Question

The total claim amount for a health insurance policy follows a distribution with density function \(f(x)=(1/1000)e^{-x/1000}\) for x>0.
The premium for the policy is set at 100 over the expected total claim amount. If 100 policies are sold, what is the approximate probability that the insurance company will have claims exceeding the premiums collected?
Please, help

Answer 1

\[EV = \int\limits_0^\infty x f(x) dx\]
\[a=\text {total prem} = 100(EV +100)\]
\[P(x>a) = \int\limits _a^\infty f(x) dx\]

Answer 2

https://en.wikipedia.org/wiki/Probability_density_function#Absolutely_continuous_univariate_distributions

Answer 3

is it not that a =1100?

Answer 4

Since f is exponential function with parameter \(\alpha =1000\), hence \(E(x) =\alpha =1000\) And the company wants exceed 100, that is 1100, right?

Answer 5

yes but 100 policies were sold so multiply that by 100 ---> 110,000 to get total prem collected

Answer 6

yes. I got this part

Answer 7

I am on Central theorem, how to turn it into formula of Central theorem?

Answer 8

not sure
is the integral for probability using density function not a good enough answer?

Answer 9

I have to use Central theorem to solve this problem to show that I understand the theorem and can apply it to solve problems.

Answer 10

you have to standardize the sum of the exponential variables
Use the following to get Z-score:
\[Z = \frac{S_n - n \mu}{\sigma \sqrt{n}}\]
\[\mu = \sigma = 1000\]
\[n = 100, S_n = 100*1100\]
Find probability using standard normal table
\[P(Z>1) \approx 0.16\]