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Chapter 8: Problem 16

A man wishes to purchase a 5 -yr term-life insurance policy that will pay the beneficiary $$\$ 20,000$$ in the event that the man's death occurs during the next 5 yr. Using life insurance tables, he determines that the probability that he will live another \(5 \mathrm{yr}\) is \(.96\). What is the minimum amount that he can expect to pay for his premium?

Short Answer

Expert verified
The minimum amount the man can expect to pay for his premium is $800.

Step by step solution

01

Calculate the probability of dying within the next 5 years

Since the probability of the man living another 5 years is 0.96, we can find the complementary probability of him dying within the next 5 years by subtracting this from 1: Probability of dying = \(1 - 0.96 = 0.04\)
02

Calculate the minimum premium amount

Since the policy will pay the beneficiary \(20,000 if the man dies within the next 5 years, the insurance company will make this payment with a probability of 0.04. To cover their potential losses, the company will collect insurance premiums from the man. The minimum amount the man can expect to pay for his premium can be found by multiplying the payout (\)20,000) by the probability of dying (0.04): Minimum premium amount = \(20000 \times 0.04 = 800\) So, the minimum amount that the man can expect to pay for his premium is $800.

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Key Concepts

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

Term-Life Insurance Policy
When discussing life insurance, a 'term-life insurance policy' is a pivotal concept that deserves attention. In essence, it is a type of life insurance that provides coverage at a fixed rate of payments for a limited period, the relevant 'term'. After that period expires, coverage is no longer guaranteed, and the client must either forgo coverage or potentially obtain further coverage with different payments or conditions.

These policies are significant in financial planning, particularly for individuals seeking peace of mind in knowing they can provide for their beneficiaries in the event of their untimely death. This type of policy is succinctly defined by its simplicity and typically lower initial costs compared to permanent life insurance.
Probability Calculation
The 'probability calculation' is a fundamental aspect of both life insurance models and broader statistical analysis. Probability, mathematically represented as a number between 0 and 1, quantifies the likelihood of an event. When an event is certain not to happen, the probability is 0, and when it is certain to occur, the probability is 1.

To calculate the probability of an individual event within a certain time frame (like death within 5 years for life insurance), actuaries use life tables that provide statistical probabilities based on factors such as age, gender, and sometimes even lifestyle choices or health status. This numerical calculation informs critical decision-making, such as premium setting in the context of life insurance.
Complementary Probability
The notion of 'complementary probability' addresses the probability of the occurrence of the opposite (complementary) event. For every event there exists another event that is the opposite, and the sum of the probabilities of these two events must equal 1.

For example, if the probability of living for 5 more years is 0.96, then the complementary probability, or the probability of dying within those 5 years, is calculable as 1 - 0.96, which equals 0.04. This concept is especially useful in life insurance calculations, where insurers need to evaluate both the likelihood of a payout event occurring and its opposite to gauge risks.
Expected Value
In the realm of statistics and probability, 'expected value' is a concept that refers to the anticipated value of a variable, calculated as the sum of all possible values each multiplied by the probability of its occurrence. It represents the average amount one can expect to be paid out or to pay in a large number of similar scenarios.

In life insurance, the expected value of a policy to the company is the payout amount multiplied by the probability of the event (the insured person's death) happening within the insured term. This figure helps insurance companies to determine the minimum premium that they should charge to ensure that they will, on average, cover the payouts they are obliged to make, thus maintaining profitability.

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Most popular questions from this chapter

The number of average hours worked per year per worker in the United States and five European countries in 2002 is given in the following table: $$\begin{array}{lcccccc}\hline & & & \text { Great } & & \text { West } & \\\\\text { Country } & \text { U.S. } & \text { Spain } & \text { Britain } & \text { France } & \text { Germany } & \text {Norway } \\ \hline \text { Average } & & & & & & \\\\\text { Hours } & 1815 & 1807 & 1707 & 1545 & 1428 & 1342 \\ \text { Werked } & & & & & & \\ \hline\end{array}$$ Find the average of the average hours worked per worker in 2002 for workers in the six countries. What is the standard deviation for these data?

Two dice are rolled. Let the random variable \(X\) denote the number that falls uppermost on the first die, and let \(Y\) denote the number that falls uppermost on the second die. a. Find the probability distributions of \(X\) and \(Y\). b. Find the probability distribution of \(X+Y\).

Records kept by the chief dietitian at the university cafeteria over a 30-wk period show the following weekly consumption of milk (in gallons). $$\begin{array}{lccccc}\hline \text { Milk } & 200 & 205 & 210 & 215 & 220 \\\\\hline \text { Weeks } & 3 & 4 & 6 & 5 & 4 \\\\\hline\end{array}$$ $$\begin{array}{lcccc}\hline \text { Milk } & 225 & 230 & 235 & 240 \\\\\hline \text { Weeks } & 3 & 2 & 2 & 1 \\ \hline\end{array}$$ a. Find the average number of gallons of milk consumed per week in the cafeteria. b. Let the random variable \(X\) denote the number of gallons of milk consumed in a week at the cafeteria. Find the probability distribution of the random variable \(X\) and compute \(E(X)\), the expected value of \(X\).

A probability distribution has a mean of 50 and a standard deviation of \(1.4\). Use Chebychev's inequality to find the value of \(c\) that guarantees the probability is at least \(96 \%\) that an outcome of the experiment lies between \(50-c\) and \(50+c .\)

The interest rates paid by 30 financial institutions on a certain day for money market deposit accounts are shown in the accompanying table: $$\begin{array}{lcccc}\hline \text { Rate, \% } & 6 & 6.25 & 6.55 & 6.56 \\ \hline \text { Institutions } & 1 & 7 & 7 & 1 \\ \hline\end{array}$$ $$\begin{array}{lcccc} \hline \text { Rate, } \% & 6.58 & 6.60 & 6.65 & 6.85 \\\\\hline \text { Institutions } & 1 & 8 & 3 & 2 \\ \hline\end{array}$$ Let the random variable \(X\) denote the interest rate paid by a randomly chosen financial institution on its money market deposit accounts and find the probability distribution associated with these data.
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