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

Answer:

Given,

X ~ Poisson

Y = 0 where X = 0

Y = 1000 for X = 1

Y = 2000 for X = 2,3,4.......

fx(x) = e^-*^x/x!

here mean = 0.6

E(Y) = 0*e^-0.6 + 1000*0.6*e^-0.6 + 2000*[e^-0.6*0.6^k/k! ]

= 1000*0.6*e^-0.6 + 2000[P(X >= 2)]

= 1000*0.6*e^-0.6 + 2000[1 - P(0) - P(1)]

= 1000*0.6*e^-0.6 + 2000[1 - e^-0.6 - 0.6*e^-0.6]

On solving we get

E(Y) = 573

Now,

E(Y^2) = 1000^2*0.6*e^-0.6 + 2000^2[1 - e^-0.6 - 0.6*e^-0.6]

= 816893

Consider,

Variance V(Y) = E(Y^2) - [E(Y)]^2

substitute values

= 816893 - 573^2

= 816893 - 328329

= 488564

Standard deviation = sqrt(variance(y))

= sqrt(488564)

= 698.9735331

= 699