The time in minutes for which a student uses a computer terminal at the center of a major university follows?

Question

an exponential probability distribution with a mean of 36 minutes. Assume a student arrives at the terminal just as another student is beginning to work on the terminal. (a) What is the probability that the wait for the second student will be 15 minutes or less? (b) What is the probability that the wait for the seond student will be between 15 and 45 minues? (c) What is the probability that the wait for the second student will have to wait one hour or more?

Answer 1

The cumulative distribution function for the exponential distribution with mean lambda = 36 minutes is:

Prob(X < a minutes) = 0 for a <= 0
Prob(X < a minutes) = 1 - e^(- a / 36) for a >=0

So letting F(a) be that function, the answers are:

(a) F(15)
(b) F(45) - F(15)
(c) 1 - F(60)

(a) 1 - e^(- 15 / 36) = 1 - e^(- 5 / 12) ~ 34.1 %
(b) (1 - e^(- 45 / 36)) - (1 - e^(- 15 / 36)) = e^(-15 / 36) - e^(- 45 / 36) = e^(- 5 / 12) - e^(- 5 / 4) ~ 37.3 %
(c) 1 - (1 - e^(- 60 / 36)) = e^(- 60 / 36) = e^(- 5 / 3) ~ 18.9 %

Dan

More interesting question: (d) Follow up to (a): Given that the second student has already waited 15 minutes, what is the probability that the remaining wait will only be 15 minutes or less. Compare the answer to (a).