a A component lasts more than 3000 hours before failure b A

(a) A component lasts more than 3000 hours before failure.
(b) A component fails in the interval from 1000 to 2000 hours.
(c) A component fails before 1000 hours.
(d) Determine the number of hours at which 10% of all components have failed.
(e) Find the mean and variance of the time till failure

Please show all work...

The probability density function of the time to failure of an electronic component in a copier (in hours) is fixxDetermine the component in a copier (in hours) is f(x)-f for x0, Determine the 1000 probabilities that

Solution

The cumulative distribution function of this is

P(X<x) = F(x) = 1 - exp(-x/1000)

Thus,

a)

P(x>3000) = 1 - F(3000) = exp(-3000/1000) = 0.049787068 [ANSWER]

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b)

P(1000<x<2000) = F(2000) - F(1000) = exp(-1000/1000) - exp(-2000/1000) = 0.232544158 [ANSWER]

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c)

P(x<1000) = F(1000) = 1 - exp(-1000/1000) = 0.632120559 [ANSWER]

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d)

Here, let a = the critical value.

P(X<a) = 0.10 = F(a)

Thus,

0.10 = 1 - exp(-a/1000)

Thus,

a = -1000*ln(0.90) = 105.3605157 [ANSWER]

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e)

For an exponential distribution, the mean is the denominator of the exponential,

u = 1000 hours

and the variance is the square of the mean,

variance = 1000^2 = 1000000 [ANSWER]