ABC corp conducted a study of service times at the drive-up window of fast-food restaurants. The average time between placing an order and receiving the order at restaurant is 2.45 minutes. Assume that the waiting time follows exponential distribution. a. What is the probability that service time is less than 1.5 minutes?

Setup: $$\lambda = 1/2.45$$ $$P(X < t=1.5) = {1-e^{-\lambda t}} = 0.46$$
is it correct?

b. A customer has been waiting for 3 minutes after placing an order, what is the expected value of remaining waiting time?

Question : how to work for remaining waiting time?

  • $\begingroup$ When asked for the probability that X is less than 1.5 minutes, you compute P(X<2). Why? $\endgroup$ – Did Aug 1 '14 at 12:59
  • $\begingroup$ @Did I have corrected my calculation.. $\endgroup$ – user269867 Aug 1 '14 at 13:14
  • $\begingroup$ Yes, the answer is about $0.458$. The expected remaining waiting time is $2.45$. There should be no need to calculate. The notation $P(X\lt t=a)$ is somewhat unusual. $\endgroup$ – André Nicolas Aug 1 '14 at 13:16
  • $\begingroup$ @AndréNicolas - how did you calculate the remaining time. the waiting time is 3 and average time is 2.45.. $\endgroup$ – user269867 Aug 1 '14 at 13:21
  • $\begingroup$ It is a property of the exponential distribution, often called memorylessness. The conditional distribution of additional waiting time, given one has been waiting for $a$, is exponential same parameter. $\endgroup$ – André Nicolas Aug 1 '14 at 13:25

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