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Consider the following reliability function.
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We have r of t equal to 1 over 0 .001 plus 1, where t is an hour's and positive.
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In question a, we want to find the reliability after 100 operating hours and after 1 ,000 operating hours.
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Essentially, we just want to evaluate r of 100 and then r of 1 ,000.
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So for t equal to 100, evaluating r, gives us.
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0 .90 .91 and when r is equal, when t is equal to 1 ,000, we obtain 0 .5.
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Next, we want to derive the hazard function.
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Hazard rate function.
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So the hazard rate function, h of t, is related to the reliability function via the following expression.
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We have that h of t is equal to minus r prime of t divided by r of t so we know that r of t is equal to one over 0 .001 t plus one our prime of t will then give us minus 0 .001 divided by 0 .001 t plus 1 squared.
02:16
And so the ratio of these two quantities and negated gives us 0 .001 divided by 0 .001 t plus 1.
02:31
Is this increasing or decreasing value rate? this is decreasing, meaning that as t increases the hazard rate decreases because the denominator increases as t increases as t increases.
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Next question, we are given the following hazard rate, h of t equal to 0 .4 where t is in years.
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And our first question, we want to find reliability function.
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So the inverse problem, as question one.
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So we find that r of t is equal to the exponential of minus the integral of h.
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Inserting h and calculating its integral, we find that the integral is equal to 0 .4 t squared over 2, simplifying to minus 0 .2 t squared.
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Next, we want to find the probability of a component failing within the first month of operation...