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Unformatted text preview: STAT 410 Spring 2008 Homework #6 (due Thursday, February 28, by 5:00 p.m.) 1. One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value 20 in. and standard deviation 0.7 in. The length of the second piece is a normal random variable with mean and standard deviation 15 in. and 0.6 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation 0.2 in. Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between 32.65 in. and 35.35 in.? Total = First + Second – Overlap. E( Total) = E(First) + E(Second) – E(Overlap) = 20 + 15 – 1 = 34 . Var( Total) = Var(First) + Var(Second) + (– 1 ) 2 Var(Overlap) = 0.7 2 + 0.6 2 + 0.2 2 = 0.49 + 0.36 + 0.04 = 0.89 . SD( Total) = 89 . = 0.9434 . Total has Normal distribution. P( 32.65 < Total < 35.35 ) = & ¡ ¢ £ ¤ ¥ < < 9434 . 34 35 . 35 Z 9434 . 34 65 . 32 P = P( – 1.43 < Z < 1.43 ) = 0.9236 – 0.0764 = 0.8472 . 2. a) At Anytown College, the heights of female students are normally distributed with mean 66 inches and standard deviation 1.5 inches. The heights of male students are also normally distributed with mean 69 inches and standard deviation 2 inches. For Homecoming, a male student and a female student are selected independently at random to be the King and the Queen. What is the probability that the female student selected to be the Queen is taller than the male student selected to be the King? student selected to be the Queen is taller than the male student selected to be the King?...
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This homework help was uploaded on 04/13/2008 for the course STAT 410 taught by Professor Alexeistepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 AlexeiStepanov
 Probability, Standard Deviation

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