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18. Life insurance

(a) It would be quite risky for you to insure the life of a 21-year-old friend under the terms of Exercise 14. There is a high probability that your friend would live and you would gain \(1250in premiums. But if he were to die, you would lose almost \)100,000. Explain carefully why selling insurance is not risky for an insurance company that insures many thousands of 21-year-old men.

(b) 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 riskier the
investment. We can measure the great risk of insuring a single person’s life in Exercise 14by computing the standard deviation of the income Y that the insurer will receive. Find σY using the distribution and mean found in Exercise 14.

Short Answer

Expert verified

(a) Risk is low because large number of policy holders are involved.

(b) The standard deviation is $9708.

Step by step solution

01

Part (a) Step 1:Given information 

Given in the question that the high probability would gain $1250in premiums. If die, would lose almost $100,000. Selling insurance is not risky for an insurance company that insures many thousands of 21-year-old men.

02

Part (a) Step 2: Explanation 

According to the information, the gain around is$1250.
Amount would lost in case of death is$100,000
The expected value with its probability is:
E(X)=∑x×P(x)=(−99750)×0.00183+….+(1250)×0.99058=303.35

03

Part (b) Step 1: Given information

The standard deviation of the investment return is used to determine the risk of an investment. The riskier the investment is, the more varied the return is.

04

Part (b) Step 2: Explanation 

The standard deviation ofY can be determined as:
σ=∑x2×P(x)-∑x×P(x)2=(−99750−303.3525)2×0.00183+……………..+(1250−303.3525)2×0.99058=9708

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Most popular questions from this chapter

14. . Life insurance A life insurance company sells a term insurance policy to a 21-year-old male that pays \(100,000if the insured dies within the next 5years. The probability that a randomly chosen male will die each year can be found in mortality tables. The company collects a premium of \)250each year a payment for the insurance. The amount Ythat the company earns on this policy is \(250per year, less the \)100,000that it must pay if the insured dies. Here is a partially completed table that shows information about risk of mortality and the values of Y=profit earned by the company:

(a) Copy the table onto your paper. Fill in the missing values of Y.
(b) Find the missing probability. Show your work.
(c) Calculate the mean μY.Interpret this value in context

21. Random numbers Let Xbe a number between 0and 1produced by a random number generator. Assuming that the random variable X has a uniform distribution, find the following probabilities:
(a) P(X>0.49)
(b) P(X≥0.49)
(c) P(0.19≤X<0.37or0.84<X≤1.27)

45. Too cool at the cabin? During the winter months, the temperatures at the Stameses' Colorado cabin can stay well below freezing 32°For 0°Cfor weeks at a time. To prevent the pipes from freezing, Mrs. Stames sets the thermostat at 50°F. She also buys a digital thermometer that records the indoor temperature each night at midnight. Unfortunately, the thermometer is programmed to measure the temperature in degrees Celsius. Based on several years' worth of data, the temperature Tin the cabin at midnight on a randomly selected night follows a Normal distribution with mean 8.5°Cand standard deviation2.25°C.
(a) Let Y=the temperature in the cabin at midnight on a randomly selected night in degrees Fahrenheit (recall that F=(9/5)C+32). Find the mean and standard deviation of Y.

(b) Find the probability that the midnight temperature in the cabin is below 40°F. Show your work.

25. Did you vote? A sample survey contacted an SRS of 663 registered voters in Oregon shortly after an election and asked respondents whether they had voted. Voter records show that 56%of registered voters had actually voted. We will see later that in repeated random samples of size 663, the proportion in the sample who voted (call this proportion V) will vary according to the Normal distribution with mean μ=0.56 and standard deviation σ=0.019.
(a) If the respondents answer truthfully, what is P(0.52≤V≤0.60)? This is the probability that the sample proportion V estimates the population
proportion 0.56 within ±0.04.
(b) In fact, 72%of the respondents said they had voted (V=0.72). If respondents answer truthfully, what is P(V≥0.72)? This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.

To introduce her class to binomial distributions, Mrs. Desai gives a 10 -item, multiple-choice quiz. The catch is, that students must simply guess an answer (A through E) for each question. Mrs. Desai uses her computer's random number generator to produce the answer key so that each possible answer has an equal chance to be chosen. Patti is one of the students in this class. Let X=the number of Patti's correct guesses.

1. Show that Xis a binomial random variable.

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