(Sec. 4.4) Suppose the number of requests for assistance received by a 24-hour t

Question

(Sec. 4.4) Suppose the number of requests for assistance received by a 24-hour towing service is on av- erage 4 per hour, and the the number of requests can be modeled by a Poisson process. Suppose further that the number of requests during any non-overlapping intervals are independent of one another. (a) On a given day, what is the probability that the towing service will receive no more than 3 requests for assistance? (b) After receiving one request for assistance, what is the probability that the towing service will not receive another request for at least 6 hours?

Explanation / Answer

Ans:

a)Poisson distribution:

On average 4 requests per hr,so for a day,i.e. 24 hrs,there will on average 4*24=96 per day.

P(x<=3)=e-96*(960/0!+961/1!+962/2!+963/3!)

=0

b)Interarrival times are exponentially distributed.

Average interarrival time=1/4=0.25

P(T>=6)=1-P(T<6)=e-6/4=e-1.5=0.2231

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