Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Convert the two recurrence intervals into annual Poisson intensities and add them.

λ1=110=0.10,λ2=150=0.02\lambda_1=\frac1{10}=0.10,\qquad \lambda_2=\frac1{50}=0.02
λannual=0.10+0.02=0.12\lambda_{\text{annual}}=0.10+0.02=0.12

Model

Model

Independent Poisson counts add, and intensity scales linearly with the observation period. Let N be the total event count.

N∼Poisson⁡(40(0.12))=Poisson⁡(4.8)N\sim\operatorname{Poisson}(40(0.12))=\operatorname{Poisson}(4.8)
T=1000NT=1000N

Compute

Compute

Locate the peak by comparing adjacent Poisson masses. The probabilities increase through count four and decrease when moving to count five.

Pr⁡(N=n)Pr⁡(N=n−1)=4.8n\frac{\Pr(N=n)}{\Pr(N=n-1)}=\frac{4.8}{n}
Pr⁡(N=4)Pr⁡(N=3)=1.20>1\frac{\Pr(N=4)}{\Pr(N=3)}=1.20>1
Pr⁡(N=5)Pr⁡(N=4)=0.96<1\frac{\Pr(N=5)}{\Pr(N=4)}=0.96<1

Answer

Answer

The count mode is four, and the one-to-one payment scale makes the payment mode 4,000.

4000(C)\boxed{4000\quad\text{(C)}}