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. Find
the proportionality constant that makes the f(t) a probability function.

Answer 1

By the definition, $f(t)=(t+10)-2$ is the probability density function if

$$\int\limits_{-\infty}^{+\infty}f(t)dt=1\to \int\limits_0^{40}\left(t+10-2\right)dt=\\[0.3cm] =\int\limits_0^{40}\left(t+8\right)dt=C\cdot\left.\left(\frac{t^2}{2}+8t\right)\right|_0^{40}=\\[0.3cm] C\cdot\left(\frac{1600}{2}+320\right)=1\to C\cdot1120=1\longrightarrow\\[0.3cm] \boxed{C=\frac{1}{1120}}$$

Comment: Perhaps the probability distribution function was incorrectly indicated and actually meant $f(t)=1/(t+10)^2$ , then

$$\int\limits_{-\infty}^{+\infty}f(t)dt=1\to \int\limits_0^{40}\left(\frac{1}{(t+10)^2}\right)dt=\\[0.3cm] -C\cdot\left.\left(\frac{1}{t+10}\right)\right|_0^{40}=-C\cdot\left(\frac{1}{50}-\frac{1}{10}\right)=1\\[0.3cm] -C\cdot\frac{1-5}{50}=1\longrightarrow C=\frac{50}{4}=\frac{25}{2}=12.5\\[0.3cm] \boxed{C=\frac{25}{2}=12.5}$$

ANSWER

$$\text{If}\,\,\,f(t)=(t+10)-2\longrightarrow C=\frac{1}{1120}\\[0.3cm] \text{If}\,\,\,f(t)=\frac{1}{(t+10)^2}\longrightarrow C=\frac{25}{2}=12.5\\[0.3cm]$$