Question

Let X = the time between two successive arrivals at the drive-up window of a local bank.

Suppose that X has an exponential distribution with an average time between arrivals of 4

minutes.

a. A car has just left the window. What is probability that it will take more than 4 minutes before the next drive-up to the window?

b. A car has just left the window. What is the probability that it will take more than 5 minutes before the next drive-up to the window?

c. Suppose it has been 2 minutes since the last car left the window. What is the probability that it will take more than 4 minutes before the next drive-up to the window?

d. What is the median amount of time between two successive arrivals at the drive-up window?

Answer #1

a.

Average time between arrivals = 4 minutes

Parameter = 1/4

Probability that it will take more than 4 minutes before the next drive-up to the window

= P(X > 4) = exp(- * 4)

= exp(- 4/4) = exp(-1) = 0.3679

b.

Probability that it will take more than 5 minutes before the next drive-up to the window

= P(X > 5) = exp(- * 5)

= exp(- 5/4) = exp(-1.25) = 0.2865

c.

Due to memoryless property of exponential distribution, probability that it will take more than 4 minutes before the next drive-up to the window = P(X > 4)

= exp(- * 4)

= exp(- 4/4) = exp(-1) = 0.3679

d.

Median amount of time between two successive arrivals at the drive-up window

= ln(2) /

= 0.693 * 4

= 2.772 minutes

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