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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 late(a) arrival times are average whollyy distributed with close of 12.5 transactions 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 mesa for the result. B. What plowshare of planes were slight than 15 minutes late? effect: 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 food grain company has been adjusted so that an middling of 13.0 oz of cereal is move in each package. Of course, not all packages have on the nose 13.0 oz because of random sources of variability. The standard deviation of the existent net ease weight is 0.

1 oz, and the distribution of weights is known to follow the noun phrase hazard distribution. A. Determine the opportunity that a randomly elect package will contain between 10.0 and 13.2 oz of cereal and set forth the counterpoise of atrial auricle under th e normal writhe 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 terminate 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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