Answer to Question #119572 in Statistics and Probability for Kyei William Frimpong

Question

A baseball team has scheduled its opening game for April 5. It is assume that if it
snows on April 5, 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 expected cost to the nearest one Ghana cedi 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: $X\sim Po(\lambda)$

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

Given $\lambda=0.6$

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

$GHS 0:X=0$

$GHS\ 1,000:X=1$

$GHS 2,000:X≥2$

Then

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

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

$$P(Y=2000)=P(X≥2)=1−P(X=0)−P(X=1)=$$

$$=1-e^{-0.6}-0.6e^{-0.6}=1-1.6e^{-0.6}$$

$$E(Y)=0⋅e^{-0.6}+1000⋅0.6e^{-0.6}+2000⋅(1−1.6e^{-0.6})=$$

$$=2000-2600e^{-0.6}\approx573.089746$$

$$E(Y^2 )=(0)^2⋅e^{-0.6}+(1000)^2⋅0.6e^{-0.6}+$$

$$+(2000)^2⋅(1−1.6e^{-0.6})=$$

$$=4000000−5800000e^{-0.6} ≈816892.510655$$

$$Var(Y)=σ^ 2 =E(Y^ 2 )−(E(Y))^ 2 ≈$$

$$≈816892.510655−(573.089746)^ 2≈488460.653685$$

$$σ=\sqrt{\sigma^2} ≈ 488460.653685≈699$$

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