BMS 511: Biostats & Statistical Analysis - The Normal Distribution PDF

BMS 511 Biostats & Statistical Analysis Chapter 11 The Normal Distribution Guang Xu, PhD, MPH Assistant Professor of Biostatistics and Public Health College of Osteopathic Medicine Marian University...

BMS 511 Biostats & Statistical Analysis Chapter 11 The Normal Distribution Guang Xu, PhD, MPH Assistant Professor of Biostatistics and Public Health College of Osteopathic Medicine Marian University Copyright © 2018 W. H. Freeman and Company  Previous Learning Objectives Define general rules of probability Independent events Conditional probability General addition rule Multiplication rule Tree diagrams Bayes’s theorem Diagnosis tests Copyright © 2018 W. H. Freeman and Company  Learning Objectives Demonstrate Normal distributions Normal distributions The 68–95–99.7 rule The standard Normal distribution Using the standard Normal table (Table B) Inverse Normal calculations Normal quantile plots Copyright © 2018 W. H. Freeman and Company  Normal distributions Normal—or Gaussian—distributions are a family of symmetrical, bell- shaped density curves defined by a mean µ (mu) and a standard deviation σ (sigma): N(µ,σ). 1 1 𝑥−𝜇 2 −2 𝜎 𝑓 𝑥 = 𝑒 2𝜋 Normal curves are used to model many biological variables. They can describe a population distribution or a probability distribution. Copyright © 2018 W. H. Freeman and Company  A family of density curves Here means are the same (µ = 15) whereas standard deviations are different (σ = 2, 4, and 6). Here means are different (µ = 10, 15, and 20) whereas standard deviations are the same (σ = 3). Copyright © 2018 W. H. Freeman and Company Normal model examples Human heights, by gender, can be modeled quite accurately by a Normal distribution. Guinea pigs survival times after inoculation of a pathogen are clearly not a good candidate for a Normal model! Copyright © 2018 W. H. Freeman and Company  The 68–95–99.7 rule All normal curves N(µ,σ) share the same properties: About 68% of all observations are within 1 standard deviation (σ) of the mean (µ). About 95% of all observations are within 2 σ of the mean µ. Almost all (99.7%) observations are within 3 σ of the mean. To obtain any other area under a Normal curve, use either technology or Table B. Copyright © 2018 W. H. Freeman and Company  Normal example (1 of 2) What percent of young adults have osteoporosis or osteopenia? World Health Organization definitions of osteoporosis based on standardized bone density levels Normal Bone density is within 1 standard deviation (z > –1) of the young adult mean or above. Low bone Bone density is 1 to 2.5 standard deviations below the young adult mass mean (z between –2.5 and –1). Osteoporosis Bone density is 2.5 standard deviations or more below the young adult mean (z ≤ – 2.5). Copyright © 2018 W. H. Freeman and Company  Normal example (2 of 2) Population of young adults N(0,1) These are bone densities of −1 or less, representing the area to the left of the middle 68% between −1 and +1. So 16%. Copyright © 2018 W. H. Freeman and Company  Another Normal example (1 of 2) Women age 70 to 79 are NOT young adults. The mean bone density in this age is about −2 on the standard scale for young adults. What is the probability that a randomly chosen woman in her 70s has osteoporosis or osteopenia (< −1)? A. 97.5% B. 95% C. 84% D. 68% E. 50% Copyright © 2018 W. H. Freeman and Company Another Normal example (2 of 2) Copyright © 2018 W. H. Freeman and Company  The standard Normal distribution We can standardize data by computing a z- 𝑥−𝜇 score: 𝑧 = 𝜎 If x has the N(µ, σ) distribution, then z has the N(0, 1) distribution. Copyright © 2018 W. H. Freeman and Company  Standardizing z-scores A z-score measures the number of standard deviations that a data value x is from the mean µ. 𝑥−𝜇 𝑧= 𝜎 When 𝑥 is 1 standard deviation larger than the mean, then 𝑧 = 1. 