Continuous probability problem

[toothpaste666]

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?

 

[Orodruin]

Yes. It is correct.

 

[toothpaste666]

thank you