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STAT 500 Homework 5 Solutions Problem 1 A major airline claims that the majority of its plan relief valves at the Atlanta drome are within 15 proceeding of scheduled arrival. If the flight juvenile arrival times are norm whollyy distributed with close of 12.5 proceeding 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 confuse for the result. B. What plowshare of planes were slight than 15 minutes late? help: 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 = 12.5 + 0.675(5.1), X = 15.94 minutes. Problem 2 The packaging process in a breakfast texture company has been adjusted so that an middling of 13.0 oz of cereal is determined in each package. Of course, not all packages have on the dot 13.0 oz because of random sources of variability. The standard deviation of the existent net sluttish weight is 0.

1 oz, and the distribution of weights is known to follow the nominative hazard distribution. A. Determine the probability that a randomly chosen package will contain between 10.0 and 13.2 oz of cereal and blow up the likeness of atrial auricle under the normal kink up which is associated with th! is 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 lav this case Answer: Since z = (13.25 13.0)/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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