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Chapter 3

Chapter 3. The Normal Distributions . 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 t he density function of the distribution. Density Curves.

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Chapter 3

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  1. Chapter 3 The Normal Distributions Chapter 3

  2. 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 Chapter 3

  3. 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 Chapter 3

  4. 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 Chapter 3

  5. Density Curves Chapter 3

  6. Mean and Median of Density Curve Chapter 3

  7. Normal Density Curves • Normal density curves are a family of bell-shaped curves • The mean of the density is denoted μ (“mu”) • The standard deviation is denoted σ (“sigma”) Chapter 3

  8. The Normal Distribution • Mean μ defines the center of the curve • Standard deviation σ defines the spread • Notation is N(µ,). Chapter 3

  9. Practice Drawing Curves! • The Normal curve is symmetrical around μ • It has infections (blue arrows) at μ ± σ Chapter 3

  10. 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 Chapter 3

  11. 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 Chapter 3

  12. 68% 68% (by 68-95-99.7 Rule) (total AUC = 100%) ? 16% 16% -1 +1 70 72.8 (height) 84% 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) Therefore, 84% of men are less than 72.8” tall. Chapter 3

  13. ? 68 70 (height values) 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) To answer this question, first determine the z-score for a value of 68 from X~N(70,2.8) Chapter 3

  14. Z score • 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: Thus, 68 is 0.71 standard deviations below μ. Chapter 3

  15. ? 68 70 (height values) Example: z score and associate value -0.71 0 (z values) Chapter 3

  16. Standard Normal Table Use Table A to determine the cumulative proportion associated with the z score See pp. 79 – 83 in your text! Chapter 3

  17. Normal Cumulative Proportions (Table A) .01 0.7 .2389 Thus, a z score of −0.71 has a cumulative proportion of .2389 Chapter 3

  18. 68 70 (height values) -0.71 0 (z scores) Normal proportions The proportion of mean less than 68” tall (z-score = −0.71 is .2389: .2389 Chapter 3

  19. 68 70 (height values) -0.71 0 (z values) 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? 1.2389 = .7611 .2389 Chapter 3

  20. 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 Chapter 3

  21. 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 Chapter 3

  22. .10 ? 70 (height) 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) Chapter 3

  23. Table AFind z score for cumulative proportion ≈.10 .08 1.2 .1003 zcum_proportion = z.1003 = −1.28 Chapter 3

  24. .10 ? 70 (height values) Visual Relationship Between Cumulative proportion and z-score -1.28 0 (Z value) Chapter 3

  25. 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 Chapter 3

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