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feral
July 18th 2008, 10:29 AM
So...I understand most of how this problem is setup, but there is one part I can't wrap my understanding around:

An auto insurance company insures an automobile worth 13,000 for one year under a policy with a 1,000 deductible. During the policy year there is a 0.08 chance of partial damage to the car and a 0.04 chance of a total loss of the car. If there is partial damage to the car, the amount X of damage (in thousands) follows a distribution with density function
f(x)={0.2601e-x4  for  0<x<13 otherwise

What is the expected claim payment.

Answer:
Y is my claim payment and its distributed
Max (0, x-1) for p=.08
12 for p=.04
0 p=.88

.88 (0) + .08 (.2601) (integral of density function * (x-1)) + (.04)(12).

Can someone explain to me why .04 is multiplied by 12? I know you need to include this .04 since we are dealing with the expected claim payment, but why 12?

NoMoreExams
July 18th 2008, 12:33 PM
So...I understand most of how this problem is setup, but there is one part I can't wrap my understanding around:

An auto insurance company insures an automobile worth 13,000 for one year under a policy with a 1,000 deductible. During the policy year there is a 0.08 chance of partial damage to the car and a 0.04 chance of a total loss of the car. If there is partial damage to the car, the amount X of damage (in thousands) follows a distribution with density function
f(x)={0.2601e-x4  for  0<x<13 otherwise

What is the expected claim payment.

Answer:
Y is my claim payment and its distributed
Max (0, x-1) for p=.08
12 for p=.04
0 p=.88

.88 (0) + .08 (.2601) (integral of density function * (x-1)) + (.04)(12).

Can someone explain to me why .04 is multiplied by 12? I know you need to include this .04 since we are dealing with the expected claim payment, but why 12?

Sounds like you need to review def. of expected value which is the probability of an outcome multiplied by the outcome summed over all the possible outcomes. You are told that there a 4% chance of totalling the car which is worth 13,000 but 1,000 of that the insurance company won't pay since that's the deductible. So if the car is totalled, the insurance company is liable for 12,000, the odds of that loss happening is .04 hence you get .04*12

.Godspeed.
July 20th 2008, 11:48 PM
This is an Exam P-type concept/question that probably ought be moved to the Exam P sub-forum. Mods, do you agree?

feral
July 21st 2008, 07:07 PM
Thanks so much for your speedy response...that was clearly a stupid question!!

Also sorry about posting this in the wrong place...:confused-: