Suppose that the probability density function (p.d.f.) of the life (in weeks) of a certain part...

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Suppose that the probability density function (p.d.f.) of the life (in weeks) of a certain part is f(x) = 3x² (500)³¹ (c) Determine P(Y ≤ 200). (a) Compute the probability the a certain part will fail in less than 200 weeks (b) To decrease the probability in part (a), four independent parts are placed in parallel. So all must fail, if the system fails. Let Y = max{X₁, X2, X3, X4} denote the lifetime of such a system, where X, denotes the lifetime of the ith component. Show that 0≤x <500. fy(y) = 12 y¹1 (500) 12 Y>0. Suppose that the probability density function (p.d.f.) of the life (in weeks) of a certain part is f(x) = 3x² (500)³¹ (c) Determine P(Y ≤ 200). (a) Compute the probability the a certain part will fail in less than 200 weeks (b) To decrease the probability in part (a), four independent parts are placed in parallel. So all must fail, if the system fails. Let Y = max{X₁, X2, X3, X4} denote the lifetime of such a system, where X, denotes the lifetime of the ith component. Show that 0≤x <500. fy(y) = 12 y¹1 (500) 12 Y>0.

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a To compute the probability that a certain part will fail in less than 200 weeks we need to integra... View the full answer