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Eunye
April 28th 2008, 02:06 AM
Question: For company A, there is a 60% chance that no claim is made during the coming year. If claims are made, the amount is normally distributed with mean 10,000 and standard deviation 2,000.

For company B, there is a 70% chance that no claim is made during the coming year. If claims are made, the amount is normally distributed with mean 9,000 and standard deviation 2,000.

Assume that the total claim amounts of two companies are independent.

What is the probability that Company B's total claim amount will exceed Company A's total claim amount?

.Godspeed.
April 28th 2008, 09:14 AM
I'm not sure if this will help, but this is the same as question #42 in the Sample Questions here:

http://soa.org/files/pdf/edu-2008-spring-exam-p.pdf

I'm posting this since you may be pulling this question from a source that does not show a suggested solution. If you scroll to the bottom and open up the Solutions, you will find one such solution. Maybe this will help.

zmkramer
April 28th 2008, 11:20 AM
Question: For company A, there is a 60% chance that no claim is made during the coming year. If claims are made, the amount is normally distributed with mean 10,000 and standard deviation 2,000.

For company B, there is a 70% chance that no claim is made during the coming year. If claims are made, the amount is normally distributed with mean 9,000 and standard deviation 2,000.

Assume that the total claim amounts of two companies are independent.

What is the probability that Company B's total claim amount will exceed Company A's total claim amount?

well let's break it down bit by bit

well if company B has a claim and company A does not then company B will have more than company A. that is .3*.6 = .18

if company A and B have a claim then we are looking at pr(B - A > 0 ) the variable B - A has Normal dist with E(B-A) = 9000-10000 and variance (B - A) = var(A) + var(B) so when you use the Normal distribution you get N(-.35) = .3632 you multiply this to the probability of both A and B having claims

.18 + (.4)*(.3)*(.3632) ~ .2235

wonderacle
April 28th 2008, 11:23 AM
B exceeds A if:
1. no claim in A, but claim in B,
2. claims in both A and B, and the mount in B exceed A,

the probability of case 1 is obvious,
since A and B are independent,
P(A=0, B>0)=(.6)*(1-.7)=.18

case 2 is a little bit complicated yet straigtforward,
first, we will compute the probability that there are claims in both A and B,
this is the same theory as case 1,
P(A>0, B>0)=(1-.6)(1-.7)=.12,
then we will calculate the probabiltiy that the amount in B exceeds A,
i will simply use A and B as the variables of amount,
what we are looking for is
P(B>A)=P(B-A>0),
since A and B are independent normal,
variable (B-A) has the following parameters,
mean(B-A)=mean(B)-mean(A)=-1000,
variance(B-A)=variance(B)+variance(A)=((2000)^2)+((2000)^2)= 8,000,000
s.d.(B-A)=2828.4271
then use z-score formula,
z=.35,
P(B-A>0)=1-.6368=.3632
thus, the probability for case 2 is
(.12)(.3632)=.04358

Therefore,
the total probability is
.18+.04358=.22358