Answer to Question #119379 in Statistics and Probability for Michael

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

We have, the lifetime of a machine is continuous on the interval (0, 40) with probability density

function f, where

f(t) $\propto$ (t+10)^-2, t is the lifetime in years

Let, k be the proportionality constant. Then,

f(t) = k(t+10)^-2, 0 < t < 40

= 0, otherwise

Now in order to f(t) become a probability function

$\int_0^{40} f(t)dt$ = 1

i.e. $\int_0^{40} k(t+10)^{-2}dt$ = 1

i.e. $k\int_0^{40} \frac{dt}{(t+10)^2}$ = 1

Let, t + 10 = u

$\implies$dt = du

When t = 0, u = 10 and when t = 40, u = 50

$\therefore k\int_{10}^{50} \frac{du}{u^2}$ = 1

i.e. $-k[\frac{1}{u}]_{10}^{50}$ = 1

i.e. $-k(\frac{1}{50}-\frac{1}{10})$ = 1

i.e. $\frac{4k}{50}$ = 1

i.e. k = 50/4 = 25/2

Answer: The proportionality constant that makes f(t) a probability function is 25/2.