Question

A baseball team has scheduled it's opening game for April 1. It is assumed that if it snows on April 1, the game is postponed and will be played on the next day that it does not snow. The team purchased insurance against snow. The policy will pay USD 1,000 for each day, up to 2 days that the game is postponed.
It is determined that the number of consecutive days of snow beginning on April 1, is a Poisson random variable with
mean 0.6.
What is the standard deviation of the amount that the insurance company
will have to pay.

Answer 1

Y - amount we need to pay

P(X=0)=P(Y=0)=e^-0.6=0.5488

P(X=1)=P(Y=1000)=0.6e^-0.6=0.3293

P$(X \geq 2)$=P(Y=2000)=1-1.6e^-0.6=0.1219

EY=0*0.5488+1000*0.3293+2000*0.1219=573.1

EY²=0²*0.5488+1000²*0.3293+2000²*0.1219=816900

VarY=EY²-(EY)²=816900-573.1²=488456

SD=$\sqrt{488456}$ =699