# Probability question involving exponential distribution.

• Hodgey8806
In summary, the conversation discusses the calculation of the approximate probability of an insurance company having claims exceeding the premiums collected for 100 policies sold. The solution involves using the expected value of the claim, expected value of the premium, and a normal approximation. However, the solution provided is incorrect. The correct solution would involve calculating the probability that a single claim exceeds the premium and using it to determine the probability that none of the 100 policies will exceed the premium.
Hodgey8806

## Homework Statement

The total claim for a health insurance policy follows a distribution with density function

f(x) = (1/1000)e$^{-x/1000}$, 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 some claims exceeding the premiums collected?

## Homework Equations

I want to understand their method a bit. Would this binomial expression be approximate?
They use the expected value of claim, expected value of premium, and a normal approximation.

## The Attempt at a Solution

My solution:
The probability of a claim being greater than 110000 = e^(-110)
Thus the probability of being less is 1-e^(-110)
P(N≥1) = 1-P(N=0) = 1 - (1)(1-e^(-110)$^{100}$(e^(-110))$^{0}$
I don't have a program to evaluate this, so would they be approximate?

Their solution:
100*E(X) = 100*1000 = 100000
100*E(X+100) = 100*1100 = 110000
σ=10000

P(X>110000) = P(z>1) = .1587

Thanks!

Hodgey8806 said:

## Homework Statement

The total claim for a health insurance policy follows a distribution with density function

f(x) = (1/1000)e$^{-x/1000}$, 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 some claims exceeding the premiums collected?

## Homework Equations

I want to understand their method a bit. Would this binomial expression be approximate?
They use the expected value of claim, expected value of premium, and a normal approximation.

## The Attempt at a Solution

My solution:
The probability of a claim being greater than 110000 = e^(-110)
Thus the probability of being less is 1-e^(-110)
P(N≥1) = 1-P(N=0) = 1 - (1)(1-e^(-110)$^{100}$(e^(-110))$^{0}$
I don't have a program to evaluate this, so would they be approximate?

Their solution:
100*E(X) = 100*1000 = 100000
100*E(X+100) = 100*1100 = 110000
σ=10000

P(X>110000) = P(z>1) = .1587

Thanks!

Your solution is incorrect, but so it the one you say "they" give.

First, for a single policy, what is the premium? The expected claim = 1000, so the premium is 1100 (\$). What is the probability that a single claim exceeds the premium? That is p = Pr{X > 1100} = exp(-1100/1000) = exp(-11/10) ≈ 0.3328710837.

So, for 100 independence policies sold, the probability that NONE of them exceed the premium is the probability that NO 'successes' occur in 100 independent trials with success probability p per trial.

Thats the way I tried it the first time. But I was thinking they might have been asking for individual claims that exceed the total expected premium.

So I wantes to verify if their solution ia actually an approximation of what I did.

Thanks!

## 1. What is the exponential distribution?

The exponential distribution is a probability distribution that is commonly used to model the time between events in a Poisson process. It is often used in situations where the probability of an event occurring is constant over time, such as radioactive decay or the failure of a machine.

## 2. How is the exponential distribution different from other probability distributions?

The exponential distribution is unique in that it only has one parameter, λ (lambda), which represents the rate parameter. Other distributions, such as the normal distribution, have multiple parameters that determine their shape and characteristics.

## 3. How is the exponential distribution related to the Poisson distribution?

The exponential distribution is closely related to the Poisson distribution, as both are used to model the time between events in a Poisson process. In fact, the exponential distribution can be derived from the Poisson distribution by taking the limit as the number of events approaches infinity.

## 4. How do you calculate probabilities with the exponential distribution?

To calculate probabilities with the exponential distribution, you can use the formula P(X>x) = e^(-λx), where x is the time period and λ is the rate parameter. This formula gives the probability that an event will occur after the specified time period.

## 5. What are some real-world applications of the exponential distribution?

The exponential distribution has many real-world applications, including modeling the lifetimes of electronic components, the time between customer arrivals in a queue, and the length of phone calls. It is also commonly used in survival analysis to estimate the time until an event, such as death or failure, occurs.

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