𝜇+𝜎−𝜇 𝜎 for 𝑥 = 𝜇 + 𝜎, 𝑧 = = =1 𝜎 𝜎 When 𝑥 is 2 standard deviations larger than the mean, then 𝑥 = 𝜇 + 𝜇+2𝜎−𝜇 2𝜎 2𝜎, 𝑧 = = =2 𝜎 𝜎 When x is larger than the mean, z is positive. When x is smaller than the mean, z is negative. Copyright © 2018 W. H. Freeman and Company  Standardizing z-scores example (1 of 2) Women’s heights follow the N(64.5",2.5") distribution. What percent of women are shorter than 67 inches tall (that’s 5'6")? mean µ = 64.5" standard deviation σ = 2.5" height x = 67" We calculate z, the standardized value of x: 𝑥−𝜇 67 − 64.5 2.5 𝑧= ,𝑧 = = = 1 → 1 stand. dev. from mean 𝜎 2.5 2.5 Given the 68–95–99.7 rule, the percent of women shorter than 67" should be, approximately, 0.68 + half of (1 – 0.68) = 0.84, or 84%. The probability of randomly selecting a woman shorter than 67" is also ~84%. Copyright © 2018 W. H. Freeman and Company Standardizing z-scores example (2 of 2) Copyright © 2018 W. H. Freeman and Company  Using Table (1 of 3) Table gives the area under the standard Normal curve to the left of any z-value. Copyright © 2018 W. H. Freeman and Company  Using Table (2 of 3) For z = 1.00, the area under the curve to the left of z is 0.8413. →84.13% of women are shorter than 67". →Therefore, 15.87% of women are taller than 67" (5'6"). Copyright © 2018 W. H. Freeman and Company  Using Table (3 of 3) For z = 1.00, the area under the curve to the left of z is 0.8413. →84.13% of women are shorter than 67". →Therefore, 15.87% of women are taller than 67" (5'6"). Copyright © 2018 W. H. Freeman and Company  Tips on using Table Because of the curve’s symmetry, there are two ways of finding the area under N(0,1) curve to the right of a z- value. Copyright © 2018 W. H. Freeman and Company  Using Table to find a middle area To calculate the area between two z-values, first get the area under N(0,1) to the left for each z-value from Table. Then subtract the smaller area from the larger area. ➔ The area under N(0,1) for a single value of z is zero. Copyright © 2018 W. H. Freeman and Company  Table middle areas example (1 of 2) The blood cholesterol levels of men age 55 to 64 are approximately Normal with mean 222 mg/dL and standard deviation 37 mg/dL. What percent of middle-age men have high cholesterol (> 240 mg/dL)? What percent have elevated cholesterol (between 200 and 240 mg/dL)? x z area left area right 240 0.49 69% 31% 200 −0.59 28% 72% Difference: 41% 41% The difference between the two areas (left or right) gives our answer. Copyright © 2018 W. H. Freeman and Company  Table middle areas example (2 of 2) The difference between the two areas (left or right) gives our answer. Copyright © 2018 W. H. Freeman and Company  Inverse Normal calculations You may also seek the range of values that correspond to a given proportion/area under the curve. For that, use technology or use Table backward. First find the desired area/proportion in the body of the table. Then read the corresponding z- value from the left column and top row. Copyright © 2018 W. H. Freeman and Company  Inverse Normal calculations example The hatching weights of commercial chickens can be modeled accurately using a Normal distribution with mean μ = 45 g and standard deviation σ = 4 g. What is the third quartile of the distribution of hatching weights? We know μ, σ, and the area under the curve; we want x. Table B gives the area left of z → look for the lower 25%. We find z ≈ 0.67. 𝑥−𝜇 𝑧= ↔𝑥 =𝜇+ 𝑧∗𝜎 𝜎 𝑥 = 45 + 0.67 ∗ 4 𝑥 = 47.68 𝑄3 ≈ 47.7g Copyright © 2018 W. H. Freeman and Company  More Normal calculation examples The blood cholesterol levels of men aged 55 to 64 are approximately normal with mean 222 mg/dL and standard deviation 37 mg/dL. What range of values corresponds to the 10% highest cholesterol levels? A) > 175 B) > 247 C) > 269 D) > 288 z Area left Area right 1.28 90% 10% Copyright © 2018 W. H. Freeman and Company  Normal quantile plots One way to assess if a data set has an approximately Normal distribution is to plot the data on a Normal quantile plot (Q-Q Plot). – The data points are ranked and the percentile ranks are converted to z-scores. The z-scores are then used for the horizontal axis and the actual data values are used for the vertical axis. Use technology to obtain Normal quantile plots. If the data have approximately a Normal distribution, the Normal quantile plot will have roughly a straight-line pattern. Copyright © 2018 W. H. Freeman and Company  Normal quantile plot examples Right skewed Roughly normal (most of the data points are (~ straight-line pattern) short survival times, while a few are longer survival times) Copyright © 2018 W. H. Freeman and Company  Learning Objectives Demonstrate Normal distributions Normal distributions The 68–95–99.7 rule The standard Normal distribution Using the standard Normal table (Table B) Inverse Normal calculations Normal quantile plots Copyright © 2018 W. H. Freeman and Company