1. (continued) Suppose the price of a box of Graham crackers, the price of a 6-p

Question

1. (continued) Suppose the price of a box of Graham crackers, the price of a 6-pack of Chocolate bars, and the price of a bag of jumbo Marshmallows (in rubles) (G, C, M) jointly follow Ng(, ) distribution with 10000 3600 1120 =13600 3600 648 1120 648 1600 480 and =| 320 265 k) What is the probability that a box of Graham crackers costs more than a 6-pack of Chocolate bars and a bag of jumbo Marshmallows on a given day? That is, find P(G> C+M). h) Alex has a sudden powerful craving for s'mores. If he only has 3525 rubles, what is the probability that he would have enough money to buy two boxes of Graham crackers, four 6-packs of Chocolate bars, and one bag of jumbo Marshmallows. That is, find P(2G +4C+ M s 3525).

Explanation / Answer

(K)P(G>C+M)=0.84

P(G>C+M)=P(G-C-M>0)

let x=G-C-M , now we use standard normal variate z=(x-mean)/sd

for E(G-C-M)=(480-320-265)=-105

Var(G-C-M)=var(G)+Var(C)+Var(M)-cov(G,C)-cov(G,M)+cov(C,M)=10000+3600+1600-3600-1120+648=11128

for x=-105, z=(-105-0)/sqrt(11128)=-0.9954

P(z>=0.9954)=1-P(z<-0.9954)=1-0.16=0.84

(h) P(2G+4C+M<=3525)=0.003

E(2G+4C+M)=2*480+4*320+265=2505

Var(2G+4C+M)=4Var(G)+16var(C)+var(M)+8cov(G,C)+2cov(G,M)+4cov(C,M)=

=4*10000+16*3600+1600+8*3600+2*1120+4*648=132832

P(2G+4C+M<=3525)=P(2G+4C+M-3525<=0)

for x=-1020, z=(-1020-0)/sqrt(132832)=-2.7986

P(z<=-2.7986)=0.003

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