In the first nine months of a year, airline consumer complaints were 0.89 per 10

ID: 3310649 • Letter: I

Question

In the first nine months of a year, airline consumer complaints were 0.89 per 100,000 passengers. Complete parts (a) through (c) below.

What is the probability that in the next 100,000 passengers, the airline will have no complaints?

B. what is the probability that in the next 100,00 passengers, the airline will have only 1 complaint

C. what is the probability that in the next 100,000 passengers the airline will have only 2 compalints

What is the probability that in the next 100,000 passengers, the airline will have no complaints?

B. what is the probability that in the next 100,00 passengers, the airline will have only 1 complaint

C. what is the probability that in the next 100,000 passengers the airline will have only 2 compalints

= 0.89 per 100,000 passengers.

P(X = x) = e- * x / x!

a) P(X = 0) = e-0.89 * 0.890 / 0! = 0.411

b) P(X = 1) = e-0.89 * 0.891 / 1! = 0.365

c) P(X = 2) = e-0.89 * 0.892 / 2! = 0.163

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