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Suppose the lifetime of a component $T_i$ 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 $0.75$ for a), $.75$ for b), $0.50$ for c), and $0.60$ 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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Please review the FAQ on How to Ask. Multiple problems and even multi-part problems are discouraged, esp. in the case as here where there is no indication of effort on the part of the poster or the actual difficulty encountered if such effort was made. –  hardmath May 16 at 14:24

1 Answer 1

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

(a) $F(s)=0$ for $0 \leq s \leq 100$ and $\mu=150$. So $$1−\Psi_A(50)= 1 − (1/\mu) \int _0^{50}(1 − F (s))ds= 1 − 50/150 = 2/3$$

(b) It is $1 −\Psi_B(50) = 1 − \Psi_A(50) = 2/3$.

(c) $F(s) = (s−100)/100$ for $100 \leq s \leq 200$. So \begin{align*} 1−\Psi_C(150) &= 1− (1/\mu)[150 F(150) − \int_0^{150} F(s)ds] \\ &= 1 − (1/150)[150(1/2) − ((150^2 /200 − 150) −(100^2/200 − 100))] \\ &= 1 − 5/12 = 7/12 \end{align*}

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

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Can someone please tell me if I answered the question correctly or not. –  user130344 Mar 2 '14 at 23:02
Would you mind formatting your answer with MathJax? meta.math.stackexchange.com/questions/5020/… –  Math1000 Feb 8 at 14:04

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