Question

The length of time (in minutes) that a certain person speaks on the telephone is found to be random phenomenon, with a probability function specified by the probability density function f(x) as 
f(x) = `{{:("Ae"^((-x)/5)",",  "for"  x ≥ 0),(0",",  "otherwise"):}`
Find the value of A that makes f(x) a p.d.f.

Answer 1

Since f(x) is a probability density Function

`int_(-oo)^oo "f"(x)  "d"x` = 1

Here `int_0^oo "Ae"^((-x)/5)  "d"x` = 1

`"A"(int "e"^((-x)/5)/(((-1)/5)))_0^oo` = 1

`- 5"A"("e"^((-x)/5))` = 1

` - 5["e"^-oo - "e"^0]` = 1

`- 5"A"(0 - 1)` = 1

5A = 1

⇒ A = `1/5`