MATH 225-week 5

MATH 225-week 5 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. 53,142 Here, 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. Id