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Sept. 2 Reading Assignment: Finish Chapter 8; skim through Chapter 7 again Try exercises 15 and 27 on pages 159–161 20 What is the best estimate below of the s.d. for these data? 10 5 0 Frequency 15 (A) 10 (B) 20 (C) 30 (D) 40 (E) 50 40 60 80 100 120 140 How much does this cow weigh? (The man in the picture weighs 165 pounds.) Histogram of Height, with Normal Curve Frequency 30 20 10 0 60 70 Height Mean = 68 inches or 5 feet 8 inches Standard deviation = 4 inches 80 Assume human heights are normally distributed with a mean of µ=68 inches and a standard deviation of σ=4 inches. Research Question 1: If I built my doors 75 inches (6 feet 3 inches) high, what percent of the people would have to duck? 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 Given the mean (68 inches), the standard deviation (4 inches), and a value (75 inches) compute 75 − mean 75 − 68 Z= = = 1.75 SD 4 This says that 75 is 1.75 standard deviations above the mean. Z-Scores: Measurement in Standard Deviations Given the mean (68 inches), the standard deviation (4 inches), and a value (60 inches) compute Z = ?? (A) Z = –2.0 (B) Z = –1.0 (C) Z = 0.0 (D) Z = 1.0 (E) Z = 2.0 Height=75 inches gives z-score=+1.75 What percent? -3 -2 -1 0 1 2 3 Use table on p. 175 to find percent to the left of Z=+1.75. Here is what p. 157 tells us: Should be p. 175 Z=+1.75 96% -3 -2 -1 0 1 2 3 What percent of people would have to duck if I built my doors 75 inches high? Ans: 4% Question 2: 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 to inches: h99 − 68 2.33 = 4 h99 = 68 + 2.33(4) = 77.3 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. What Z? 99% p. 175 tells us that the Z-score at the 99th percentile is 2.33. In inches, 2.33 standard deviations above the mean is 77.3. 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. What is your Z-Score within your sex? What is your percentile within your sex? 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. There are roughly 317 million people in US. About 49% are over the age of 20 (Census Bureau). That is about 155 million. Hence, there should be roughly .000011 times 155 million or 1705 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! Table 8.1 practice (page 175) Match the percentages of a normal distribution with the following standardized (z-) scores: 1. below z = −1.00 2. below z = 1.96 3. above z = 0.84 4. above z = 0 5. below z = –0.50 (A) 20% (B) 97.5% (C) 50% (D) 16% (E) 31% Table 8.1 practice (page 175) Match the percentages of a normal distribution with the following standardized (z-) scores: 1. below z = −1.00 2. below z = 1.96 3. above z = 0.84 4. above z = 0 5. below z = –0.50 (A) 20% (B) 97.5% (C) 50% (D) 16% (E) 31% Table 8.1 practice (page 175) Match the percentages of a normal distribution with the following standardized (z-) scores: 1. below z = −1.00 2. below z = 1.96 3. above z = 0.84 4. above z = 0 5. below z = –0.50 (A) 20% (B) 97.5% (C) 50% (D) 16% (E) 31% Table 8.1 practice (page 175) Match the percentages of a normal distribution with the following standardized (z-) scores: 1. below z = −1.00 2. below z = 1.96 3. above z = 0.84 4. above z = 0 5. below z = –0.50 (A) 20% (B) 97.5% (C) 50% (D) 16% (E) 31% Table 8.1 practice (page 175) Match the percentages of a normal distribution with the following standardized (z-) scores: 1. below z = −1.00 2. below z = 1.96 3. above z = 0.84 4. above z = 0 5. below z = –0.50 (A) 20% (B) 97.5% (C) 50% (D) 16% (E) 31% Table 8.1 practice (page 175) Match the percentages of a normal distribution with the following standardized (z-) scores: 1. below z = −1.00 2. below z = 1.96 3. above z = 0.84 4. above z = 0 5. below z = –0.50 (A) 20% (B) 97.5% (C) 50% (D) 16% (E) 31% Table 8.1 practice (page 175) Match the standardized (z-)scores with the following percentiles of a standard normal distribution: 1. 25th %ile 2. 75th %ile 3. 42nd %ile 4. 98th %ile 5. 10th %ile (A) –0.67 (B) 0.67 (C) 2.05 (D) 0.20 (E) –1.28 Table 8.1 practice (page 175) Match the standardized (z-)scores with the following percentiles of a standard normal distribution: 1. 25th %ile 2. 75th %ile 3. 42nd %ile 4. 98th %ile 5. 10th %ile (A) –0.67 (B) 0.67 (C) 2.05 (D) 0.20 (E) –1.28 Table 8.1 practice (page 175) Match the standardized (z-)scores with the following percentiles of a standard normal distribution: 1. 25th %ile 2. 75th %ile 3. 42nd %ile 4. 98th %ile 5. 10th %ile (A) –0.67 (B) 0.67 (C) 2.05 (D) 0.20 (E) –1.28 Table 8.1 practice (page 175) Match the standardized (z-)scores with the following percentiles of a standard normal distribution: 1. 25th %ile 2. 75th %ile 3. 42nd %ile 4. 98th %ile 5. 10th %ile (A) –0.67 (B) 0.67 (C) 2.05 (D) 0.20 (E) –1.28 Table 8.1 practice (page 175) Match the standardized (z-)scores with the following percentiles of a standard normal distribution: 1. 25th %ile 2. 75th %ile 3. 42nd %ile 4. 98th %ile 5. 10th %ile (A) –0.67 (B) 0.67 (C) 2.05 (D) 0.20 (E) –1.28 Table 8.1 practice (page 175) Match the standardized (z-)scores with the following percentiles of a standard normal distribution: 1. 25th %ile 2. 75th %ile 3. 42nd %ile 4. 98th %ile 5. 10th %ile (A) –0.67 (B) 0.67 (C) 2.05 (D) 0.20 (E) –1.28