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!
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.