The number of requests for assistance received by a towing service is a Poisson
ID: 3441364 • Letter: T
Question
The number of requests for assistance received by a towing service is a Poisson random variable with rate parameter = 4 per hour.
(a) Compute the probability that exactly 4 requests are received during a particular hour.
(b) What is the probability of receiving exactly 15 requests during a particular 4-hour period?
(c) If the operators of the towing service take a 45-min break for lunch, what is the probability
that they do not miss any calls for assistance? How many calls would your expect during their
break?
Explanation / Answer
a)
Note that the probability of x successes out of n trials is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 4
x = the number of successes = 4
Thus, the probability is
P ( 4 ) = 0.195366815 [answer]
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b)
The mean for a 4-hr period is 4 calls/hr * 4 hrs = 16.
Note that the probability of x successes out of n trials is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 16
x = the number of successes = 15
Thus, the probability is
P ( 15 ) = 0.099217532 [answer]
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c)
For a 45-min time, the mean is 4 call/hr * (3/4 hr) = 3 calls.
Note that the probability of x successes out of n trials is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 3
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 0.049787068 [answer]
Again, we expect 3 calls in this period. [answer,3]
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