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STAT 500 Homework 5 Solutions Problem 1 A major airline claims that the majority of its schedule relief valves at the Atlanta drome are within 15 proceeding of scheduled arrival. If the flight ripe arrival times are norm whollyy distributed with close of 12.5 minutes with a standard deviation of 5.1 minutes: A. What division of planes where between 12.5 and 19.6 minutes late? wait on: P(12.5 ? X ? 19.6) = 0.4177, z = (19.6 -12.5)/5.1 = 1.39, check the Standard Normal flurry for the result. B. What plowshare of planes were slight than 15 minutes late? termination: Since z = (15 - 12.5)/5.1 = 0.49, P(X < 15) = P(Z < 0.49) = 0.6879 C. What percentage of planes were more than 30 minutes late? Answer: Since z = (30 - 12.5)/5.1 = 3.43, P(X > 30) = P(Z > 3.43) = 0.0003, from Standard Normal table D. 75% of the planes were less than how many minutes late? Answer: Since z = 0.675 = (X - 12.5)/5.1, X - 12.5 = 0.675(5.1), so X = 1 2.5 + 0.675(5.1), X = 15.94 minutes. Problem 2 The packaging process in a breakfast caryopsis company has been adjusted so that an middling of 13.0 oz of cereal is set(p) in each package. Of course, not all packages have precisely 13.0 oz because of random sources of variability. The standard deviation of the existent net absolve weight is 0.1 oz, and the distribution of weights is known to follow the nominal hazard distribution. A. Determine the fortune that a randomly chosen package will contain between 10.0 and 13.2 oz of cereal and work out the equipoise of atrial auricle under the normal stoop which is associated with this probability value. Answer: Since z = (13.2 13.0)/0.1 = 2.0, P(13.0 < X < 13.2) = P(0 < Z < 2.0) = 0.4772 B. What is the probability that the weight of the cereal will exceed 13.25 oz? Illustrate the proportion of ear under the normal curve which is relevant give notice this case Answer: Since z = (13.25 1 3.0)/0.1 = 2.5, P(X > 13.25) = P(Z > 2! .5) = 0.0062 C. What is the...If you want to get a full essay, browse it on our website: OrderCustomPaper.com

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