SOLUTION: Industry standards suggest that 16% of new vehicles require warranty service within the first year. Jones Nissan, sold 8 Nissans yesterday. (Round the Mean answer to 2 decimal plac

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Question 1118074: Industry standards suggest that 16% of new vehicles require warranty service within the first year. Jones Nissan, sold 8 Nissans yesterday. (Round the Mean answer to 2 decimal places and the other answers to 4 decimal places.)

a. What is the probability that none of these vehicles requires warranty service?

Probability

b. What is the probability that exactly one of these vehicles requires warranty service?

Probability

c. Determine the probability that exactly two of these vehicles require warranty service.

Probability

d. What is the probability that less than three of these vehicles require warranty service?

Probability

e. Compute the mean and standard deviation of this probability distribution.

Mean µ =
Standard deviation σ =
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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Industry standards suggest that 16% of new vehicles require warranty service within the first year. Jones Nissan, s
old 8 Nissans yesterday. (Round the Mean answer to 2 decimal places and the other answers to 4 decimal places.) 

a.  What is the probability that none of these vehicles requires warranty service?

 
    Probability  = %281-0.16%29%5E8 = 0.84%5E8 = 0.2479 = 24.79%.

 
b.  What is the probability that exactly one of these vehicles requires warranty service?

 
     Probability that exactly one of the 8 cars requires service, while 7 others do not = 8%2A0.16%2A0.84%5E7 = 0.3777 = 33.77%.

 
c.   Determine the probability that exactly two of these vehicles require warranty service.

 
     Probability = C%5B8%5D%5E2%2A0.16%5E2%2A0.84%5E6 = %28%288%2A7%29%2F%281%2A2%29%29%2A0.16%5E2%2A0.84%5E6 = 0.2518 = 25.18%.	 

 
d.   What is the probability that less than three of these vehicles require warranty service?

 
     Probability = 0.84%5E8+%2B+C%5B8%5D%5E1%2A0.16%2A0.84%5E7+%2B+C%5B8%5D%5E2%2A0.16%5E2%2A0.84%5E6 = 0.2479 + 0.3777 + 0.2518 = 0.8774 = 87.74%.	 

 
e.   Compute the mean and standard deviation of this probability distribution.

 
  Mean  µ = 	 
  Standard deviation σ =