Chapter 2 The Normal Distribution
Lesson 2-1 Density Curve
Review Graph the data Calculate a numerical summary of the data Describe the shape, center, spread and outliers of the data
Histogram with Curve
Area Under the Curve area = proportion (or percent) = probability
Density Curve Is always on or above the horizontal axis, and The area underneath is exactly 1.
Median and Mean of a Density curve Median is the equal-areas point, the point that divides the area under the curve in half Mean is the balance point, at which the curve would balance if made of solid material. Median and mean are the same for a symmetric density curve Mean of a skewed curve is pulled away from the median in the direction of the long tail.
Mean and Median Symmetric Skewed to the Right
Notation Actual Observation Mean x Standard Deviation s Idealized Distribution Mean Standard Deviation
Example Page 83, #2.2 The density curve of a uniformed distribution. The curve takes the constant value 1 over the interval from 0 to 1 and is zero outside the range of values. This means that the data describe by this distribution takes values that are uniformly spread between 0 and 1. Use the areas under this density curve to answer the following questions 0 1
Example Page 83, #2.2 A) Why is the total area under this curve equal to 1? The area under the curve is a rectangle with height 1 and width 1. 1 1 0 1
Example Page 83, #2.2 B) What percent of the observation lie above 0.8?.20 (1 0.80) 0.20 A (0.20)(1) 0.20 20% 0 1
Example Page 83, #2.2 C) What percent of the observation lies below 0.60?.60 (.60 0).60 60% 0 1
Example Page 83, #2.2 D) What percent of the observation lie between 0.25 and 0.75.25.75 (.75.25).50 50% 0 1
Example Page 83, #2.2 E) What is the mean μ of this distribution? Mean = ½ or 0.50, the balance point of the density curve. 0 1
Example, Page 84, #2.4 The following figure displays three density curves, each with three points indicated. At which of these points on each curve do the mean and median fall?
Example, Page 84, #2.4 Median B Mean C Median A Mean A Median B Mean A
Example Density Curve I m thinking about a density curve that consists of a straight line segment from the point (0, 2/3) to the point (1,4/3) in the x-y plane. A). Sketch this density curve 4 3 2 3 1 1 3 1 1 3 4 2 1 4
Example Density Curve B). What percent of the observation lie below ½? Area of the Rectangle A bh 1 2 1 2 3 3 Area of the Triangle 1 1 bh A 2 3 1 2 2 12 A 1 1 5 T 41.67% 3 12 12 4 3 2 3 1 1 3 1 1 3 4 2 1 4
Example Density Curve C). What percent of the observation lie below 1? AT 1 100% 4 3 2 3 1 1 3 1 1 3 4 2 1 4
Example Density Curve C). What percent of the observation lie between ½ and 1? Area of the Rectangle 1 A bh 1 1 2 2 Area of the Triangle 1 1 bh A 3 2 1 2 2 12 A 1 1 7 T 58.3% 2 12 12 4 3 2 3 1 1 3 1 1 3 4 2 1 4
Lesson 2-1 Normal Distribution
Normal Distribution Symmetric, single peak and bell shape Tails fall quickly so we do no expect outliers Mean and median lie together at the peak in the center of the curve Standard deviation determines the shape of the curve
Standard Devotion Inflection Point Inflection Point
Why the Normal Distribution? Are good descriptions for some distribution of real data SAT Score Characteristics of a biological population Tossing coins many times Works well for other roughly symmetric distributions
68 95 99.7 Rule
68 95 99.7 Rule
Example, Page 89, #2.6 The distribution of heights of adult American men is approximately normal with mean 69 inches and standard deviation 2.5 inches. Draw a normal curve on which the mean and standard deviation are correctly located. 69 2.5
Example, Page 89, #2.6 69 2.5 2.5 66.5 69 71.5
Example, Page 89, #2.8 Scores on the Wechsler Adult Intelligence Scale (WAIS, a standard IQ test ) for 20 to 34 age group are approximately normally distributed with μ = 110 and σ = 25. Use the 68 95 99.7 rule to answer these questions.
Example, Page 89, #2.8 A) About what percent of people in this age group have scores about 110? about 68% 85 110 135
110 135 160 Example, Page 89, #2.8 B) About what percent have scores above 160? about 2.35%
60 85 110 135 160 Example, Page 89, #2.8 C) In what range do the middle 95% of IQ scores lie? between 60 to 160 or 110 50
60 85 110 135 160 Example, Page 91, #2.14 Wechsler Adult Intelligence Scale (WAIS) scores for young adults are N(110, 25).
