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Question
Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function
F(x) = `{{:(0",", "for" x ≤ 0),(x/2",", "for" 0 ≤ x < 1),(1/2",", "for" ≤ x < 2),(x/4",", "for" 2 ≤ x < 4),(1",", "for" x ≥ 4):}`
What is the probability that a person will have to wait (i) more than 3 minutes, (ii) less than 3 minutes and (iii) between 1 and 3 minutes?
Solution
The probability that a person will have to wait
(i) more than 3 minutes
P(x > 3) = `int_2^4 "f"(x) "d"x`
= `int_3^4 (1/4) "d"x`
=`1/4(x)_3^4`
= `1/4 (4 - 3)`
= `1/4 (1)
∴ P(x > 3) = `1/4`
(ii) Less than 3 Minutes
P(x < 3) = `int_0^1 "f"(x) "d"x int_1^2 "f"(x) "d"x + int_2^3 "f"(x) "d"x`
= `int_0^1 1/2 "d"x + int_1^2 (0) "d"x + int_2^3 1/4 "d"x`
= `1/2 [x]_0^1 + 0 + 1/4 [x]2^3`
= `12 [1 - 0] + 1/4 [3 - 2]`
= `1/2 + 1/4`
=`(2 + 1)/4`
∴ P(x < 3) = `3/4`
(iii) between 1 and 3 minutes
P(1 < x < 3) = `int_1^2 "f"(x) "d"x + int_2^3 "f"(x) "d"x`
= `int_1^2 (0) "d"x + int_2^3 (1/4) "d"x`
= `0 + 1/4 (x)_2^3`
= `1/4 [3 - 2]`
= `1/4(1)`
`"P"(1 < oo < 3) = 1/4`
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