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Chapter 7: Normal Probability Distributions

Chapter 7: Normal Probability Distributions. 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. § 7.1: Normal Distributions.

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Chapter 7: Normal Probability Distributions

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  1. Chapter 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. §7.1: Normal Distributions • Normal random variables are the most common type of continuous random variable • First described de Moivre in 1733 • Laplace elaborated the mathematics in 1812 • Describe some (not all) natural phenomena • More importantly, describe the behavior of means

  4. This is the age distribution of a pediatric population. The overlying curve represents its Normal pdf model Normal Probability Density Function • Recall the continuous random variables are described with smooth probability density functions (pdfs) – Ch 5 • Normal pdfs are recognized by their familiar bell-shape

  5. Area Under the Curve • The darker bars of the histogram correspond to ages less than or equal to 9 (~40% of observations) • This darker area under the curve also corresponds to ages less than 9 (~40% of the total area)

  6. μ controls location σ controls spread Parameters μ and σ • Normal pdfs are a family of distributions • Family members identified by parameters μ(mean)and σ(standard deviation)

  7. σ μ Mean and Standard Deviation of Normal Density

  8. Standard Deviation σ • Points of inflections (where the slopes of the curve begins to level) occur one σbelow and above μ • Practice sketching Normal curves to feel inflection points • Practice labeling the horizontal axis of curves with standard deviation markers (figure)

  9. 68-95-99.7 Rule forNormal Distributions • 68% of the AUC falls within ±1σ of μ • 95% of the AUC falls within ±2σ of μ • 99.7% of the AUC falls within ±3σ of μ

  10. Wechsler adult intelligence scores are Normally distributed with μ = 100 and σ = 15; X ~ N(100, 15). Using the 68-95-99.7 rule: 68% of scores fall in μ ± σ= 100 ± 15 = 85 to 115 95% of scores fall in μ ± 2σ= 100 ± (2)(15) = 70 to 130 99.7% of scores in μ ± 3σ= 100 ± (3)(15) = 55 to 145 Example: 68-95-99.7 Rule

  11. Because of the Normal curve is symmetrical and the total AUC adds to 1… … we can determine the AUC in tails, e.g., Because 95% of curve is in μ ± 2σ, 2.5% is in each tail beyond μ ± 2σ 95% Symmetry in the Tails

  12. Example: Male Height • Male height is approximately Normal with μ= 70.0˝ and σ= 2.8˝ • Because of the 68-95-99.7 rule, 68% of population is in the range 70.0˝ 2.8˝ = 67.2 ˝ to 72.8˝ • Because the total AUC adds to 100%, 32% are in the tails below 67.2˝ and above 72.8˝ • Because of symmetry, half of this 32% (i.e., 16%) is below 67.2˝ and 16% is above 72.8˝

  13. Example: Male Height 64% 16% 16% 70 67.2 72.8

  14. Reexpression of Non-Normal Variables • Many biostatistical variables are notNormal • We can reexpress non-Normal variables with a mathematical transformationto make them more Normal • Example of mathematical transforms include logarithms, exponents, square roots, and so on • Let us review the logarithmic transformation

  15. Logarithms • Logarithms are exponents of their base • There are two main logarithmic bases • common log10(base 10) • natural ln (base e) • Landmarks: • log10(1) = 0 (because 100 = 1) • log10(10) = 1 (because 101 = 10)

  16. Example: Logarithmic Re-expression Since only 2.5% of population has values greater than 3.67 → use this as cut-point for suspiciously high results • Prostate specific antigen (PSA) not Normal in 60 year olds but the ln(PSA) is approximately Normal with μ = −0.3 and σ = 0.8 • 95% of ln(PSA) falls in μ ± 2σ= −0.3± (2)(0.8) = −1.9 to 1.3 • Thus, 2.5% are above ln(PSA) 1.3; take anti-log of 1.3: e1.3 = 3.67

  17. §7.2: Determining Normal Probabilities To determine a Normal probability when the value does not fall directly on a ±1σ, ±2σ, or ±3σ landmark, follow this procedure: 1. State the problem 2. Standardize the value (z score) 3. Sketch and shade the curve 4. Use Table B to determine the probability

  18. Example: Normal ProbabilityStep 1. Statement of Problem • We want to determine the percentage of human gestations that are less than 40 weeks in length • We know that uncomplicated human pregnancy from conception to birth is approximately Normally distributed with μ = 39 weeks and σ = 2 weeks. [Note: clinicians measure gestation from last menstrual period to birth, which adds 2 weeks to the μ.] • Let X represent human gestation: X ~ N(39, 2) • Statement of the problem: Pr(X ≤ 40) = ?

  19. Standard Normal (Z) Variable • Standard Normal variable≡ a Normal random variable with μ = 0 and σ = 0 • Called “Z variables” • Notation: Z ~ N(0,1) • Use Table B to look up cumulative probabilities • Part of Table B shown on next slide…

  20. Example: A Standard Normal (Z) variable with a value of 1.96 has a cumulative probability of .9750.

  21. Normal ProbabilityStep 2. Standardize To standardize, subtract μ and divide by σ. The z-scoretells you how the number of σ-units the value falls above or below μ

  22. Steps 3 & 4. Sketch and Use Table B 3. Sketch andlabel axes 4. Use Table B to lookup Pr(Z ≤ 0.5) = 0.6915

  23. Probabilities Between Two Points Let a represent the lower boundary and b represent the upper boundary of a range: Pr(a ≤ Z ≤ b) = Pr(Z ≤ b) − Pr(Z ≤ a) Use of this concept will be demonstrate in class and on HW exercises.

  24. §7.3 Finding Values Corresponding to Normal Probabilities • State the problem. • Use Table B to look up the z-percentile value. • Sketch 4. Unstandardize with this formula

  25. Looking up the z percentile value • Use Table B to look up the z percentile value, i.e., the z score for the probability in questions • Look inside the table for the entry closest to the associated cumulative probability. • Then trace the z score to the row and column labels.

  26. Suppose you wanted the 97.5th percentile z score. Look inside the table for .9750. Then trace the z score to the margins. Notation: Let zp represents the z score with cumulative probability p, e.g., z.975= 1.96

  27. Finding Normal Values - Example Suppose we want to know what gestational length is less than 97.5% of all gestations? Step 1. State the problem! Let X represent gestations length Prior problem established X ~ N(39, 2) We want the gestation length that is shorter than .975 of all gestations. This is equivalent to the gestation that is longer than.025 of gestations.

  28. Example, cont. Step 2. Use Table B to look up the z value. Table B lists only “left tails”. “less than 97.5%” (right tail) = “greater than 2.5%” (left tail). z lookup in table shows z.025 = −1.96

  29. 3. Sketch 4. Unstandardize “The 2.5th percentile gestation is 35 weeks.”

  30. Normal “Q-Q” Plot of same distribution 7.4 Assessing Departures from Normality The best way to assess Normality is graphically Approximately Normal histogram A Normal distribution will adhere to a diagonal line on the Q-Q plot

  31. Negative Skew A negative skew will show an upward curve on the Q-Q plot

  32. Positive Skew A positive skew will show an downward curve on the Q-Q plot

  33. Same data as previous slide but with logarithmic transform A mathematical transform can Normalize a skew

  34. Leptokurtotic A leptokurtotic distribution (skinny tails) will show an S-shape on the Q-Q plot

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