Question

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

0.91 per 100,000 passengers..

What is the probability that in the next 100,000 passengers, the airline will have at least two complaints?

The probability that the airline will have at least two complaints is ( ).

(Round to four decimal places as needed.)

Answer 1

Use Poisson distribution with mean m = 0.91, and formula
P[x] = (e^-m *m^x)/x!

where x is the actual number of successes that result from the experiment.

Probability that the airline will have at least two complaints can be represented as P[x>2]

P[x>2] = 1 - P[x=0] - P[x=1]

P[x=0] = (e^-0.91*0.91⁰)/0! = 0.4025

P[x=1] = (e^-0.91*0.91¹)/1! = 0.3663

P[x>2] = 1 - 0.4025 - 0.3663 = 1 - 0.7688

P[x>2] = 0.2312

Thus, the probability that the airline will have at least two complaints is 0.2312