Question

The time in minutes for which a student uses a computer terminal at the computer center of a major university follows an exponential probability distribution with a mean of 31 minutes. Assume a student arrives at the terminal just as another student is beginning to work on the terminal.

1. What is the probability that the wait for the second student will be 15 minutes or less (to 4 decimals)?
   
   
2. What is the probability that the wait for the second student will be between 15 and 45 minutes (to 4 decimals)?
   
   
3. What is the probability that the second student will have to wait an hour or more (to 4 decimals)?

Answer 1

EXPONENTIAL DISTRIBUTION : The cumulative distribution function (CDF) for the exponential distribution with mean lambda = 31 minutes is:

Prob(X < a minutes) = 0 for a <= 0
Prob(X < a minutes) = 1 - e^(- a / 31) 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 / 31) = .384
(b) (1 - e^(- 45 / 31)) - (1 - e^(- 15 / 31)) = e^(-15 / 31) - e^(- 45 / 31) = .7658 - .384 = .3818
(c) 1 - (1 - e^(- 60 / 31)) = e^(- 60 / 31) = .1443