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Unformatted text preview: AMS 310.01 FALL 2003 Homework #7 Solutions
5.38 In this case, n=84, p=0.3, = 84 0.3 = 25.2 , 2 = 84 0.3 0.7 = 17.64 , =4.2 P (20 X 30) = P ( X 30)  P ( X < 20) = 30.5 19.5  25.2  25.2 F  = F F (1.26)  F (1.36) 4.2 4.2 = 0.8962  0.0869 = 0.8093 5.46 This is the Uniform distribution with  = 0.025  (0.025) = 0.05 (a) P (0.010 error 0.015) = (0.015  0.010) / 0.050 = 0.1 (b) P (0.012 error 0.012) = (0.012 + 0.012) / 0.050 = 0.48 5.58 The exponential density is
1 1  x e f ( x) = 0 x > 0, > 0 elsewhere and the exponential distribution is F ( x) = f ( s )ds = 1  e  x / 0 x for x > 0 20 / 50 = 1  e 2 / 5 = 0.3297 (a) P ( X < 20) = 1  e 60 / 50 ) = e 6 / 5 = 0.3012 (b) P ( X 60) = 1  P ( X < 60) = 1  (1  e 5.60 Assuming that the time between calls is distributed as an exponential random variable 6 6 1 6 10 t e Therefore F (t ) = 1  e 10 t , thus, f (t ) = 10 P (T < 0.5) = 1  e  (0.6)(0.5) = 1  e 0.3 = 0.2592 (a)  (0.6)(3) ) = e 1.8 = 0.1653 (b) P (T > 3) = 1  P (T 3) = 1  (1  e with = 0.6 = ...
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This homework help was uploaded on 01/31/2008 for the course AMS 310.01 taught by Professor Mendell during the Fall '03 term at SUNY Stony Brook.
 Fall '03
 Mendell

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