Problem 1. The lifetime of a device of type A is Exponentially distributed with mean 10 (years) while the lifetime of a device of type B has a Gamma distribution with parameters (cid:11) = 2 and (cid:21) = 1=10. It is known that 40% of the devices in the factory are of type A and the remaining 60% are of type B. Let X be the lifetime of a randomly chosen device. (a) Compute the mean and the variance of X. Solution 2.Exponentially distributed MTBF = 500 h t = 600 h MTBF = 1/? ? = 1/MTBF = 2*10^-3 fr/hr R(t) = e^-?t R(600) = e^-2*10^-3*600 R(600) =0.30119 = 30.119% --> Probability of successfull/survive R(T+t) = R(t) R(T+100) = R(100) R(100) = e^-2*10^-3*100 R(100) = 0.81873 = 81.873 Probability of failure, U(100) = 1 - R(100) U(100) = 0.18127 = 18.873 %.