first nine months of a year, airline consumer complaints were 0.89 per 100,000 p
ID: 3066475 • Letter: F
Question
first nine months of a year, airline consumer complaints were 0.89 per 100,000 passengers. Complete parts (al) through (o) below a. What is the probability that in the next 100,000 passengers, the airline will have no The probability that the airline will have no complaints is complaints? (Round to four decimal places as needed.) b. What is the probability that in the next 100,000 passengers, The probability that the airline will have at least one complaint is the airline will have at least one complaint? (Round to four decimal places as needed.) c. What is the probability that in the The probability that the airline will have at least two complaints is next 100,000 passengers, the airline will have at least two complaints? Round to four decimal places as needed.)Explanation / Answer
Solution:-
a) The probabbility that the airline will have no complaints is 0.411
= 0.89
x = 0
By applying poisons distribution:-
P(x; ) = (e-) (x) / x!
P(x = 0) = 0.411
b) The probabbility that the airline will have atleast one complaints is 0.5893.
= 0.89
x = 1
By applying poisons distribution:-
P(x; ) = (e-) (x) / x!
P(x > 1) = 0.5893
c) The probabbility that the airline will have atleast two complaints is 0.2239.
= 0.89
x = 2
By applying poisons distribution:-
P(x; ) = (e-) (x) / x!
P(x > 2) = 0.2239
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