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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.
Solution
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`
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