Question 4.154

4.154 Risk for one versus many life insurance policies.

It would be quite risky for an insurance company to insure the life of only one 25-year-old man under the terms of Exercise 4.153. There is a high probability that person would live and the company would gain $875 in premiums. But if he were to die, the company would lose almost $100,000. We have seen that the risk of an investment is often measured by the standard deviation of the return on the investment. The more variable the return is (the larger σ is), the riskier the investment.

  1. Suppose only one person's life is insured. Compute standard deviation of the income that the insurer will receive. Find , using the distribution and mean you found in Exercise 4.153.
  2. Suppose that the insurance company insures two men. Define the total income as where is the income made from man . Find the mean and standard deviation of .
  3. You should have found that the standard deviation computed in part (b) is greater than that found in part (a). But this does not necessarily imply that insuring two people is riskier than insuring one person. What needs to be recognized is that the mean income has also gone up. So, to measure the riskiness of each scenario we need to scale the standard deviation values relative to the mean values. This is simply done by computing , which is called the coefficient of variation (CV). Compute the coefficients of variation for insuring one person and for insuring two people. What do the CV values suggest about the relative riskiness of the two scenarios?
  4. Compute the mean total income, standard deviation of total income, and the CV of total income when 30 people are insured.

    242

  5. Compute the mean total income, standard deviation of total income, and the CV of total income when 1000 people are insured.
  6. There is a remarkable result in probability theory that states that the sum of a large number of independent random variables follows approximately the Normal distribution even if the random variables themselves are not Normal. In most cases, 30 is sufficiently “large.” Given this fact, use the mean and standard deviation from part (d) to compute the probability that the insurance company will lose money from insuring 30 people—that is, compute . Compute now the probability of a loss to the company if 1000 people are insured. What did you learn from these probability computations?