Answer to Question #115916 in Statistics and Probability for Nii

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. Hint: Show that f(t) is legitimate and find the proportionality constant.

Answer 1

$\text {Let b is a constant, then}\\ f(t)=b[(t+10)-2]=b(t+8), 0<t<40\\ b\int _0^{40}(t+8)dt=1\\ b\left[\frac{t^2}{2}+8t\right]^{40}_0=1\\ b[\frac{40^2}{2}+8(40)]=1 b(1120)=1\\ \therefore b=\frac{1}{1120}\\ P(t<10)=\frac{1}{1120}\int _0^{10}t+8dt\\ =\frac{1}{1120}\left[\frac{t^2}{2}+8t\right]^{40}_0\\ =\frac{1}{1120}[\frac{10^2}{2}+8(10)]\\ =\frac{1}{1120}\times 130\\ =\frac{13}{112}$