Math 225 Week 5 Test 2020_100% | Math225 Week 5 Test _A Grade

Math 225 Week 5 Test 2020 – Chamberlain College of Nursing 1. An organization has members who possess IQs in the top 4% of the population. If IQs are normally distributed, with a mean of 100 and a standard deviation of 15, what is the minimum IQ required for admission into the organization? Use Excel, and round your answer to the nearest integer: 126 2. The top 5% of applicants on a test will receive a scholarship. If the test scores are normally distributed with a mean of 600 and a standard distribution of 85, how low can an applicant score to still qualify for a scholarship? Use Excel, and round your answer to the nearest integer. 740 -Here, the mean, μ, is 600 and the standard deviation, σ, is 85. Let x be the score on the test. As the top 5% of the applicants will receive a scholarship, the area to the right of x is 5%=0.05. So the area to the left of x is 1−0.05=0.95. Use Excel to find x. -Open Excel. Click on an empty cell. Type =NORM.INV(0.95,600,85) and press ENTER. -The answer rounded to the nearest integer, is x≈740. Thus, an applicant can score a 740 and still be in the top 5% of applicants on a test in order to receive a scholarship. 3. The weights of oranges are normally distributed with a mean of 12.4 pounds and a standard deviation of 3 pounds. Find the minimum value that would be included in the top 5% of orange weights. Use Excel, and round your answer to one decimal place. 17.3 -Here, the mean, μ, is 12.4 and the standard deviation, σ, is 3. Let x be the minimum value that would be included in the top 5% of orange weights. The area to the right of x is 5%=0.05. So, the area to the left of x is 1−0.05=0.95. Use Excel to find x.-1. Open Excel. Click on an empty cell. Type =NORM.INV(0.95,12.4,3) and press ENTER. -The answer, rounded to one decimal place, is x≈17.3. Thus, the minimum value that would be included in the top 5% of orange weights is 17.3 pounds 4. Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean of μ=81 points and a standard deviation of σ=4 points. The middle 50% of the exam scores are between what two values? Use Excel, and round your answers to the nearest integer. 78, 84 The probability to the left of x1 is 0.25. Use Excel to find x1. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.25,81,4) and press ENTER. Rounding to the nearest integer, x1≈78. The probability to the left of x2 is 0.25+0.50=0.75. Use Excel to find x2. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.75,81,4) and press ENTER. Rounding to the nearest integer, x2≈84. Thus, the middle 50% of the exam scores are between 78 and 84. 5. The number of walnuts in a mass-produced bag is modeled by a normal distribution with a mean of 44 and a standard deviation of 5. Find the number of walnuts in a bag that has more walnuts than 80% of the other bags. Use Excel, and round your answer to the nearest integer. 48 Here, the mean, μ, is 44 and the standard deviation, σ, is 5. Let x be the number of walnuts in the bag. The area to the left of x is 80%=0.80. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.80,44,5) and press ENTER. The answer, rounded to the nearest integer, is x≈48. Thus, there are 48 walnuts in a bag that has more walnuts than 80% of the other bags.6. A firm’s marketing manager believes that total sales for next year will follow the normal distribution, with a mean of $3.2 million and a standard deviation of $250,000. Determine the sales level that has only a 3% chance of being exceeded next year. Use Excel, and round your answer to the nearest dollar.$3,670,198 Here, the mean, μ, is 3.2 million =3,200,000 and the standard deviation, σ, is 250,000. Let x be sales for next year. To determine the sales level that has only a 3% chance of being exceeded next year, the area to the right of x is 0.03. So the area to the left of x is 1−0.03=0.97. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.97,,) and press ENTER. The answer, rounded to the nearest dollar, is x≈3,670,198. Thus, the sales level that has only a 3% chance of being exceeded next year is $3,670,198. 7. Suppose that the weight of navel oranges is normally distributed with a mean of μ=6 ounces and a standard deviation of σ=0.8 ounces. Find the weight below that one can find the lightest 90% of all navel oranges. Use Excel, and round your answer to two decimal places. 7.03 Here, the mean, μ, is 6 and the standard deviation, σ, is 0.8. The area to the left of x is 90%=0.90. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.90,6,0.8) and press ENTER. The answer, rounded to two decimal places, is x≈7.03. Thus, navel oranges that weigh less than 7.03 ounces compose the lightest 90% of all navel oranges. 8. A tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 47,500 miles and a standard deviation of 3,000 miles. What mileage would correspond to the the highest 3% of the tires? Use Excel, and round your answer to the nearest integer. 53,142Here, the mean, μ, is 47,500 and the standard deviation, σ, is 3,000. Let x be the minimum number of miles for a tire to be in the top 3%. The area to the right of x is 3%=0.03. So, the area to the left of x is 1−0.03=0.97. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.97,47500,3000) and press ENTER. The answer, rounded to the nearest integer, is x≈53,142. Thus, the approximate number of miles for the highest 3% of the tires is 53,142 miles. 1. The average credit card debt owed by Americans is $6375, with a standard deviation of $1200. Suppose a random sample of 36 Americans is selected. Identify each of the following: 1. 6375 2. 1200 3. n=36 4. 6375 5. 200 2. The heights of all basketball players are normally distributed with a mean of 72 inches and a population standard deviation of 1.5 inches. If a sample of 15 players are selected at random from the population, select the expected mean of the sampling distribution and the standard deviation of the sampling distribution below. σx¯=0.387 inches μx¯=72 inches The standard deviation of the sampling distribution σx¯=σn√=1.51√5=0.387inches. Likewise, when the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=72 inches.1. After collecting the data, Peter finds that the standardized test scores of the students in a school are normally distributed with mean 85 points and standard deviation 3 points. Use the Empirical Rule to find the probability that a randomly selected student's score is greater than 76 points. Provide the final answer as a percent rounded to two decimal places. 99.85% Notice that 76 points is 3 standard deviations less than the mean. Based on the Empirical Rule, approximately 99.7% of the scores are within 3 standard deviations of the mean. Since the normal distribution is symmetric, this implies that 0.15% of the scores are less than the score that is 3 standard deviations below the mean. Alternatively, 99.85% of the scores are greater than the score that is 3 standard deviations below the mean. Therefore, the probability that a randomly selected student's score is greater than 76 points is approximately 99.85%. 2. After collecting the data, Christopher finds that the total snowfall per year in Reamstown is normally distributed with mean 94 inches and standard deviation 14 inches. Which of the following gives the probability that in a randomly selected year, the snowfall was greater than 52 inches? Use the empirical rule Provide the final answer as a percent rounded to two decimal places. 99.85% 3. The College Board conducted research studies to estimate the mean SAT score in 2016 and its standard deviation. The estimated mean was 1020 points out of 1600 possible points, and the estimated standard deviation was 192 points. Assume SAT scores follow a normal distribution. Using the Empirical Rule, about 95% of the scores lie between which two values? 636 to 1404 4. After collecting the data, Kenneth finds that the body weights of the forty students in a class are normally distributed with mean 140 pounds and standard deviation 9 pounds. Use the Empirical Rule to find the probability that a randomly selected student has a body weight of greater than 113 pounds. Provide the final answer as a percent rounded to two decimal places - - -