# Probability calculation with Poisson Distribution

Between 2:00 and 6:00 in the morning there are on average 16 phone calls arriving in a certain hospital. The responsible leaves his room (with the phone) for 10 minutes
a) What is the probability that he doesn't miss a call
b) Assume he has to pay a fine of 5$for each missed call. What is the expectation and the variance of the amount he has to pay as fine. I'm thinking this is a Poisson problem for the first part a) I tried subtracting 0 calls from 1 $$1-((e^{-16})*(16^{10})/0! = 0.9999999$$ but the actual answer is $$0.5134171$$ for b) For this part I don't have much idea how to compute but thought if he is gone for 10 minutes in 4 hours(240 minutes) then $$(1/24)*5$$ should give me the mean and variance since they should be equal for Poisson distibution. The answer for b is $$3.333333$$ I do not have much experience with these types of problems any help here would be appreciated tremendously! • In 4 hours there are an average of 16 calls so 4 calls per hour, 4/6= 2/3 call per 10 minutes. Your$\lambda\$ is 2/3, not 16. Also, in the numerator you take k= 16, in the denominator k= 0. – user247327 Feb 14 at 11:53
• Welcome to MSE. Please edit and use MathJax to properly format math expressions – Lee David Chung Lin Feb 14 at 12:49

For part a), the average number of phone calls from 2 to 6 is 16, so $$\lambda = 16$$ for 4 hours. So for one hour, $$\lambda = 4$$, then he only stays in the room for 50 minutes, so for the remaining 10 mins, $$\lambda = 2/3$$. Thus the probability of him not missing a call is $$P(X=0)=0.5134171$$, when $$\lambda = 2/3$$.
For part b), since for a Poisson distribution, $$E(X)=\lambda$$ (average rate = mean), thus from part a), $$E(X)=\lambda=2/3$$ so to find the expected amount of fine he has to pay, you just need to multiply 2/3 to 5 and get 10/3, which is 3.333333.