Question

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 ​$​, a ​$ gas​ card, or a ​$ 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 ​, the chance of winning the gas card was 1 out of and the chance of winning the shopping card was out of 31,866​, the chance of winning the gas card was 1 out of 31,866, and the chance of winning the shopping card was 31,864 out of 31,866.
Required:
a. How many fliers do you think the automobile dealership sent​ out?
b. 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?
c. 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?

Answer 1

Answer:

(a) 31101

(b) $5.65

(c) $113.38

Step-by-step explanation:

The complete question is:

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 $20,000, a $125 gas card, or a $5 shopping card. To claim his or her prize, a prospective print on the back of the flier listed the probabilities of winning. The chance of winning the car was 1 out of 31,101, the chance of winning the gas card was 1 out of 31,101, and the chance of winning the shopping card was 31, 099 out of 31,101.

Solution:

The information provided is as follows:

$P(\text{Car})=\frac{1}{31101}\\\\P(\text{Gas Card})=\frac{1}{31101}\\\\P(\text{Shopping Card})=\frac{31099}{31101}$

(a)

The number of fliers the automobile dealership sent​ out is, n = 31,101.

This is because the probability of winning any of the three prize is out of 31,101.

(b)

Compute the expected value of the prize won by a prospective customer receiving a​ flier as follows:

$E(X)=\sum x\times P(X=x)$

         $=[20000\times \frac{1}{31101}]+[125\times \frac{1}{31101}]+[5\times \frac{31099}{31101}]\\\\=0.6431+0.0040+4.9997\\\\=5.6468\\\\\approx 5.65$

Thus, the expected value of the prize won by a prospective customer receiving a​ flier is $5.65.

(c)

Compute the standard deviation of the prize won by a prospective customer receiving a​ flier as follows:

$SD(X)=\sqrt{V(X)}=\sqrt{E(X^{2}-(E(X))^{2}}$

           $=\sqrt{[(20000)^{2}\times \frac{1}{31101}+(125)^{2}\times \frac{1}{31101}+(5)^{2}\times \frac{31099}{31101}]-(5.65)^{2}}\\\\=\sqrt{12854.9011}\\\\=113.37947\\\\\approx 113.38$

Thus, the standard deviation of the prize won by a prospective customer receiving a​ flier is $113.38.