SOLUTION: Industry standards suggest that 16% of new vehicles require warranty service within the first year. Jones Nissan, sold 11 Nissans yesterday. (Round the Mean answer to 2 decimal pla
Algebra ->
Probability-and-statistics
-> SOLUTION: Industry standards suggest that 16% of new vehicles require warranty service within the first year. Jones Nissan, sold 11 Nissans yesterday. (Round the Mean answer to 2 decimal pla
Log On
Question 1147973: Industry standards suggest that 16% of new vehicles require warranty service within the first year. Jones Nissan, sold 11 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 =
You can put this solution on YOUR website! .
Industry standards suggest that 16% of new vehicles require warranty service within the first year.
Jones Nissan, sold 11 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?
b. What is the probability that exactly one of these vehicles requires warranty service?
c. Determine the probability that exactly two of these vehicles require warranty service.
d. What is the probability that less than three of these vehicles require warranty service?
e. Compute the mean and standard deviation of this probability distribution.
Mean µ =
Standard deviation σ =
~~~~~~~~~~~~~~~~~~~~~
(a) Probability P(0) = = = 0.1469.
(b) Probability P(1) = = = 0.3078.
(c) Probability P(2) = = = 0.2932.
(d) Probability P = P(0) + P(1) + P(2) = the sum of probabilities of (a), (b) and (c) above = 0.1469 + 0.3078 + 0.2932 = 0.7479.