Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

no text concepts found

Transcript

Chapter 3 The Normal Distributions 5/24/2017 Chapter 3 1 Density Curves • Here is a histogram of vocabulary scores of n = 947 seventh graders • The smooth curve drawn over the histogram is a mathematical model which represents the density function of the distribution 5/24/2017 Chapter 3 2 Density Curves • The shaded bars on this histogram corresponds to the scores that are less than 6.0 • This area represents is 30.3% of the total area of the histogram and is equal to the percentage in that range 5/24/2017 Chapter 3 3 Area Under the Curve (AUC) • This figure shades area under the curve (AUC) corresponding to scores less than 6 • This also corresponds to the proportion in that range: AUC = proportion in that range 5/24/2017 Chapter 3 4 Density Curves 5/24/2017 Chapter 3 5 Mean and Median of Density Curve 5/24/2017 Chapter 3 6 Normal Density Curves • Normal density curves are a family of bellshaped curves • The mean of the density is denoted μ (“mu”) • The standard deviation is denoted σ (“sigma”) 5/24/2017 Chapter 3 7 The Normal Distribution • Mean μ defines the center of the curve • Standard deviation σ defines the spread • Notation is N(µ,). 5/24/2017 Chapter 3 8 Practice Drawing Curves! • The Normal curve is symmetrical around μ • It has infections (blue arrows) at μ ± σ 5/24/2017 Chapter 3 9 The 68-95-99.7 Rule • • • • 68% of AUC within μ ± 1σ 95% fall within μ ± 2σ 99.7% within μ ± 3σ Memorize! This rule applies only to Normal curves 5/24/2017 Chapter 3 10 Application of 68-95-99.7 rule • Male height has a Normal distribution with μ = 70.0 inches and σ = 2.8 inches • Notation: Let X ≡ male height; X~ N(μ = 70, σ = 2.8) 68-95-99.7 rule • 68% in µ = 70.0 2.8 = 67.2 to 72.8 • 95% in µ 2 = 70.0 2(2.8) = 64.4 to 75.6 • 99.7% in µ 3 = 70.0 3(2.8) = 61.6 to 78.4 5/24/2017 Chapter 3 11 Application: 68-95-99.7 Rule What proportion of men are less than 72.8 inches tall? μ + σ = 70 + 2.8 = 72.8 (i.e., 72.8 is one σ above μ) 68% ?68% 16% -1 (by 68-95-99.7 Rule) 16% (total AUC = 100%) +1 70 72.8 (height) 84% Therefore, 84% of men are less than 72.8” tall. 5/24/2017 Chapter 3 12 Finding Normal proportions What proportion of men are less than 68” tall? This is equal to the AUC to the left of 68 on X~N(70,2.8) ? 68 70 (height values) To answer this question, first determine the z-score for a value of 68 from X~N(70,2.8) 5/24/2017 Chapter 3 13 Z score z x • The z-score tells you how many standard deviation the value falls below (negative z score) or above (positive z score) mean μ • The z-score of 68 when X~N(70,2.8) is: z x 68 70 0.71 2.8 Thus, 68 is 0.71 standard deviations below μ. 5/24/2017 Chapter 3 14 Example: z score and associate value ? 68 70 (height values) -0.71 5/24/2017 0 Chapter 3 (z values) 15 Standard Normal Table Use Table A to determine the cumulative proportion associated with the z score See pp. 79 – 83 in your text! 5/24/2017 Chapter 3 16 Normal Cumulative Proportions (Table A) z .00 .01 .02 0.8 .2119 .2090 .2061 .2420 .2389 .2358 .2743 .2709 .2676 0.7 0.6 Thus, a z score of −0.71 has a cumulative 5/24/2017 Chapter 3of .2389 proportion 17 Normal proportions The proportion of mean less than 68” tall (z-score = −0.71 is .2389: .2389 68 70 -0.71 5/24/2017 0 Chapter 3 (height values) (z scores) 18 Area to the right (“greater than”) Since the total AUC = 1: AUC to the right = 1 – AUC to left Example: What % of men are greater than 68” tall? .2389 1.2389 = .7611 68 70 -0.71 5/24/2017 0 Chapter 3 (height values) (z values) 19 Normal proportions “The key to calculating Normal proportions is to match the area you want with the areas that represent cumulative proportions. If you make a sketch of the area you want, you will almost never go wrong. Find areas for cumulative proportions … from [Table A] (p. 79)” Follow the “method in the picture” (see pp. 79 – 80) to determine areas in right tails and between two points 5/24/2017 Chapter 3 20 Finding Normal values We just covered finding proportions for Normal variables. At other times, we may know the proportion and need to find the Normal value. Method for finding a Normal value: 1. State the problem 2. Sketch the curve 3. Use Table A to look up the proportion & z-score 4. Unstandardize the z-score with this formula x z 5/24/2017 Chapter 3 21 State the Problem & Sketch Curve Problem: How tall must a man be to be taller than 10% of men in the population? (This is the same as asking how tall he has to be to be shorter than 90% of men.) Recall X~N(70, 2.8) .10 ? 70 5/24/2017 (height) Chapter 3 22 Table A Find z score for cumulative proportion ≈.10 z .07 .08 .09 1.3 .0853 .0838 .0823 .1020 .1003 .0985 .1210 .1190 .1170 1.2 1.1 zcum_proportion = z.1003 = −1.28 5/24/2017 Chapter 3 23 Visual Relationship Between Cumulative proportion and z-score .10 ? 70 -1.28 5/24/2017 0 Chapter 3 (height values) (Z value) 24 Unstandardize • x = μ + z∙σ = 70 + (1.28 )(2.8) = 70 + (3.58) = 66.42 • Conclude: A man would have to be less than 66.42 inches tall to place him in the lowest 10% of heights 5/24/2017 Chapter 3 25