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Part 1: One more trip to the Rock and Roll Hall of Fame In 2005, the Irish band U2 was inducted into the Rock and

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Part 1: One more trip to the Rock and Roll Hall of Fame In 2005, the Irish band U2 was inducted into the Rock and Roll Hall of Fame. When we look at the distribution of song lengths, in seconds, across all of U2's albums, we see a distribution that is approximately Normal in shape, with a mean of 263.2 seconds and a standard deviation of 65.1 seconds. Please use this information to answer Questions I through 10. Hint: A very useful first step in these types of problems involves drawing the distribution and marking it out three standard deviations on either side of the mean. 1. The Empirical Rule for the 68-95-99.7 Rule) tells us that approximately 68% of U2's songs are between seconds and seconds long- 2. The Empirical Rule (or the 68-95-99.7 Rule) tells us that approximately 95% of U2's songs are between seconds and seconds lang. 3. The Empirical Rule for the 68-95-99.7 Rule) tells us that approximately 99.7% of UT's songs are between seconds and seconds long. 4. The song "Beautiful Day" is 246 seconds long. What percentage of U2 songs are shorter than for less than) 246 seconds? 5. The song "With or Without You" is 296 seconds long. This means "With or Without You" is at the percentile.6. The song "Where the Streets Have no Name" is 337 seconds long. What percentage of U2 songs are longer than (or greater than) 137 seconds? 7. The song "Zooropa," which is 390 seconds long, is at the percentile. 8. What percentage of U2 songs are between 246 and 137 seconds long? 9. A U2 song that is at Q, in the distribution would be approximately seconds long- 10. The U2 song that cuts off the top (or upper) 3% of the distribution is approximately seconds long- 3Part 2: Even more practice! I I. The number of dill pickle spears in a gallon container is Normally distributed, with a standard deviation of 2.5 spears. Suppose you learn that a gallon container with 120 spears is at the N8"percentile in this distribution. From this information, we know the mean of the distribution would therefore need to be equal to approximately what value? For each of the following multiple-choice questions, please select what you believe to be the best answer, There is no need to explain the reason for your answer choices. 12. The number of mistakes made on a particular typing test is Normally distributed. Five office employees- Stanley, Meredith, Kelly, Oscar, and Toby each take this typing best, and the number of mistakes made by each of these employees is converted to a Escore. The E-scores are presented in the table below. Based on this information, who made the least mistakes on the typing test? Employee Stanley Meredith Kelly Oscar Toby B-score 21 A. Stanley B. Meredith C. Kelly D. Oscar E. Toby 13. In a Normal distribution, approximately how many standard deviations above the mean is Q.T A. 17 B. 3.0 C. 0.7 D. 0.3 E. It's impossible to answer this question without knowing the mean and standard deviation of the distribution.14. It turns out that the average retirement age of National Football League (or NFL) players is Normally distributed, with a mean of 33 years and a standard deviation of 2 years. Eddie George is a former NFL player who retired at the age of 30 years. When asked to determine the percentage of NFL players who retire at an age younger than 30 years, Justin did the following. He first converted 30 to a z-score using the following formula: 2 13-10 -- 15 Next, Justin looked up the z-score of 1.5 in Table B and found that this z-score corresponded with a percentile of 93.32. He then reported that "93.32% of NFL players retire at an age younger than 30 years." What is wrong with the way Justin solved this problem? A. Nothing is wrong. B. Justin should have multiplied 1.5 by 100%% to get a final answer of 15%. C. Justin should have reported his final answer as 1.5%. D. Justin should have subtracted I from 1.5 to get a z-score of 0.5, and he then should have reported the final answer as 69.15% E. Justin should have obtained a negative z-score and a final answer of 6.68%. 15. A distribution of exam scores is Normal, with a mean of 79 points and a standard deviation of 5 points. Tiffany is told that her exam score is at the 50 percentile, Ricky is told that his exam score is equal to the mean exam score, and Bianca is told that her exam score is equal to the median exam score. Who has the higher exam score! A. Tiffany B. Ricky C. Bianca D. Tiffany, Ricky, and Bianca all scored exactly the same on the exam. E. More information is needed in order to be able to answer this question. 5IMPORTANT: If any problem requires a calculation to answer, please attempt to write out or type out how you arrived at your answer so we can see your thought process. We cannot give you full credit if no work is shown. A copy of Table B will be necessary to complete this assignment, and you can find a copy among the Weeks 6 and 7 materials on Carmen (and in the back for your textbook).Area represents percentile Standard score -1.800 2000 -1/000 1.000 2000 3000 Table B Percentiles of the normal distributions Standard Percentile Standard Score (2) Percentile Standard Score (z) Percentile -3.4 Score (z) 0.03 -1.1 13.57 -3.3 1.2 0.05 88.49 -1.0 15.87 -3.2 1.3 0.07 90.32 0.9 18.41 -3.1 1.4 0.10 91.92 0.8 21.19 3.0 0.13 1.5 93.32 -0.7 24.20 -2.9 1.6 0.19 94.52 -0.6 27.42 -2.8 1.7 0.26 95.54 0.5 30.85 2.7 1.8 0.35 96.41 -0.4 34.46 -2.6 1.9 D.47 97.13 -0.3 38.21 -2.5 2.0 0.62 97.73 -0.2 42.07 -2.4 2.1 0.82 98.21 -0.1 46.02 -2.3 2.2 1.07 98.61 0.0 50.00 -2.2 2.3 1.39 98.93 D.1 53.98 -2.1 2.4 1.79 99.18 0.2 57.93 -2.0 2.5 2.27 99.38 0.3 61.79 -1.9 2.6 2.87 99.53 0.4 65.54 -1.8 2.7 3.59 0.5 99.65 69.15 -1.7 2.8 4.46 99.74 0.6 72.58 2.9 -1.6 5.48 99.81 0.7 75.80 -1.5 3.0 6.68 99.87 0.8 78.81 -1.4 3.1 8.08 99.90 0.9 81.59 -1.3 3.2 9.68 99.93 1.0 84.13 3.3 -1.2 11.51 99.95 1.1 86.43 3.4 99.97

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