One year, airline A had 5.01 mishandled bags per 1,000 passengers. complete part

Question

One year, airline A had 5.01 mishandled bags per 1,000 passengers. complete parts a. through c. below. what is the probability that in the next 1,.000 passengers, the airline will have no mishandled bags? the probability that the airline will have no mishandled bags is what is the probability that in the next 1,000 passengers, the airline will have at least one mishandled bag? the probability that the airline will have at least one mishandled bags is what is the probability that in the next 1,000 passengers, the airline will have at least two mishandled bags? the probability that the airline will have at least two mishandled bags is

Explanation / Answer

It follows a binomial distribution with p = 0.00501 and n = 1000 as p is very small and n is very large we will limit it to poisson distibution with lamda = n*p = 5.01

P(Airlines will have no mishandled bags i.e. x = 0) = e -5.01 = 0.00667

P(Airlines will have atleast one mishandled bag) = P( X >= 1) = 1 - p(X<1) = 1 - P(X=0) = 1 - 0.00667 = 0.9933

P(Airlines will have atleast two mishandled bag) = P( X >= 2) = 1 - [p(X<2) ]= 1 - [P(X=0) + P(X=1)

                                                   = 1 - [0.00667 + 0.0334] = 0.9599

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