A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and 10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. Calculate the expected amount paid to the company under this policy during a one-year period.

answer option:

(A) 2,769 (B) 5,000 (C) 7,231 (D) 8,347 (E) 10,578

I will denote X as the event when a snowstorm occurs. where x follows parameter of poisson 1.5 mean.

I will now make the distribution payment of the snowstorm (y) = 0, x<=1, and y=10,000, x>1

So E[X] = 0 * p(x<=1) + 10,000*p(x>1)

so E[X]= 10,000 * p(x>1)

p(x>1)= ....? * this would be an easy problem if the mean was an exponential *lol*. So since this is a discrete case:

would I do: 10,000* p(x=1) + 10,000 * p(x<=1). <---- complement of p(x>1) = p(x<=1), and I know for discrete cases you have to evaluate these things at p(x=1),

for some reason I believe I am missing something here, such as a series of some sort.

answer option:

(A) 2,769 (B) 5,000 (C) 7,231 (D) 8,347 (E) 10,578

I will denote X as the event when a snowstorm occurs. where x follows parameter of poisson 1.5 mean.

I will now make the distribution payment of the snowstorm (y) = 0, x<=1, and y=10,000, x>1

So E[X] = 0 * p(x<=1) + 10,000*p(x>1)

so E[X]= 10,000 * p(x>1)

p(x>1)= ....? * this would be an easy problem if the mean was an exponential *lol*. So since this is a discrete case:

would I do: 10,000* p(x=1) + 10,000 * p(x<=1). <---- complement of p(x>1) = p(x<=1), and I know for discrete cases you have to evaluate these things at p(x=1),

for some reason I believe I am missing something here, such as a series of some sort.

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