60 85 110 135 160 Example, Page 91, #2.14 A. If someone s score were reported as the 16 th percentile about what score would that individual have. approximately 85
60 85 110 135 160 Example, Page 91, #2.14 B. If someone s score were reported as the 84 th percentile and 97.5 th percentile about what score would that individual have. approximately 135 and 160
Lesson 2-2 The Standard Normal Distribution
How do I find the area under the curve? Use calculus (find the integral). Too advance for the class. Use a series of tables to find areas for every possible mean and standard deviation. Infinitely many tables. Z-Scores
Z-Scores Assume that your variable is normally distributed. Use your mean and standard deviation to convert your data into z-scores such that the new distribution has a mean of 0 and a standard deviation of 1
Standard Normal Distribution The standard normal distribution is the normal distribution N(0,1) with mean 0 and standard deviation 1. If a variable x has any normal distribution N(μ, σ) with mean μ and standard deviation σ then the standardized variable is z x
Standard Normal Distribution
Example Page 95, #2.20 Three landmarks of baseball achievement are Ty Cobb s batting average of.420 in 1911, Ted Williams s.406 in 1941, and George Brett s 0.390 in 1980. These batting averages cannot be compared directly because the distribution of major league batting averages has changed over the years. The distributions are quite symmetric and (except for outliers such as Cobb, Williams and Brett) reasonably normal. While the mean batting average has been held roughly constant by rule changes and the balance between hitting and pitching, the standard deviation has dropped over time. Here are the facts:
Example Page 95, #2.20 Decade Mean Std. Dev. 1910s.266.0371 1940s.267.0326 1970s.261.0317 Compute the standardized batting averages for Cobb, Williams, and Brett to compare how far each stood above his peers.
Example Page 95, #2.20 z x Decade Mean Std. Dev. 1910s.266.0371 1940s.267.0326 Cobb: 1970s.261.0317 z.420.266.0371 Williams 4.15 Brett.390.261 z 4.07.0317 z.406.267.0326 4.26 Williams s z-score is the highest.
Example, Page 109, #2.28 Use Table A to find the proportion of observation from a Standard normal distribution that falls in each of the following regions. In each case sketch a standard normal Curve and shade the area representing the region a) b) c) d) z 2.25 z 2.25 z 1.77 2.25 z 1.77
Example, Page 109, #2.28 a) z 2.25-2.25 0
Example, Page 109, #2.28
Example, Page 109, #2.28 a) z 2.25 The area to the left of -2.25 is 0.0122-2.25 0
Example, Page 109, #2.28 b) z 2.25 Find the area to the left of -2.25 z 2.25 0.0122 1 0.0122 0.9878-2.25 0 Area greater than -2.25 = 1 area below -2.25
Example, Page 109, #2.28 c) A z 1.77 1 z 1.77 0 1.77
Example, Page 109, #2.28
Example, Page 109, #2.28 c) A z 1.77 1 z 1.77 1 0.9616 0.0384 0 1.77 Area greater than 1.77 = 1 area below 1.77
Example, Page 109, #2.28 d) 2.25 z 1.77 1.77 2.25 A z z A 0.9616 0.0122 0. 9494-2.25 0 1.77 Area between -2.25 and 1.77 = area below 1.77 area below -2.25
TI 83 (Area) z 2.25 2 nd Vars
TI 83 (Area) z 2.25 2.25 z 1.77
Example Page 103, #2.22 Use Table A to find the value of z of a standard normal Variable that satisfies each of the following conditions. Use the value of z from Table A that comes closet to satisfying the condition. In each case sketch a standard curve with your value of z marked on the axis. a) The point z with 25% of the observation falling below it. b) The point z with 40% of the observation falling above it.
Example Page 103, #2.22 a) The point z with 25% of the observation falling below it. A 0.25 z?
Example Page 103, #2.22
Example Page 103, #2.22 a) The point z with 25% of the observation falling below it. A 0.25 z 0.67
Example Page 103, #2.22 b) The point z with 40% of the observation falling above it. Find the area below 0.40 1 0.40.60 A 0.40 z?
