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Chapter 3: Problem 206

Accident records collected by an automobile insurance company give the following information. The probability that an insured driver has an automobile accident is \(.15 .\) If an accident has occurred, the damage to the vehicle amounts to \(20 \%\) of its market value with a probability of \(.80,\) to \(60 \%\) of its market value with a probability of \(.12,\) and to a total loss with a probability of .08. What premium should the company charge on a 12,000 dollar car so that the expected gain by the company is zero?

Short Answer

Expert verified
The premium should be $561.60.

Step by step solution

01

Calculate the Expected Damage Cost

Let's calculate the expected damage cost if an accident occurs. Each damage level is assessed as follows:- Damage to 20%: Probability of 0.80, so the cost is \(0.20 \times 12,000 = 2,400\).- Damage to 60%: Probability of 0.12, so the cost is \(0.60 \times 12,000 = 7,200\).- Total loss (100% damage): Probability of 0.08, so the cost is \(1.00 \times 12,000 = 12,000\).Now, calculate the expected damage cost:\[\text{Expected Damage Cost} = (0.80 \times 2,400) + (0.12 \times 7,200) + (0.08 \times 12,000)\]Calculate the values:\[= 1920 + 864 + 960 = 3744\]
02

Calculate the Expected Cost for the Insurance Company

With a 15% probability of an accident occurring, the expected cost resulting from an accident is:\[\text{Probability of Accident} \times \text{Expected Damage Cost} = 0.15 \times 3744 = 561.6\]This is the expected cost the insurance company will incur per driver insured.
03

Determine the Insurance Premium

To ensure the expected gain for the company is zero, the premium charged must counterbalance the expected cost. Hence, the premium should be equal to the expected cost of 561.6 dollars.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
Expected value is a fundamental concept in probability and statistics. When dealing with uncertain outcomes, the expected value gives us a way to compute a single number that summarizes the average outcome over many trials or instances.

In the context of our exercise, each potential damage scenario鈥攄amage to 20%, damage to 60%, and a total loss鈥攊s associated with a probability and a corresponding cost. The expected damage cost is computed by multiplying each damage cost by its respective probability and summing up all these products.

This is how we calculate the expected damage cost:
  • 20% damage chances result in a cost of 2,400 dollars with probability 0.80.
  • 60% damage leads to 7,200 dollars of cost with probability 0.12.
  • Total loss (100%) means 12,000 dollars cost, occurring with probability 0.08.
The expected value of damage cost thus is the sum of these probability-weighted outcomes, which turned out to be 3,744 dollars.

By using expected value, insurance companies can predict average expenses and set premiums accordingly.
Insurance Mathematics
Insurance mathematics is a specialized branch of mathematics where concepts of probability and statistics are applied to ensure financial viability and fair pricing in the insurance industry.

When setting an insurance premium, companies aim to balance potential payouts with the premiums collected to ensure an average net gain close to zero.

The exercise involved determining a fair insurance premium. Here, the company calculated how much, on average, they expect to pay for accidents per insured driver (561.6 dollars in this case).
  • This includes understanding not just individual incidents, but the spread and probability of these costs across all policyholders.
  • The goal is to ensure that premiums cover the expected losses due to accidents.

This way, the insurance company remains financially solvent while providing meaningful coverage to their customers. Insurance mathematics, therefore, deals with risk assessment and premium calculation, essential for both the company and the insured.
Damage Probability
The concept of damage probability is crucial to understanding how insurance companies assess risk and cost. It relates to the likelihood of different damage scenarios occurring and their impacts on insurance policies.

In this problem, the company considers three main damage scenarios, each with a defined probability:
  • 20% damage with an 80% probability.
  • 60% damage with a 12% probability.
  • Total loss with an 8% probability.

The company uses these probabilities to predict average damages across all insured cars.

By calculating the expected damage cost (3744 dollars), these damage probabilities help the company understand how much they are likely to pay out per accident.

This understanding will help the company set appropriate premiums to cover these costs while providing necessary risk coverage to the policyholder. By accurately assessing damage probabilities, insurers can maintain competitiveness and solvency.

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Most popular questions from this chapter

It is known that \(5 \%\) of the members of a population have disease \(A,\) which can be discovered by a blood test. Suppose that \(N\) (a large number) people are to be tested. This can be done in two ways: 1\. Each person is tested separately, or 2\. the blood samples of \(k\) people are pooled together and analyzed. (Assume that \(N=n k\), with \(n\) an integer.) If the test is negative, all of them are healthy (that is, just this one test is needed). If the test is positive, each of the \(k\) persons must be tested separately (that is, a total of \(k+1\) tests are needed). a. For fixed \(k,\) what is the expected number of tests needed in option \(2 ?\) b. Find the \(k\) that will minimize the expected number of tests in option 2 . c. If \(k\) is selected as in part (b), on the average how many tests does option 2 save in comparison with option 1?

If \(Y\) is a geometric random variable, define \(Y^{*}=Y-1 .\) If \(Y\) is interpreted as the number of the trial on which the first success occurs, then \(Y^{*}\) can be interpreted as the number of failures before the first success. If \(Y^{*}=Y-1, P\left(Y^{*}=y\right)=P(Y-1=y)=P(Y=y+1)\) for \(y=0,1,2, \ldots .\) Show that $$P\left(Y^{*}=y\right)=q^{y} p, \quad y=0,1,2, \dots$$ The probability distribution of \(Y^{*}\) is sometimes used by actuaries as a model for the distribution of the number of insurance claims made in a specific time period.

Would you rather take a multiple-choice test or a full-recall test? If you have absolutely no knowledge of the test material, you will score zero on a full-recall test. However, if you are given 5 choices for each multiple-choice question, you have at least one chance in five of guessing each correct answer! Suppose that a multiple-choice exam contains 100 questions, each with 5 possible answers, and guess the answer to each of the questions. a. What is the expected value of the number \(Y\) of questions that will be correctly answered? b. Find the standard deviation of \(Y\). c. Calculate the intervals \(\mu \pm 2 \sigma\) and \(\mu \pm 3 \sigma\) d. If the results of the exam are curved so that 50 correct answers is a passing score, are you likely to receive a passing score? Explain.

Sampling for defectives from large lots of manufactured product yields a number of defectives, \(Y\), that follows a binomial probability distribution. A sampling plan consists of specifying the number of items \(n\) to be included in a sample and an acceptance number a. The lot is accepted if \(Y \leq a\) and rejected if \(Y>a\). Let \(p\) denote the proportion of defectives in the lot. For \(n=5\) and \(a=0\), calculate the probability of lot acceptance if a. \(p=0\) b. \(p=.1\) C. \(p=.3\) d. \(p=.5\) e. \(p=1.0\) A graph showing the probability of lot acceptance as a function of lot fraction defective is called the operating characteristic curve for the sample plan. Construct the operating characteristic curve for the plan \(n=5, a=0 .\) Notice that a sampling plan is an example of statistical inference. Accepting or rejecting a lot based on information contained in the sample is equivalent to concluding that the lot is either good or bad. "Good" implies that a low fraction is defective and that the lot is therefore suitable for shipment.

A corporation is sampling without replacement for \(n=3\) firms to determine the one from which to purchase certain supplies. The sample is to be selected from a pool of six firms, of which four are local and two are not local. Let \(Y\) denote the number of nonlocal firms among the three selected. a. \(P(Y=1)\) b. \(P(Y \geq 1)\) c. \(P(Y \leq 1)\)
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