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**getback0****Member**- Registered: 2010-10-10
- Posts: 25

An insurance company designates 10% of its customers as high risk and 90% as low risk.

The number of claims made by a customer in a calendar year is Poisson distributed with

mean θ and is independent of the number of claims made by a customer in the previous

calendar year. For high risk customers, θ = 0.6, while for low risk customers θ = 0.1.

Calculate the expected number of claims made in calendar year 1998 by a customer who

made one claim in calendar year 1997.

My solution:

Given that claims made by a customer in a year are independent of those made in the previous year, I disregarded the 1997 claim filing.

For any year, the probability that a customer makes one claim would be -

f(k) = [0.1 * (e ^ -0.6) * (0.6 ^ k) / k! ] + [0.9 * (e ^ -0.1) * (0.1 ^ k) / k! ]

E(x) = ∑ xf(x)

works out to 0.1*0.6 + 0.9*0.1 = 0.15

doesnt match the key

I am way off here, i guess!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,680

Hi getback();

I have this in my notes and .1 and .9 are not the probabilities of a high and low risk. Use Bayes theorem there.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**getback0****Member**- Registered: 2010-10-10
- Posts: 25

Got it - so use baye's theorem to get the probability for high risk and low risk drivers in 1998 given that he had one claim and then use the logic I did

thanks

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,680

Try that and see what you get.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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