Question

The length of satisfactory service (years) provided
by a certain model of laptop computers is a random variable having the probability density:

f(x)=1/4e^-x/4.5 for x>0
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

Answer 1

So for part (a) from guessing that I have to convert
the function from f(x) to a distribution function F(x)
and then plug in 2.5 for F(x) to get my answer.

Answer 2

So far I have:

\[F(x)=\int\limits_{0}^{x}\frac{ 1 }{ 4.5 } e ^{-x/4.5}=.2222e^{-.2222x}dx\]

Answer 3

So F(2.5)= \[.2222e^{-.2222*(2.5)}=-.5738\]

Answer 4

nevermind question solved.