Question 1025634: The US department of transportation maintains statistics for consumer complaints for 100,000 airline passengers. In the first nine months of 2009 consumer complaints were 0.91 per 100,000 passengers. What is the probability that in the next 100,000 passengers that will be
A) no complaints?
B) at least one complaint?
C) at least two complaints?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Poisson test
infrequent events that occur randomly
number could be infinite theoretically
Discrete number
lambda=0.91
for 0 complaints exp(-0.91)(0.91)^0/0!=exp(-0.91)=0.4025
for 1 or more complaints, use complement, and it is 0.5975.
for exactly 1 complaint, it is exp (-0.91) lambda^1/1!
=0.4025*(0.91)=0.3663
Therefore, for 2 or more complaints, it is 1-(0.4025+0.3663)=0.2312
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