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Jan 21 Statistic for the day: The width of train tracks is 4 feet 8.5 inches. Why? Assignment: Read Chapter 9 Exercises from Chapter 8: 16, 18 These slides were created by Tom Hettmansperger and in some cases modified by David Hunter Research Question 1: How high should I build my doorways so that 99% of the people will not have to duck? Secondary Question 2: If I built my doors 75 inches (6 feet 3 inches) high, what percent of the people would have to duck? Histogram of Height, with Normal Curve Frequency 30 20 10 0 60 70 80 Height Question 2 Question 1 Find the value at Question 1 so that 99% of the distribution is below it. The value at Question 2 is 75; find the amount of distribution above it. Z-Scores: Measurement in Standard Deviations Given the mean (68), the standard deviation (4), and a value (height say 75) compute Z = (75-mean) / SD = (75-68) / 4 = 1.75 This says that 75 is 1.75 standard deviations above the mean. Answer to Question 2: What percent of people would have to duck if I built my doors 75 inches high? Recall: 75 has a Z-score of 1.75 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. So 4% of the distribution is above 75 inches. Histogram of Height, with Normal Curve Frequency 30 20 4% in here 96% in here (FIND IN TABLE) 10 0 60 70 Height 75 (GIVEN) 80 Question 2 The value at Question 2 is 75; find the amount of distribution above it. Convert 75 to Z = 1.75 and use Table 8.1 in book. Question 1: What is the value so that 99% of the distribution is below it? Called the 99th percentile. 1. Look up the Z-score that corresponds to the 99th percentile. From the table: Z = 2.33. 2. Now convert it over to inches: 2.33 = (h – 68)/4 h = 68 + 2.33(4) = 77.3 Since 77 inches is 6 feet 5 inches, 99% of the distribution is shorter than 77 inches and they will not have to duck. Histogram of Height, with Normal Curve Frequency 30 20 99% in here (GIVEN) 10 0 60 70 80 Height (FIND): Question 1 77.3 inches is the 99th percentile Find the value at Question 1 so that 99% of the distribution is below it. Look up Z-score for 99th percentile and convert it back to inches. Compare Heights of Females and Males Stat 100 students Sp01 Height 80 70 60 Female Male Sex Heights in Inches ( red circle is my doorway 77 inches) 85 78 75 (6-3) 65 Lakers n=14 PSU BB n=12 Steelers n=30 Stat 100 Males n=78 Shaquille O’Neal is 7 feet 1 inch or 85 inches tall. How many people in the country are taller? 1. We will assume that heights are normally distributed with mean 68 inches and standard deviation 4 inches. 2. 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! We would generally consider him from a different population. 3. There is .000011 above 4.25 standard deviations. 3. There are roughly 250 million people in US. 48.8% are over the age of 20. That is 122 million. 4. Hence, there should be roughly .000011 times 122 million or 1342 people taller than Shaquille O’Neal Who Is the Tallest Person in Class? What is your Z-Score? What is your percentile? How many PSU students are taller? Suppose someone claims to have tossed a coin 100 times and got 70 heads. Would you believe them? We need to know what the distribution of the number of heads in 100 tosses looks like for a fair coin. We need the mean and standard deviation for this distribution. Toss a coin 100 times Repeat 500 times and form a histogram 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 bell shaped. So the distribution of the number of heads in 100 tosses of a fair coin is: Roughly normal, mean about 50, SD about 5 What is the Z-score of 70? Ans: 4 What is the percentile? Ans: .999968 or 99.9968% Now do you believe them? NO Weighted coin is a BETTER explanation