Example Page 103, #2.22
Example Page 103, #2.22 b) The point z with 40% of the observation falling above it. Find the area below 0.40 1 0.40.60 A 0.40 z 0.25
TI 83 (Z Value) 25% below 2 nd Vars
TI 83 (Z Value) 40% Above
Example, Page 103, #2.24 Scores on the Wechsler Adult Intelligence Scale (a standard IQ test) for the 20 to 34 age group are approximately normally distributed with μ = 110 and σ = 25 A. What percent of people age 20 to 34 have IQ scores above 100? B. What percent have scores above 150? C. How high an IQ score is needed to be the highest 25%?
Example, Page 103, #2.24 μ = 110 and σ = 25 A. What percent of people age 20 to 34 have IQ scores about 100? z x μ 100 110 0.40 σ 25 65.5% 100 110-0.40 0 Z
Example, Page 103, #2.24 μ = 110 andσ = 25 B. What percent have scores above 150? z x μ 150 110 1.6 σ 25 5.5% 110 0 150 1.60 Z
Example, Page 103, #2.24 μ = 110 andσ = 25 C. How high an IQ score is needed to be the highest 25%? x μ z σ x 110 0.68 25 x 127 A = 0.25 110?
Example, Page 110, #2.30 The annual rate of return on stock indexes (which combine many individual stocks) is approximately normal. Since 1945, the Standard & Poor s 500 Index has had a mean yearly return of 12%, with a standard deviation of 16.5%. Take this normal distribution to be the distribution of yearly returns over a long period. A. In what range do the middle 95% of all yearly returns lie? B. The market is down for the year if the return on the index is less than zero. In what proportion of years is the market down?
Example, Page 110, #2.30 μ = 12% and σ = 16.5% A. In what range do the middle 95% of all yearly returns lie? 12% 2(16.5%) -21% to 45% -21% 12% 45%
Example, Page 110, #2.30 μ = 12% and σ = 16.5% B. The market is down for the year if the return on the index is less than zero. In what proportion of years is the market down? z x μ 0 12 0.727 σ 16.5 0.2335 0% -0.73 12% 0 Z
Example, Page 110, #2.30 μ = 12% and σ = 16.5% C. In what proportion of years does the index gain 25% or more? z x μ 25 12 0.7878 σ 16.5 0.2154 12 25 0 0.79 Z
Lesson 2-2 Accessing Normality
Assessing Normality Construct a frequency histogram or a stemplot or boxplot Check to see if graph is approximately bell-shape and symmetric about the mean Construct a normal probability plot Use TI to see if plotted points lie close to a straight line
Example, Page 108, #2.26 Repeated careful measurements of the same physical quantity often have a distribution that is close to normal. Here are Henry Cavendish s 29 measurements of the density of the earth, made in 1798. (The data gives the density of the earth as multiple of the density of water.) 5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.65 5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.39 5.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85
Example, Page 108, #2.26 5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.65 5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.39 5.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85 A. Construct a stemplot to show that the data are reasonably symmetric
Example, Page 108, #2.26 A. Construct a stemplot to show that the data are reasonably symmetric 4.88 48 8 48 8 49 50 7 51 0 52 6 7 9 9 53 0 4 4 6 9 54 2 4 6 7 55 0 3 5 7 8 56 1 2 3 5 8 57 5 9 58 5
Example, Page 108, #2.26 B. Now check how closely they follow the 68-95-99.7 rule. Find x and s, then count the number of observations that fall between x s, between x 2s, between x 3s. Compare the percents of the 29 observations in each of these intervals with the 68-95-99.7 rule. 48 8 49 50 7 51 0 52 6 7 9 9 53 0 4 4 6 9 54 2 4 6 7 55 0 3 5 7 8 56 1 2 3 5 8 57 5 9 58 5 4.88 48 8
Example, Page 108, #2.26 1 0% 5.01 x 2s 2 6.9% x 5.23 11 37.9% s 11 37.9% 5.45 4 13.8% 5.67 5.89 0 0% x x s x 2s 48 8 49 50 7 51 0 52 6 7 9 9 53 0 4 4 6 9 54 2 4 6 7 55 0 3 5 7 8 56 1 2 3 5 8 57 5 9 58 5 4.88 48 8
Example, Page 108, #2.26 C. Use your calculator to construct a normal probability plot for Cavendish s density of the earth data, write a brief statement about the normality of the data. Does the normal probability reinforce your findings in (a).
Example, Page 108, #2.26 The linearity of the normal probability indicates an approximately normal distribution.