# Normal Distributions and Claim Amounts

Suppose for company A, there is a $60 \%$ chance that no claim is made in the coming year. If one or more claims are made, then the total amount is normally distributed with mean $1000$ and standard deviation $2000$.

Suppose for company B, there is a $70 \%$ chance that no claim is made during the coming year. If one or more claims are made, then the total amount is normally distributed with mean $9000$ and standard deviation $2000$.

Assume the total claim amounts are independent. What is the probability that in the coming year, company B's total claim amount will exceed company A's total claim amount?

Let $X_A$ be company A's total claim amount and $X_B$ likewise. We want to find $P(X_B - X_A >0)$. Now $E[X_B-X_A] = 8000$ and $\text{Var}[X_B-X_A] = 2000^2+2000^2$. So what we want to find is:

$\displaystyle P \left[Z \geq \frac{-8000}{\sqrt{2000^2+2000^2}} \right]$?

I feel that we are given two conditional distributions: $P(X_A|I_A)$ and $P(X_B|I_B)$ where $I_A$ and $I_B$ indicate that more than one claim is made for company A and B respectively.

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 I do state the question explicitly. – PEV Oct 26 '10 at 20:31 You are right, sorry 'bout that. I have removed the remark. – Ross Millikan Oct 26 '10 at 20:46 So what we want to find is the following: $P\{[I_{A}^{C} \cap I_B] \cup [(I_A \cap I_B) \cap (X_A < X_B)] \}$? – PEV Oct 26 '10 at 23:34 Exactly. And the two sides of your union are nicely disjoint. – Ross Millikan Oct 27 '10 at 0:06