Question

A review of an airplane's operations revealed that, historically, the airlane had an average of 6.42 mishandled bags per 1,000 passengers. A "mishandling bag" is luggage that was not sent together with passenger's plane. i.e., it was either lost or arrived late.

a) What is the chance that for the next 2,000 passengers, the airline will have more than 11 mishandling bags?

b) After a consulting company evaluated the airline's operations, the airline overhauled its bag-monitoring computer system. As a result, the number of mishandling bags per 1,000 passengers decreased by 30%. Given that new computer system, what is the probability that for the next 3,000 passengers, the airline will have fewert than 10 mishandling bags?

Answer #1

**Answer:**

a)

Given,

= 6.42*2

= 12.84

Poisson distribution P(X) = e^-*^x/x!

P(X > 11) = 1 - P(X <= 11)

= 1 - [e^-12.84*12.84^0/0! + e^-12.84*12.84^1/1! + e^-12.84*12.84^2/2! + .............+ e^-12.84*12.84^11/11!]

= 1 - 0.3696

= 0.6304

b)

= 6.42 - (6.42*30/100) = 4.494

Now for 3000, = 4.494*3 = 13.482

P(X < 10) = P(0) + P(1) + P(2) + ........ + P(9)

= e^-13.482*13.482^0/0! + e^-13.482*13.482^1/1! + e^-13.482*13.482^2/2! + .......... + e^-13.482*13.482^9/9!

= 0.1363

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