STAT 230 Final with Solution, Summer 2010

Final solution, Summer 2010 Exercise Lori just bought a new set of 4 tires for her car. The life of each tire is normally distributed with a mean of 45000 miles and a standard deviation of 3200 miles. Find the probability that all 4 tires will last at least 46000 miles. Assume the life of each of these tires is independent of the lives of other tires. Final solution, Summer 2010 Exercise Lori just bought a new set of 4 tires for her car. The life of each tire is normally distributed with a mean of 45000 miles and a standard deviation of 3200 miles. Find the probability that all 4 tires will last at least 46000 miles. Assume the life of each of these tires is independent of the lives of other tires. Solution The probability that one tire lives more than 46000 miles is 0.377, and the probability that all 4 tires live more than 46000 each is (0.377)4 = 0.02 . Exercise Suppose that in a community the distributions of heights of men and women (in centimeters) are N(173, 40) and N(160, 20), respectively. Calculate the probability that the average height of 10 randomly selected men is at least 5 centimeters larger than the average height of six randomly selected women. Exercise Suppose that in a community the distributions of heights of men and women (in centimeters) are N(173, 40) and N(160, 20), respectively. Calculate the probability that the average height of 10 randomly selected men is at least 5 centimeters larger than the average height of six randomly selected women. Solution Let X1, . . . , X10 be the heights of the ten men, and Y1, . . . , Y6 be the heights of the six women. Exercise Suppose that in a community the distributions of heights of men and women (in centimeters) are N(173, 40) and N(160, 20), respectively. Calculate the probability that the average height of 10 randomly selected men is at least 5 centimeters larger than the average height of six randomly selected women. Solution Let X1, . . . , X10 be the heights of the ten men, and Y1, . . . , Y6 be the heights of the six women. we want to nd P(X − Y > 5) Exercise Suppose that in a community the distributions of heights of men and women (in centimeters) are N(173, 40) and N(160, 20), respectively. Calculate the probability that the average height of 10 randomly selected men is at least 5 centimeters larger than the average height of six randomly selected women. Solution Let X1, . . . , X10 be the heights of the ten men, and Y1, . . . , Y6 be the heights of the six women. we want to nd P(X − Y > 5) X has a normal distribution with mean 173 and variance 4 and Y has a normal distribution with mean 160 and variance 10/3, and then X − Y has a normal distribution with mean 13 and variance 22/3 Exercise Suppose that in a community the distributions of heights of men and women (in centimeters) are N(173, 40) and N(160, 20), respectively. Calculate the probability that the average height of 10 randomly selected men is at least 5 centimeters larger than the average height of six randomly selected women. Solution Let X1, . . . , X10 be the heights of the ten men, and Y1, . . . , Y6 be the heights of the six women. we want to nd P(X − Y > 5) X has a normal distribution with mean 173 and variance 4 and Y has a normal distribution with mean 160 and variance 10/3, and then X − Y has a normal distribution with mean 13 and variance 22/3 P(X − Y > 5) = 0.86 (from the standard normal table) Exercise Otto is trying out for the javelin throw to compete in the olympics. The lengths of his javelin throws is normally distributed with a mean of 290 feet and a standard deviation of 10 feet. Find the probability that the longest of three of his throws is 320 feet or more. Exercise Otto is trying out for the javelin throw to compete in the olympics. The lengths of his javelin throws is normally distributed with a mean of 290 feet and a standard deviation of 10 feet. Find the probability that the longest of three of his throws is 320 feet or more. Solution Let X1, X2, X3 be the lengths of his three throws. Exercise Otto is trying out for the javelin throw to compete in the olympics. The lengths of his javelin throws is normally distributed with a mean of 290 feet and a standard deviation of 10 feet. Find the probability that the longest of three of his throws is 320 feet or more. Solution Let X1, X2, X3 be the lengths of his three throws. P(max(X1, X2, X3) > 320) = 1 − (P(X1 < 320))3 Exercise Otto is trying out for the javelin throw to compete in the olympics. The lengths of his javelin throws is normally distributed with a mean of 290 feet and a standard deviation of 10 feet. Find the probability that the longest of three of his throws is 320 feet or more. Solution Let X1, X2, X3 be the lengths of his three throws. P(max(X1, X2, X3) > 320) = 1 − (P(X1 < 320))3 = 1 − (P(Z < 3))3 = 1 − (0.998)3 = 0.004 . Exercise he time it takes for a student to nish an aptitude test (in hours) has the pdf f (x) = 6(x − 1)(2 − x) , 1 < x < 2 Approximate the probability that the average length of time it takes for a random sample of 15 students to complete the test is less than 1 hour and 25 minutes. Exercise he time it takes for a student to nish an aptitude test (in hours) has the pdf f (x) = 6(x − 1)(2 − x) , 1 < x < 2 Approximate the probability that the average length of time it takes for a random sample of 15 students to complete the test is less than 1 hour and 25 minutes. Solution Let X1, . . . , X15 be the exam time of the 15 students. Exercise he time it takes for a student to nish an aptitude test (in hours) has the pdf f (x) = 6(x − 1)(2 − x) , 1 < x < 2 Approximate the probability that the average length of time it takes for a random sample of 15 students to complete the test is less than 1 hour and 25 minutes. Solution Let X1, . . . , X15 be the exam time of the 15 students. we want to nd P(X < 1.42) (time in hours!!) Exercise he time it takes for a student to nish an aptitude test (in hours) has the pdf f (x) = 6(x − 1)(2 − x) , 1 < x < 2 Approximate the probability that the average length of time it takes for a random sample of 15 students to complete the test is less than 1 hour and 25 minutes. Solution Let X1, . . . , X15 be the exam time of the 15 students. we want to nd P(X < 1.42) (time in hours!!) E(X1) = 3/2 and Var(X1) = 1/20 Exercise he time it takes for a student to nish an aptitude test (in hours) has the pdf f (x) = 6(x − 1)(2 − x) , 1 < x < 2 Approximate the probability that the average length of time it takes for a random sample of 15 students to complete the test is less than 1 hour and 25 minutes. Solution Let X1, . . . , X15 be the exam time of the 15 students. we want to nd P(X < 1.42) (time in hours!!) E(X1) = 3/2 and Var(X1) = 1/20 by the central limit theorem, X−3/2 1/ √ 300 has an approximate standard normal distribution Exercise he time it takes for a student to nish an aptitude test (in hours) has the pdf f (x) = 6(x − 1)(2 − x) , 1 < x < 2 Approximate the probability that the average length of time it takes for a random sample of 15 students to complete the test is less than 1 hour and 25 minutes. Solution Let X1, . . . , X15 be the exam time of the 15 students. we want to nd P(X < 1.42) (time in hours!!) E(X1) = 3/2 and Var(X1) = 1/20 by the central limit theorem, X−3/2 1/ √ 300 has an approximate standard normal distribution P(X < 1.42) ≃ P(Z < −1.38) = 0.083