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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"):}`
What is the probability that the number of minutes that person will talk over the phone is (i) more than 10 minutes, (ii) less than 5 minutes and (iii) between 5 and 10 minutes
Solution
(i) more than 10 minutes
`int_10^00 "f"(x) "d"x`
= `1/5 int_10^oo "e"^(x/5) "d"x`
= `1/5 ("e"^((-x)/5)/(((-1)/5)))^oo`
= `- ["e"^((-x)/5)]_10^oo`
= `- ["e"^-oo - "e"^((-10)/5)]`
= `- [0 - "e"^-2]`
= `"e"^-2`
= `1/"e"^2`
(ii) less than 5 minutes
`int_0^5 f(x) "d"x = int_0^5 "Ae"^((x)/5)`
= `1/5 int_0^5 "e"^((-x)/5) "d"x`
= `1/5 ["e"^((-x)/5)/((-1)/5)]_0^5`
= `- ["e"^((-x)/5)]_0^5`
= `- ["e"^((-5)/5) - "e"^0]`
= `- ("e"^-1 - 1)`
= `1 - "e"^-1`
= `1 - 1/"e"`
= `("e" - 1)/"e"`
(iii) between 5 and 10 minutes
`int__5^10 "f"(x) "d"x = int_5^10 "Ae"^((-x)/5) "d"x`
= `int_5^10 1/5 "e"^((-x)/5) "d"x`
= `1/5 ["e"^((-x)/5)/((-1)/5)]_5^10`
= `- ["e"^((-x)/5)]_5^10`
= `- ["e"^((-10)/5) - "e"^((-5)/5)]`
= `[-"e"^-2 - "e"^-1]`
= `"e"^-1 - "e"^-2`
= `1/"e"- 1/"e"^2`
= `("e" - 1)/"e"^2`
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