Question 1066663: Industry standards suggest that 12% of new vehicles require warranty service within the first year. Jones Nissan, sold 9 Nissans yesterday
What is the probability that none of these vehicles requires warranty service?
What is the probability that exactly one of these vehicles requires warranty service?
Determine the probability that exactly two of these vehicles require warranty service.
What is the probability that less than three of these vehicles require warranty service?
Compute the mean and standard deviation of this probability distribution.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! a. None requires is .88^9=0.3165
b. for 1 is 9*(.12)(.88^8)=0.3884
c.for 2 is 36*(.12)^2(.88^7)=0.2119
fewer than 3 means 0,1,2 or the sum of the above 3, which is 0.9168.
This is a binomial
The mean is np=9*0.12=1.08 vehicles
the sd is the sqrt of the variance, which is np(1-p)=1.08*.88=.9504; sqrt (.9504)=0.9749.
The above 0,1,2 are calculated from nCx(.12^x)(.88^(n-x))
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