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Suppose the lifetime of a component Ti in hours is uniformly distributed on [100, 200]. Components are replaced as soon as one fails and assume that this process has been going on long enough to reach equilibrium. (a) What is the probability that the current component has been in operation for at least 50 hours? (b) What is the probability that the current component will last for at least 50 hours more? (c) What is the probability that the total lifetime of the current component will be at least 150 hours? (d) Suppose that it is known that the current component has been in operation for exactly 90 hours. What is the probability that it will last at least 50 more hours?

I am not sure how to really start this. Intuitively, I would think the answers would be .75 for a), .75 for b), .50 for c), and .60 for d)?

Also, how would you do this for an exponential distribution with mean of 150? In this case I would not know how to approach this.

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1 Answer 1

In case of uniform distribution, this is how you answer this question.

(a)F(s)=0 for 0≤s≤100 andμ=150. So 1−ΨA(50)= 1 − (1/μ)* integrate (1 − F (s))ds from 0 to 50

= 1 − 50/150 = 2/3.

(b) It is 1 − ΨB(50) = 1 − ΨA(50) = 2/3.

(c) F(s) = (s−100)/100 for 100 ≤ s ≤ 200. So 1−ΨC(150) = 1− (1/μ)[150*F(150) − integrate F(s)ds from 0 to 150] = 1 − (1/150)[150(1/2) − ((150^2 /200 − 150) −(100^2/200 − 100))]

= 1 − 5/12 = 7/12.

(d) 1 − F(140) = (200 − 140)/200 = 3/5 ̸ which is not equal to 1 − ΨC(140).

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Can someone please tell me if I answered the question correctly or not. –  user130344 Mar 2 at 23:02

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