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

$\text {Let c is a constant, then}\\ f(t)=c[(t+10)-2]=c(t+8), 0<t<40\\ \text{Its known that if f(t) is a density function,}\\ Then,\; \int _0^{40}f(t)dt=1\, so,\\ c\int _0^{40}t+8dt=1\\ c\left[\frac{t^2}{2}+8t\right]^{40}_0=1\\ c[\frac{40^2}{2}+8(40)]=1\\ c(1120)=1\\ \therefore c=\frac{1}{1120}\\ P(t<10)=\frac{1}{1120}\int _0^{10}t+8dt\\ =\frac{1}{1120}\left[\frac{t^2}{2}+8t\right]^{40}_0\\ =\frac{1}{1120}[\frac{10^2}{2}+8(10)]\\ =\frac{1}{1120}\times 130\\ =\frac{13}{112}$