Answer to Question #115403 in Statistics and Probability for Popcy

Question

A baseball team has scheduled its opening game for April 1. It is assume that if it snows on April 1, the game is postponed and will be play on the next day that, it does not snow. The team purchased insurance against snow. The policy will pay GHS 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

Let $X=$ the number of days of consecutive days of snow beginning on April 1: $X\sim P(\lambda)$

$$P(X=x)={e^{-\lambda}\lambda^x\over x!}$$

Given $\lambda=0.6.$

Let $Y=$ the amount that the insurance company must pay. The possible values for $Y$ would be:

$GHS\ 0:X=0$

$GHS\ 1,000:X=1$

$GHS\ 2,000: X\geq2$

Then

$$P(Y=0)=P(X=0)={e^{-0.6}0.6^0\over 0!}=e^{-0.6}$$

$$P(Y=1000)=P(X=1)={e^{-0.6}0.6^1\over 1!}=0.6e^{-0.6}$$

$$P(Y=2000)=P(X\geq2)=1-P(X=0)-P(X=1)=$$ $$=1-e^{-0.6}-0.6e^{-0.6}=1-1.6e^{-0.6}$$

$$E(Y)=0\cdot e^{-0.6}+1000\cdot 0.6e^{-0.6}+2000\cdot (1-1.6e^{-0.6})=$$ $$=2000-2600e^{-0.6}\approx573.089746$$

$$E(Y^2)=0^2\cdot e^{-0.6}+1000^2\cdot 0.6e^{-0.6}+2000^2\cdot (1-1.6e^{-0.6})=$$ $$=4000000-5800000e^{-0.6}\approx816892.510655$$

$$Var(Y)=\sigma^2=E(Y^2)-(E(Y))^2\approx$$ $$\approx816892.510655-(573.089746)^2\approx488460.653685$$

$$\sigma=\sqrt{\sigma^2}\approx\sqrt{488460.653685}\approx699$$

The standard deviation of the amount that the insurance company will have to pay is GHS 699.