SOLUTION: I DEARLY NEED HELP WITH THIS PROBLEM..I HOPE SOMEBODY COULD HELP ME..TANX AND GOODBLESS A large manufacturer has purchased an insurance policy for $500,000 per year to insure i

Question 615052: I DEARLY NEED HELP WITH THIS PROBLEM..I HOPE SOMEBODY COULD HELP ME..TANX AND GOODBLESS
A large manufacturer has purchased an insurance policy for $500,000 per year to insure itself against 4 specific types of losses. The costs associated with each type of loss and their probabilities are listed in the accompanying table. The probability that no loss will occur is .63 and it is assumed that at most one type of loss will occur in a given year. Of the total cost of policy 20% is to cover administrative expenses.
Cost --- Probability
$ 100,000 -- .15
$ 800,000 -- .10
$ 1,500,000 -- .08
$ 2,500,000 -- .04
a) What is the expected profit to the insurance company on this policy?
b) What is the standard deviation of the profits to the insurance company?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A large manufacturer has purchased an insurance policy for $500,000 per year to insure itself against 4 specific types of losses. The costs associated with each type of loss and their probabilities are listed in the accompanying table. The probability that no loss will occur is .63 and it is assumed that at most one type of loss will occur in a given year.
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Of the total cost of policy 20% is to cover administrative expenses.
That would be $100,000
Cost --- Probability
$ 100,000 -- .15
$ 800,000 -- .10
$ 1,500,000 -- .08
$ 2,500,000 -- .04
a) What is the expected profit to the insurance company on this policy?
E(x) = sum of products of (payments*propability)
Note: A payment of 100,000 is really a loss of 400,000 to the purchaser
and a gain of 400,000-100,000(overhead) = 300,000 to the insurance company.
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E(gain to the company) = 300,000*0.15 -400,000*0.10 - 1.1 mill*0.08 -2.1 mil*0.04 + 400,000*0.63
= 45000 - 40000 - 88000 - 84000 + 252000 = $85000
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b) what is the standard deviation?
Ans: $629503.77
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I used my TI-84 to get that standard deviation number.
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Cheers,
Stan H.