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Daniel S. Yates The Practice of Statistics Third Edition Chapter 2: Describing Location in a Distribution Section 2.2 Normal Distributions Copyright © 2008 by W. H. Freeman & Company Essential Questions • What are the properties of the Normal Curve? • How do you explain the 68-95-99.7 empirical rule? • What does N(µ, σ) mean? • What is the standard Normal distribution? The mathematical model for the Normal Density Curve 1 f ( x) e 2 1 x 2 2 Why the Normal Distribution? Many measurements distributions in the real world can be modeled by the Normal Distribution • Scores on Test (SAT, ACT, psychological test, etc.) • Repeated careful measurements of the same quantity • Characteristics of biological population Beware- many real world distributions many not follow the nomal distributions. The Normal Curve Point of Inflection Normal curves are defined by the means and the standard deviation. Short Notation for Normal Density Curves Since the normal density curves are completely defined by the means μ and the standard deviation σ, we can use the abbreviation: N(μ, σ) Characteristics of a Normal Density Curve • Symmetric • Single peaked – bell shaped • Means and median are located at the center of symmetry for the curve • The standard deviation is the distance from the Means to the point of inflection on the curve • The curve is described by means, μ, and standard deviation, σ • There are many normal curves • All curves follow the 68-95-99.7 rule The 68-95-99.7 Rule The Normal Curve Applet • Turn to page 137 Activity 2C • Export-9-4-2008\NormalCurve.html 68-95-99.7 Rule 68.2-95.4-99.7 Rule We will use the simplified version in the book: 68-95-99.7 rule. 68-95-99.7 Percentile • Percentile score is the percent of individuals who scored less than or equal to your score. • Percentile is used for comparing an individual observation relative to other individuals in the distribution. • In practice, when observations are quite large then the percentile is reported using density curves. Percentile Using Density Curve 0.15% 2.5% 16% 50% 84% 97.5% 99.85% Example The distribution of heights for women is N(64.5”, 2.5”). • μ = ____ σ = _____ • The middle 68% of women are between _______________ • The middle 95% are between ________ • The 50th percentile of women are ____ or lower. • The 97.5th percentile of women are ____ or lower. • ___% of women are 67” or shorter. • ___% of women are 59.5” or taller. • The ____ percentile would be women about 71.5” or shorter. The Standard Normal Curve Since all normal density curves, N(μ, σ) have the same characteristics we can use one curve to do all of our calculations. This curve is the Standard Normal Curve N(0,1). μ = 0 and σ = 1 When we change any N(µ, σ) to N(0, 1) we can find areas for all N(µ, σ) using one single table called a Standard Normal Table. • Turn to Table A in front of the Textbook. • Notice: one side is for negative z-scores and one side for positive z’s. Example Using the Standard Normal Table A Find the proportion of the data from the standard Normal distribution that will have a z-score of less than -2.15. Standard Normal Calculations Export-9-4-2008\NormalCurve.html • Find the proportion of observations from a standard Normal distribution that falls in each of the following region. Ans: 0.0122 • z ≤ -2.25 • z ≥ -2.25 Ans: 1 – 0.0122 = 0.9878 • z > 1.77 Ans: 1 – 0.9616 = 0.0384 • -2.25 < z < 1.77 Ans: 1 – ( 0.0384 + 0.0122) = 0.9494 Working in Reverse • Use Table A to find the value z of a standard Normal variable that satisfies each of the following conditions. In each case sketch a standard Normal curve with your value of z marked on the axis. • The point z with 70% of the observations falling below it. • Ans: z = 0.52. • The point with 85% of the observations falling above it. • Ans: z = -1.04. • Find the number z such that the proportion of observations are less than z is 0.8. • Ans: z = 0.84. • Find the number z such that 90% of all observations are greater than z. • Ans: z = -1.28. Procedure for Normal Distribution Calculations Example During World War II, physical training was required for male students in many colleges, as preparation for military service. That provided an opportunity to collect data on physical performance on a large scale. A study of 12,000 ablebodied male students at the University of Illinois found that their times for the mile run had a mean 7.11 minutes and a standard deviation of 0.739. The distribution of times was roughly Normal. That seems pretty fast; presumably the times were taken after the required training. µ = 7.11 minutes σ = 0.739 n = 12,000 (a) About how many of the 12,000 male students ran a mile in less than 5 minutes? Using format in Exercise 1.17 on page 65. Example N(7.11 min, 0.739) a). About how many of the 12,000 male students ran a mile less than 5 minutes? Step 1 State the problem. z X < 5 minutes Step 2 Standardize and draw. 5 7.11 z 2.86 0.739 Step 3. Use Table A. Proportion = 0.0021 Step 4 (0.0021)(12000) = 25.2 Approximately 25 students run less than 5 minutes per mile. Example • N(7.11, 0.739) • What is the 93rd percentile in the mile run time distribution? z Step 1 Proportion = .93 What is z? Step 2 From table A z = 1.48 Step 3. Unstandardizing. X 7.11 1.48 0.739 X (1.48)(0.739) 7.11 8.20 minutes Step 4 Conclusion. The 93rd percentile is 8.20 minutes. Checking for Normal Two method • Method I Construct a histogram or a stemplot and compare to the 68-95-99.7 empirical rule. • Method 2 Construct a Normal probability plot. Method 1 • Construct a histogram or stem-plot. • Locate x, x s, x 2s, x 3s On the x axis. • Count the data that falls in each interval and compare how well it compares to the 68-95-97.5 rule. Manual Procedure for Constructing a Normal Probability Plot • Arrange the observed data in ascending order. • Change each observed data to a z-score • Plot each data point x against the corresponding z. • If the data distribution is close to Normal the plotted points will lie close to a straight line. Normal Probability Plots Approximately Normal Skewed Right distribution