A representative from the National Football League\'s Marketing Division randoml

Question

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Kansas until he finds a person who attended the last home football game. Let p, the probability that he succeeds in finding such a person, equal 0.20. And, let X denote the number of people he selects until he finds his first success.

a) What is the probability that the marketing representative must select 4 people before he finds one who attended the last home football game?

b) Explain which distribution you use and why?

c) How many people should we expect (that is, what is the average number) the marketing representative needs to select before he finds one who attended the last home football game?

Explanation / Answer

Sol)

A) To find the desired probability, we need to find P(X = 4), which can be determined readily using the p.m.f. of a geometric random variable with p = 0.20, 1p = 0.80, and x = 4:

The pmf of geometric distribution is P(x) =p*qx-1

P(X=4)=0.8034-1×0.20=0.1024

There is about a 10% chance that the marketing representative would have to select 4 people before he would find one who attended the last home football game.

B) Let X denote the number of people he selects until he finds his first success. So we use geometric distribution.

C)

The average number is:

=E(X)=1/p=10 / 0.20= 5

That is, we should expect the marketing representative to have to select 5 people before he finds one who attended the last football game

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