Let X and Y be the number of hours that a randomly selected person watches home

Question

Let X and Y be the number of hours that a randomly selected person watches home shopping and football, respectively, during a three-month period. The following information is known about X and Y:

E[X] = 45

E[Y] = 20

Var[X] = 55

Var[Y] = 35

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 home shopping or football during this three-month period.

Approximate the value of P(T < 6610).

Explanation / Answer

her let W=100(X+Y) is total number of hours that these one hundred people watch home shopping or football during this three-month period.

mean of W =100(E(X)+E(Y)) =100*(45+20)=6500

and std deviation of W =(100*(55+35+2*10))1/2 =104.8809

hence P(T<6610)=P(Z<(6610-6500)/104.8809)=P(Z<1.0488)=0.8529

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.