Consider the operation of a carnival ride which starts every five minutes. An an

Question

Consider the operation of a carnival ride which starts every five minutes. An analysis of past data reveals that the time between successive customer arrivals can be described by an exponential distribution with a mean of 30 seconds. The ride has 12 seats available each time it leaves. Suppose that a ride has just begun and there are no customers waiting. What is the probability that the next ride will be Ipaded to capacity? Now suppose that there are 11 customers present, the last one of which arrived 20 seconds ago, and you would like to be on the next ride when it begins in two minutes. However, you must first put your coat in a public locker nearby which will require one minute. What is the probability that you will be on the next ride?

Explanation / Answer

One important thing to remember while solving this is

When the waiting time to see successive customers followsexponential distribution then X: # of customers in a certain timefollows Poisson distribution.

P(x=k,t) = (t)-k e- t/k!

P(X=0,t) = e- t which is exponentialdistribution.

Therefore

a)

average number of customers in 5 mins =5*60sec /30sec = 10customers per 5 mins

therefore P(the next ride will be loaded to capacity)

= P(x >=12)

=1-P(x<=11)

= 1-? (10)-k e- 10 /k! where k = 0 to11

= 1- 69.67%

=30.32%

b)

we know that last customers have come 20 sec back and we willtake 1 min more . So to be on the ride we don’t want anycustomer to come in 80 seconds time.

P(x=0,80) =(80/30)0 e(-1*80/30) /0!=6.94%

Hope that helps. Feel free to ask for any clarifications

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