Solutions05 - ENGRD 2700, Spring 09 Homework 5 Solutions...

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ENGRD 2700, Spring ’09 Homework 5 Solutions Homework 5 Solutions Problem 1 (Devore 4.36) Let X be the size of a droplet squeezed through the nozzle. We are told that X is a normal random variable with μ = 1050 = 150. (a) We have P ( X < 1500) = P ± X - 1050 150 < 1500 - 1050 150 = P ( Z < 3) = Φ(3) = 0 . 9987 . Similarly, P ( X 1000) = P ± X - 1050 150 1000 - 1050 150 = P ( Z ≥ - 0 . 33) = 1 - Φ( - 0 . 33) = 0 . 6293 . (b) P (1000 < X < 1500) = P ( X < 1500) - P ( X < 1000) = 0 . 9987 - 0 . 3707 = 0 . 6280. (c) We are looking for the 2 nd percentile of a normal (1050, 150) distribution. From Table A.3 in the textbook, we see that the 2 nd percentile of a standard normal distribution is about - 2 . 05. By the boxed proposition on p. 151, our answer is 1050 + ( - 2 . 05) · 150 = 742 . 5 , i.e. the smallest 2% of all droplets have a size of 742.5 μ m or less. (d) Since each of the five droplets has probability 0.9987 of being less than 1500 μ m in size, P (at least one exceeds 1500 μ m) = 1 - P (none exceeds 1500 μ m) = 1 - (0 . 9987) 5 = 0 . 0065 . Problem 2 (Devore 4.38) Let X 1 be the diameter of a cork produced by the first machine, and let X 2 be the diameter of a cork produced by the second machine. We have 1
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ENGRD 2700, Spring ’09 Homework 5 Solutions P (2 . 9 < X 1 < 3 . 1) = P ± 2 . 9 - 3 0 . 1 < X 1 - 3 0 . 1 < 3 . 1 - 3 0 . 1 = P ( - 1 < Z < 1) = Φ(1) - Φ( - 1) = 0 . 6826 and P (2 . 9 < X 2 < 3 . 1) = P ± 2 . 9 - 3 . 04 0 . 02 < X 2 - 3 . 04 0 . 02 < 3 . 1 - 3 . 04
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This note was uploaded on 04/05/2009 for the course ENGRD 2700 taught by Professor Staff during the Spring '05 term at Cornell University (Engineering School).

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Solutions05 - ENGRD 2700, Spring 09 Homework 5 Solutions...

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