Question

The lifetime of a machine is continuous on the interval (0 ,40) with probability density function f,where f(t) is proportional to (t+10)^-2 and t is the lifetime in years.Calculate the probability that the lifetime of the machine part is less than 10 years

Answer 1

$\text {Let a is a constant, then}\\ f(t)=a(t+10)^{-2}, 0<t<40\\ a\int _0^{40}\left(t+10\right)^{-2}dt=1\\ a\left[-(t+10\right)^{-1}]^{40}_{0}=1\\ -a(50^{-1}-10^{-1})=1\\ -a(\frac{-2}{25})=1\\ a=\frac{25}{2}\\ \therefore f(t)=f(t)=\frac{25}{2}(t+10)^{-2}, 0<t<40\\ P(t<10)=\frac{25}{2}\int _0^{10}\left(t+10\right)^{-2}dt\\ =\frac{25}{2}\left[-(t+10\right)^{-1}]^{10}_{0}\\ =\frac{-25}{2}(20^{-1}-10^{-1})\\ =\frac{-25}{2}\times\frac{-1}{20}\\ =\frac{5}{8}$