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Unformatted text preview: 36225  Homework 4 SOLUTIONS1.4 points: 2 for E(Y), 2 for V(Y)The random variable of interest isY, the number of successful explorations inn= 10 explorations. ThenYhas a binomial distribution withp=.1. HenceE(Y) =np= 10(.1) = 1 andV(Y) =np(1p) = 10(.1)(.9) =.92.4 points: 2 for realizing that it is Bin(32,p=.9), 2 for probability calculationLetYbe the number of passengers who arrive for the flight.⇒Y∼Bin(n= 32,p=.9).P(more people show up than seats)=P(Y= 31) +P(Y= 32) =(3231)(.9)31(.1)1+(3232)(.9)32(.1)= 32×.931×.1 +.932=.1564.3.6 points: 2 for setup, 3 for work, 1 for final answerE(4Y) =5Xy=04y·PY(y) =5Xy=04y5y!13y235y=5Xy=05y!43y235yBy hintz}{=43+235= 25= 32.4.8 points: 4 for each part(a)P(Y > a) =P(Y≥a+ 1) =∞Xy=a+1(1p)y1p=p1p∞Xy=a+1(1p)ygeometric seriesz}{=p1p·(1p)a+11(1p)= (1p)a[ =qa](b)P(Y > a+bY > a) =P({Y > a+b} ∩ {Y > a})P(Y > a)=P(Y > a+b)P(Y > a)part (a)z}{=(1p)a+b(1p)a=(1p)b=P(Y > b)5.8 points: (a) 2 points, (b) 3 points (c) 3 pointsLet Y be the number of calls until first “prochoice” person.=⇒Y∼Geometric(p=.62).(a)P(Y= 5) = (1.62)(51)·.62 =.013(b)P(Y >2) = (problem4(a)) = (1.62)2=.1444(c)P(Y≥8Y >5) =P(Y >7Y >5) = (memoryless) =P(Y >2) = (1.62)2=.144416.12 points: (a) 3 points [1 for NB, 1 for r 1 for p] (b) 3 points (c) 6 points [2 forX=Y6, 2 for E(Y6), 2 for V(Y6)]LetYbe the number of people that the telemarketer needs to contact until 6 subscriptionsare sold.(a)⇒Y∼NB(r= 6, p= 1/3)⇒PY(y) =y161!13623y6y= 6,7,8,...(b)P(Y≥8) = 1P(Y≤7) = 1[P(Y= 6) +P(Y= 7)]= 1" 6161!13623+715!136231#= 1.007 =.993(c) LetXbe the be the number of people who decline before the quota is reached....
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This note was uploaded on 07/15/2011 for the course STAT 36225 taught by Professor Tom during the Summer '09 term at Carnegie Mellon.
 Summer '09
 TOM
 Binomial

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