410Hw05ans.pdf - STAT 410 Homework#5(due Friday March 29 by 4:00 p.m Spring 2019 A Stepanov Please include your name with your last name underlined your

# 410Hw05ans.pdf - STAT 410 Homework#5(due Friday March 29 by...

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STAT 410 Homework #5 (due Friday, March 29, by 4:00 p.m.) Spring 2019 A. Stepanov Please include your name ( with your last name underlined ) , your NetID, and your section number at the top of the first page. No credit will be given without supporting work. 1. Every month, the government of Neverland spends G million dollars purchasing guns, B million dollars purchasing butter, and P million dollars purchasing pants. Assume that ( G , B , P ) jointly follow a N 3 ( , ) , a 3- dimensional multivariate normal distribution with μ = 315 175 151 , = 1600 200 136 200 625 136 136 136 400 . a) Find the probability that the government of Neverland spends more on guns than twice the amount it spends on butter during a given month. That is, find P ( G > 2 B ) . Want P ( G > 2 B ) = P ( G – 2 B > 0 ) = ? G – 2 B has Normal distribution, E ( G – 2 B ) = G – 2 B = 315 – 2 175 = 35 , Var ( G – 2 B ) = 1600 200 136 1 1 2 0 200 625 136 2 136 136 400 0           = 1 2000 1450 136 2 0 = 4900 . P ( G – 2 B > 0 ) = P ( Z > 0 35 4900   ) = P ( Z > 0.50 ) = 0.3085 .
b) Find the probability that the government of Neverland spends more on guns than it spends on butter and pants together during a given month. That is, find P ( G > B + P ) . Want P ( G > B + P ) = P ( G – B – P > 0 ) = ? G – B – P has Normal distribution, E ( G – B – P ) = G B P = 315 – 175 – 151 = 11 , Var ( G – B – P ) = 1600 200 136 1 1 1 1 200 625 136 1 136 136 400 1           = 1 1936 689 400 1 1 = 3025 . P ( G – B – P > 0 ) = P ( Z > 0 11 3025   ) = P ( Z > 0.20 ) = 0.4207 . c) Find the probability that the government of Neverland exceeds the \$600 million spending limit during a given month. That is, find P ( G + B + P > 600 ) . G + B + P has Normal distribution, E ( G + B + P ) = G + B + P = 315 + 175 + 151 = 641 , Var ( G + B + P ) = 1600 200 136 1 1 1 1 200 625 136 1 136 136 400 1           = 1 1264 289 128 1 1 = 1681 . P ( G + B + P > 600 ) = P ( Z > 600 641 1681 ) = P ( Z > – 1.00 ) = 0.8413 .
2 – 5. The Weibull distribution has many applications in reliability engineering, survival analysis, and general insurance. Let > 0, > 0. Consider the probability density function δ 1 β δ , δ δ ; β β x e x x f , x > 0, zero otherwise.

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