The number of requests for assistance received by a towing s

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?

Solution

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]