Respuesta :
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:
[tex]P(\text{Car})=\frac{1}{31101}\\\\P(\text{Gas Card})=\frac{1}{31101}\\\\P(\text{Shopping Card})=\frac{31099}{31101}[/tex]
(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:
[tex]E(X)=\sum x\times P(X=x)[/tex]
[tex]=[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[/tex]
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:
[tex]SD(X)=\sqrt{V(X)}=\sqrt{E(X^{2}-(E(X))^{2}}[/tex]
[tex]=\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[/tex]
Thus, the standard deviation of the prize won by a prospective customer receiving a flier is $113.38.
