A dealership sent out fliers to prospective customers, indicating that they had already won one of three different prizes. The three prices

Question

A dealership sent out fliers to prospective customers, indicating that they had already won one of three different prizes. The three prices are either a new car valued at 5,000, a $500 gas card, or a $55 gift 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 13,320, the chance of winning the gas card was 1 out of 13,320, and the chance of winning the gift card was 13,318 out of 13,320. Complete parts (a) and (b).

a). Using the probabilities listed on the flier, what is the expected value of the prize won by a prospective customer receiving a flier?
b). Using 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?

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Ruby 2 weeks 2021-09-13T03:36:27+00:00 1 Answer 0

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    2021-09-13T03:37:34+00:00

    Answer:

    The expected value of the prize won by a prospective customer receiving a flier os $55.4.

    The standard deviation of the value of the prize won by a prospective customer receiving a flier is $43.

    Step-by-step explanation:

    The expected value of the prizes is the sum ofthe prizes, weighted by their probability of happening.

    E(X)=\frac{1}{13,320}*5000+ \frac{1}{13,320}*500+ \frac{13,318}{13,320}*55\\\\E(X)=\frac{5000+500+732,490}{13,320} =\frac{737,990}{13,320}= 55,4

    The expected value of the prizes is $55,4.

    The standard deviation will be calculated as the square root of the variance:

    V(X)=\frac{1}{13,320}*(5000-55.4)^2+ \frac{1}{13,320}*(500-55.4)^2+ \frac{13,318}{13,320}*(55-55.4)^2\\\\V(X)=\frac{4944.6^2+444.6^2+13,318*0.4^2}{13,320} =\frac{24,648,869.2}{13,320}= 1,850.52

    \sigma=\sqrt{V(X)} =\sqrt{1850.52}=43

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