Question 1000130: Please help me solve this problem
Suppose the number of mechanical failures occurring in an industrial plant follows a Poisson distribution with an average of 1.4 failures per week.
(a) What is the probability of no mechanical failures in a given week?
(b) What is the probability that four or more mechanical failures will occur during a three-week period?
Thanks
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! For (a) f(x)=exp^(-1.4)1.4^0/0!
that is equal to exp(-1.4)=0.2466. That is the probability of no mechanical failures in a given week.
For four or more, I need 1,2,3
exp(-1.4)1.4/1=0.3452
exp(-1.4)(1.4)^2/2=0.2417
for 3 exp(-1.4)1.4^3/6=0.1128
These four sum to 0.9463.
The probability of more than 4 is 1-0.9463=0.0537
exp(-lambda)lambda^x/x!, where lambda is the average of failures or whatever is being counted.
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