Question

The lifetime T in hours of a system (with some redundancy) has pdf fr(t) Cte 110)(t). What is the probability that the system will last for at least 5 hours?

Answer 1

here for above to be valid: $\int_{0}^{\infty }$ f(t) dt =1

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

therefore C =1/4

hence P(system last for at least 5 hours) =P(T>5)

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