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?
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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