Question

Question 9 Let X and Y be the number of hours that a randomly selected person watches movies and sporting events, respectively, during a three-month period. The following information is known about X and Y: E(X) = 50 E(Y) = 20 Var(X) = 50 Var(Y) = 30 Cov(X,Y) = 10 One hundred people are randomly selected and observed for these three months. Let T be the total number of hours that these one hundred people watch movies or sporting events during this three-month period. Approximate the value of Pr(T < 7100). Answer: 0.8413

Answer 1

here expected number of hours spend by a person in watching movies or sporting events

=P(A) =E(X+Y) =50+20 =70

and standard deviation sqrt(Var(X+Y))=sqrt(Var(X)+Var(Y)+2*Cov(X,Y))=sqrt(50+30+2*10) =10

for 100 people, expected total time E(T ) =100*70 =7000

and standard deviation =10*sqrt(100) =100

therefore from Normal Approximation:

P(T<7100) =P(Z<(7100-7000)/100) =P(Z<1) =0.8413