- #1
toothpaste666
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Homework Statement
The length of satisfactory service (years) provided by a certain model of laptop computer is a random variable having the probability density
f(x) = (1/4.5)e^(-x/4.5) for x > 0
and 0 for x <= 0
find the probabilities that one of these laptops will provide satisfactory service for
a) at most 2.5 years
b) anywhere from 4 to 6 years
c) at least 6.75 years
The Attempt at a Solution
first we evaluate the integral
∫(1/4.5)e^(-x/4.5)dx = (1/4.5) ∫ e^(-x/4.5)dx = (1/4.5)(-4.5)e^(-x/4.5) = -e^(-x/4.5)
part a) evaluate from 0 to 2.5
-e^(-2.5/4.5) - (-e^0) = 1 - e^(-2.5/4.5) = .4262
b) evaluate from 4 to 6
-(e^(-6/4.5) - e^(-4/4.5)) = .1475
c) from 6.75 to ∞
-(e^(-∞) - e^(-6.75/4.5)) = -(0 - .2231) = .2231
is my reasoning correct?