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.

- What is the probability that the wait for the second student
will be 15 minutes or less (to 4 decimals)?

- What is the probability that the wait for the second student
will be between 15 and 45 minutes (to 4 decimals)?

- What is the probability that the second student will have to wait an hour or more (to 4 decimals)?

Answer #1

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

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