 0 2 39 40 has ) 2 , 39 ( ~ from 40 value the example, For z N X z-score = no. of σ-units above (positive z) or below (negative z) distribution mean μ Turn value into z score:
- 22. 7: Normal Probability Distributions 22 3. Sketch 4. Use Table B to lookup Pr(Z ≤ 0.5) = 0.6915 Steps 3 & 4: Sketch & Table B
- 23. 7: Normal Probability Distributions 23 a represents a lower boundary b represents an upper boundary Pr(a ≤ Z ≤ b) = Pr(Z ≤ b) − Pr(Z ≤ a) Probabilities Between Points
- 24. 7: Normal Probability Distributions 24 Pr(-2 ≤ Z ≤ 0.5) = Pr(Z ≤ 0.5) − Pr(Z ≤ -2) .6687 = .6915 − .0228 Between Two Points See p. 144 in text .6687 .6915 .0228 -2 0.5 0.5 -2
- 25. 7: Normal Probability Distributions 25 §7.3 Values Corresponding to Normal Probabilities 1. State the problem 2. Find Z-score corresponding to percentile (Table B) 3. Sketch 4. Unstandardize: p z x
- 26. 7: Normal Probability Distributions 26 z percentiles zp ≡ the Normal z variable with cumulative probability p Use Table B to look up the value of zp Look inside the table for the closest cumulative probability entry Trace the z score to row and column
- 27. 7: Normal Probability Distributions 27 Notation: Let zp represents the z score with cumulative probability p, e.g., z.975 = 1.96 e.g., What is the 97.5th percentile on the Standard Normal curve? z.975 = 1.96
- 28. 7: Normal Probability Distributions 28 Step 1: State Problem Question: What gestational length is smaller than 97.5% of gestations? • Let X represent gestations length • We know from prior research that X ~ N(39, 2) • A value that is smaller than .975 of gestations has a cumulative probability of.025
- 29. 7: Normal Probability Distributions 29 Step 2 (z percentile) Less than 97.5% (right tail) = greater than 2.5% (left tail) z lookup: z.025 = −1.96 z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 –1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233
- 30. 7: Normal Probability Distributions 30 35 ) 2 )( 96 . 1 ( 39 p z x The 2.5th percentile is 35 weeks Unstandardize and sketch
- 31. 7: Normal Probability Distributions 31 7.4 Assessing Departures from Normality Same distribution on Normal “Q-Q” Plot Approximately Normal histogram Normal distributions adhere to diagonal line on Q-Q plot
- 32. 7: Normal Probability Distributions 32 Negative Skew Negative skew shows upward curve on Q-Q plot
- 33. 7: Normal Probability Distributions 33 Positive Skew Positive skew shows downward curve on Q-Q plot
- 34. 7: Normal Probability Distributions 34 Same data as prior slide with logarithmic transformation The log transform Normalize the skew
- 35. 7: Normal Probability Distributions 35 Leptokurtotic Leptokurtotic distribution show S-shape on Q-Q plot