Question

The lifetime T in hours of a system (with some redundancy) has pdf

$f_{T}(t) = Cte^{-\frac{t}{2}}1_{[0,\infty )}(t)$

What is the probability that the system will last for at least 5 hours?

Answer 1

here for this to be valid:

$\int_{0}^{\infty }$ f(t) dt should be equal to 1

$\int_{0}^{\infty }$ f(t) dt =$\int_{0}^{\infty }$Cte^-t/2 dt =C*(-2te^-t/2-4e^-t/2) |^$\infty$₀ =C*4 =1

C=1/4

therefore P( system will last for at least 5 hour )=P(X>5)=1-P(X<5)

=1-$\int_{0}^{5 }$ f(t) dt =1-$\int_{0}^{5 }$(1/4)te^-t/2 dt =1-(1/4)*(-2te^-t/2-4e^-t/2) |⁵₀ =1-0.7127=0.2873