The failure times of a class of items are found to be uniformly distributed betw

Question

The failure times of a class of items are found to be uniformly distributed between 1,000 and 1, 500 hours. (a) Find the probability of an item (i) surviving for 1,000 hours, (ii) failing in 1, 100 hours, (iii) surviving beyond 1, 500 hours, and (iv) failing within 1, 200 hours. (b) Given that a particular item has survived for 1, 200 hours, what is the probability that it will fail during the next 100 hours? (c) If the component has survived for 1, 498 hours, what is the probability of its failing during the next hour?

Explanation / Answer

(a) failure times of a class of items are uniformly distributed

f(t) = 1/ (1500 - 1000) = 1/500; 1000 < t < 1500

F(t) = (t - 1000)/ 500 ; 1000 < t < 1500

(i) surviving for 1000 hours = 1 - Pr (failure before 1000 hours) = 1 - 0 = 1

(ii) Falling in 1100 Hours = 1 - Pr (failure in 1100 hours) = 1 - (1100 - 1000)/ 500 = 1 - 0.2 = 0.8

(iii) surviving beyond 1500 hours = 1 - Pr(failure beyond 1500 hours) = 1 - 1 = 0

(iv)  Falling in 1200 Hours = 1 - Pr (failure in 1200 hours) = 1 - (1200 - 1000)/ 500 = 1 - 0.4 = 0.6

(b) The perticular item has survived for 1200 hours so we have to find that it will fail during next 100 hours

so Pr (1200 <t < 1300 hrs l t > 1200 hrs) = {(1300 - 1200)/500} / { 1- (1200 - 1000)/ 500} = 0.2/ 0.6 = 1/3

(c) If component has survived for 1498 hours, probability of its falling during next hour =

Pr (Component failure in next Hour l Component survived in 1498 Hours) = (1/500)/ [ 1 - 498/ 500] = 1/2

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