4) Mechanical failures of a certain type of aircraft are governed by a Poisson p

Question

4) Mechanical failures of a certain type of aircraft are governed by a Poisson process. Based on past records, an aircraft has a failure once every 6000 hours of flight time. If aircraft are scheduled for inspection and maintenance after every 3000 flight hours, what is the probability that an aircraft will fail between inspections?

5) Referring to the aircraft problem: In a fleet of twelve of these aircraft, what is the probability that not more than two aircraft will have mechanical failures within the 3000-hour interval? (Assume the failures between aircraft are statistically independent.)

6) Referring to the aircraft problem, you are an engineer in charge of aircraft safety. You want to ensure that the probability of failure between inspections is no more than 1% for each aircraft. How would you revise the inspection and maintenance schedule to ensure this level of safety? Please respond on paper only. Your response should be as specific as possible.

PLEASE SHOW WORK, NOT JUST ANSWERS. I ACTUALLY WANT TO UNDERSTAND THE PROBLEM.

Explanation / Answer

4) We are given = 1/6000 failures/hour.

Since the interested probability of failure is for anything greater than or equal to 1 aircraft, we need to find,

P(Xt 1) = 1 - P(Xt = 0)

Poisson probability function is given by,

P(x) = {(e-)* (x)} / x!, where

Therefore,

Here, = 1/6000 failure/hour * (t = 3000hrs) = 1/2

P(Xt = 0) = {(e-1/2)* (1/20)} / 0! = e-1/2

P(Xt 1) = 1 - e-1/2 = 0.3935

5) The number of airplanes with mechanical failures Y with p = 1 e1/2 is given,

P(Y12 2) = P(Y12 = 0) + P(Y12 = 1) + P(Y12 = 2)

   = (1 p) 12 + 12p(1 p) 11 + 12C2 p2 (1 p)10

   = e6 + 12(1 e 1/2) e 11/2 + 66(1 e 1/2) 2 e 5

   = 0.00247875 + 12*0.393469* 0.00408677 + 66*0.3934692 * 0.00673795

   = 0.09062

6) We want

P(X 1) = 1 P(X = 0) = 1 et 0.01

t ln(0.99)/=60.302 hrs There should be an inspection every 60.3 hours or less

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