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Sept. 3 Statistic for the day: Compared to same month the previous year, average monthly drop in U.S. traffic fatalities from May 2007 to Feb. 2008: 4.2% Drop in March 2008: 22.1% Drop in April 2008: 17.9% Smoothing the histogram: The Normal Curve (Chapter 8) A histogram tends to be rough. To replace it with a bell shaped curve: Center the “bell” at the mean. The “bell” should be just wide enough so that the middle 95% of the bell is 4 standard deviations wide. This makes systematic, accurate predictions of all sorts possible, provided the bell shape is appropriate for the underlying population. Recall this dataset on handspans from the previous lecture: Histogram of HandSpan, with Normal Curve 23.5 21.0 21.5 23.0 21.0 22.0 22.5 23.0 21.0 22.5 23.0 22.0 22.5 24.5 24.5 20.5 22.5 22.5 23.0 21.5 22.0 20.5 22.0 24.5 21.5 21.0 23.5 22.0 22.0 22.0 24.0 20.5 24.0 22.0 24.5 22.0 19.0 23.5 20.5 22.0 18.0 24.0 22.5 22.0 20.0 21.5 21.0 23.5 21.5 21.5 22.0 22.5 20.5 21.0 24.0 22.0 Women (n = 89) 21.5 18.0 20.5 18.5 19.0 18.5 20.0 20.0 20.0 21.0 19.5 19.0 16.0 17.5 18.5 20.0 19.0 19.0 20.0 18.0 20.5 21.5 20.5 20.5 19.5 22.0 20.0 21.0 19.5 19.5 18.5 21.5 20.0 22.5 17.0 17.0 20.5 21.0 20.0 23.0 21.5 21.5 20.0 18.5 18.5 18.5 20.0 20.0 20.5 21.0 19.0 20.5 21.0 20.0 21.0 20.0 18.5 16.5 19.0 19.0 19.0 20.5 17.0 19.5 19.0 20.0 19.5 20.5 20.0 19.5 19.0 18.5 20.0 17.0 21.0 18.5 20.5 19.5 19.0 20.5 21.0 18.5 17.5 19.5 18.5 19.5 20.0 20.0 18.5 Frequency 20 Men (n = 78) 23.5 22.5 24.5 24.5 20.0 21.0 24.0 21.0 21.0 22.5 23.0 24.0 24.5 23.0 10 0 15 20 25 HandSpan Mean = 20.86 Standard deviation = 1.927 Histogram of HandSpan, with Normal Curve Histogram of Height, with Normal Curve 20 Frequency Frequency 30 10 20 10 0 0 15 20 HandSpan Mean = 20.86 Standard deviation = 1.927 25 60 70 80 Height Mean = 68 inches or 5 feet 8 inches Standard deviation = 4 inches 1 Research Question 2: How high should I build my doorways so that 99% of the people will not have to duck? Z-Scores: Measurement in Standard Deviations 75 − mean 75 − 68 Z= = = 1.75 SD 4 Histogram of Height, with Normal Curve 30 Frequency Research Question 1: If I built my doors 75 inches (6 feet 3 inches) high, what percent of the people would have to duck? 20 10 0 60 70 80 Height Question 1 (x=75) Question 2 (x=??) Q1: The value of x is 75; find the amount of distribution above it. Q2: Find the value of x so that 99% of the distribution is below it. Compute your Z-score. 1. How many standard deviations are you above or below the mean. Use: Mean = 68 inches Standard deviation = 4 inches 2. Now use the table from the book (p. 157) to determine what percentile you are. Compare Heights of Females and Males Stat 100 students Sp01 80 Assume male heights have a normal distribution with mean 70 inches and st dev 3 inches. Assume female heights have a normal distribution with mean 64 inches and st dev 3 inches. Height What is your Z-Score within your sex? 70 What is your percentile within your sex? 60 Female Male Sex 2 Histogram of Height, with Normal Curve Answer to Question 1: What percent of people would have to duck if I built my doors 75 inches high? From the standard normal table in the book: .96 or 96% of the distribution is below 1.75. Hence, .04 or 4% is above 1.75. Frequency Recall: 75 has a Z-score of 1.75 30 20 4% in here 10 0 60 So 4% of the distribution is above 75 inches. 70 75 80 Height Question 1 (x=75) The value at x is 75; find the amount of distribution above it. Convert 75 to Z = 1.75 and use Table 8.1 on p. 157. Question 2: What is the value so that 99% of the distribution is below it? (called the 99th percentile.) 3. Now convert it over to inches: 2.33 = h99 − 68 4 h99 = 68 + 2.33(4) = 77.3 30 Frequency 1. Look up the Z-score that corresponds to the 99th percentile. From the table: Z = 2.33. Histogram of Height, with Normal Curve 20 10 99% in here 0 60 70 80 Height Question 2 77.3 inches is the 99th percentile Therefore, 99% of the distribution is shorter than 77.3 inches (6 foot 5.3 inches) and that’s how high the door should be built. To find the value so that 99% of the distribution is below it: Look up the Z-score for the 99th percentile and convert it back to inches. Answer these questions: To answer question 1, first convert 72 inches to a z-score: (Assume that adults’ heights are normally distributed with mean 68 inches and standard deviation 4 inches.) .84 or 84% 3 Answer to Question 2: What is the first quartile of heights? Translation: “First quartile” means 25th percentile, which means .25 are below that height. From p. 157: Find the z-score corresponding to the 25th percentile. –.67 Now convert this z-score into a height: Z − score = h − 68 4 h = 68 + 4( Z − score) There are roughly 305 million people in US. About 49% are over the age of 20 (Census Bureau). That is about 150 million. Shaquille O’Neal is 7 feet 1 inch or 85 inches tall. How many people in the country are taller? We will assume that heights are normally distributed with mean 68 inches and standard deviation 4 inches. O’Neal’s Z-score is Z = (85-68)/4 = 4.25. In other words O’Neal is 4.25 standard deviations above the mean(!) There is only 0.000011 of the normal distribution above 4.25 standard deviations. Page 157 Hence, there should be roughly .000011 times 150 million or 1650 people taller than Shaquille O’Neal. Note: This is an extremely rough calculation, since the normal distribution approximation is less accurate at the extremes. Also, cutting off at age 20 might miss some tall teens! Page 158 Suppose someone claims to have tossed a fair coin 100 times and got 70 heads. Would you believe them? 4 Toss a coin 100 times Repeat 500 times and form a histogram So the distribution of the number of heads in 100 tosses of a fair coin is: 90 80 Frequency 70 60 50 40 30 20 10 0 35 45 55 Number of heads 65 1. What is the mean? 2. What is the standard deviation? 3. Let’s suppose the smooth version is normal. 5