Question

A regional automobile dealership sent out fliers to prospective customers indicating that they had already won...

A regional automobile dealership sent out fliers to prospective customers indicating that they had already won one of three different​ prizes: an automobile valued at ​$25,000​ a ​$100 gas​ card, or a ​$5 shopping card. To claim his or her​ prize, a prospective customer needed to present the flier at the​ dealership's showroom. The fine print on the back of the flier listed the probabilities of winning. The chance of winning the car was 1 out of 31,070​, the chance of winning the gas card was 1 out of 31,070 and the chance of winning the shopping card was 31,068 out of 31,070. Complete parts​ (a) through​ (d).

How many fliers do you think the automobile dealership sent​ out?

Using your answer to​ (a) and the probabilities listed on the​ flier, what is the expected value of the prize won by a prospective customer receiving a​ flier? (Round to nearest cent)

Using your answer to​ (a) and the probabilities listed on the​ flier, what is the standard deviation of the value of the prize won by a prospective customer receiving a​ flier? (Round to nearest cent)

Do you think this is an effective​ promotion? Why or why​ not?

A. No. The promotion will not turn a profit regardless of how many customers redeeming fliers make a purchase.

B. Yes. The promotion will turn a profit if any of the customers redeeming fliers make a​ purchase, which is likely.

C. Yes. The promotion will turn a profit if the customers redeeming fliers make an average purchase greater than the expected value of the prize​ won, which is likely.

D. No. The promotion will only turn a profit if the customers redeeming fliers make an average purchase less than the expected value of the prize​ won, which is unlikely.

Homework Answers

Answer #1

a)

number of fliers =31070

b)

x P(X=x) xP(x) x2P(x)
25000 1/31070 0.80463 20115.86740
100 1/31070 0.00322 0.32185
5 31068/31070 4.99968 24.99839
total 5.8075 20141.1876
E(x) =μ= ΣxP(x) = 5.8075
E(x2) = Σx2P(x) = 20141.1876
Var(x)=σ2 = E(x2)-(E(x))2= 20107.460220
std deviation=         σ= √σ2 = 141.80078

expected value =5.81

standard deviation of the value =141.80

B. Yes. The promotion will turn a profit if any of the customers redeeming fliers make a​ purchase, which is likely.

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