Insurance claims follow a Poisson process with a rate 100 cl

Insurance claims follow a Poisson process with a rate =100 claims per year.

What is the mean and standard deviation of the number of claims per year?

What is the probability of receiving more than 100 claims per year?

What is the probability of receiving exactly 100 claims in a year?

Why is the probability of receiving exactly 100 claims in a year small?

What is the probability of no claims in a year?

Solution

a)

mean = lambda = 100
standard deviation = sqrt(lambda) = 10 [answers]

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b)

Note that P(more than x) = 1 - P(at most x).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    100      
          
x = our critical value of successes =    100      
          
Then the cumulative probability of P(at most x) from a table/technology is          
          
P(at most   100   ) =    0.526562199
          
Thus, the probability of at least   101   successes is  
          
P(more than   100   ) =    0.473437801 [answer]

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c)

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 =    100      
          
x = the number of successes =    100      
          
Thus, the probability is          
          
P (    100   ) =    0.039860997 [answer]

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d)

It is relatively small, just around 4%.

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e)

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 =    100      
          
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    3.72008*10^-44 [answer]