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

Let $c=$ consr, then

$$f(t) = \begin{cases} c(t+10)^{-2} & 0\leq t\leq 40 \\ 0 &otherwise \end{cases}$$

$$1=\displaystyle\int_{-\infin}^{\infin}f(t)dt=\displaystyle\int_{0}^{40}{c\over (t+10)^2}dt=$$

$$=-c\bigg[{1\over t+10}\bigg]\begin{matrix} 40 \\ 0 \end{matrix}=-c\bigg({1\over 40+10}-{1\over 0+10}\bigg)=0.08c$$

$$c=12.5$$

$$f(t) = \begin{cases} 12.5(t+10)^{-2} & 0\leq t\leq 40 \\ 0 &otherwise \end{cases}$$

Calculate the probability that the lifetime of the machine part is less than 10 years. 

$$P(T<10)=\displaystyle\int_{-\infin}^{10}f(t)dt=\displaystyle\int_{0}^{10}{12.5\over (t+10)^2}dt=$$

$$=-12.5\bigg[{1\over t+10}\bigg]\begin{matrix} 10 \\ 0 \end{matrix}=-12.5\bigg({1\over 10+10}-{1\over 0+10}\bigg)=0.625$$