Expected Value for Life Insurance There is a 0.9968 probability that a randomly selected 50-year-old female lives through the year (based on data from the U.S. Department of Health and Human Services). A Fidelity life insurance company charges \(226 for insuring that the female will live through the year. If she does not survive the year, the policy pays out \)50,000 as a death benefit.

a. From the perspective of the 50-year-old female, what are the values corresponding to the two events of surviving the year and not surviving.

b. If a 50-year-old female purchases the policy, what is her expected value?

c. Can the insurance company expect to make a profit from many such policies? Why

Short Answer

Expert verified

a. From the perspective of the 50-year oldfemale, the monetary value of the event of surviving through the year is equal to -$226, and the monetary value for the event not surviving is equal to $49,774.

b. The expected value of purchasing an insurance policy is equal to -66 dollars.

c. The insurance company makes a lot of profit by selling many such policies because the insurance company earns 66 dollars on each policy sold.

Step by step solution

01

Given information

The probability of a 50-year oldfemale living through the year is equal to 0.9968.

The insurance company charges $226 for one year and gives $50,000 on the death of a woman.

02

Monetary values

a.

The monetary value of the event of surviving through the year is equal to -$226 as the charges of the insurance policy need to be paid.

Corresponding to the event of not surviving, any particular woman will receive $50,000.

Therefore, the corresponding monetary value is equal to the following:

50,000-226=49,774dollars

Thus, the monetary value of the event not surviving is equal to $49,774.

03

Expected value

b.

The expected value of purchasing the insurance policy is equal to the following:

Expectedvalue=Netamountreceived×Probabilityofnotsurviving-Netamountgiven×Probabilityofsurviving=497741-0.9968-2260.9968=-66dollars

Thus, the expected value of purchasing an insurance policy is equal to -66 dollars.

04

Profit for the insurance company

c.

Since on each policy sold, the insurance company earns 66 dollars, the insurance company makes a lot of profit by selling many such policies.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.

a. For a randomly selected subject, find the probability of a bone density test score less than 1.54.

b. For a randomly selected subject, find the probability of a bone density test score greater than -1.54.

c. For a randomly selected subject, find the probability of a bone density test score between -1.33 and 2.33.

d. Find \({Q_1}\), the bone density test score separating the bottom 25% from the top 75%.

e. If the mean bone density test score is found for 9 randomly selected subjects, find the probability that the mean is greater than 0.50.

The accompanying table lists probabilities for the corresponding numbers of girls in four births. What is the random variable, what are its possible values, and are its values numerical?

Number of Girls x

P(x)

0

0.063

1

0.250

2

0.375

3

0.250

4

0.063

Currently, an average of 7 residents of the village of Westport (population 760) die each year (based on data from the U.S. National Center for Health Statistics).

a. Find the mean number of deaths per day.

b. Find the probability that on a given day, there are no deaths.

c. Find the probability that on a given day, there is more than one death.

d. Based on the preceding results, should Westport have a contingency plan to handle more than one death per day? Why or why not?

Binomial Probability Formula. In Exercises 13 and 14, answer the questions designed to help understand the rationale for the binomial probability formula.

Guessing Answers Standard tests, such as the SAT, ACT, or Medical College Admission Test (MCAT), typically use multiple choice questions, each with five possible answers (a, b, c, d, e), one of which is correct. Assume that you guess the answers to the first three questions.

a. Use the multiplication rule to find the probability that the first two guesses are wrong and the third is correct. That is, find P(WWC), where W denotes a wrong answerand C denotes a correct answer.

b.Beginning with WWC, make a complete list of the different possible arrangements of two wrong answers and one correct answer, then find the probability for each entry in the list.

c. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made?

In Exercises 7–14, determine whether a probability

distribution is given. If a probability distribution is given, find its mean and standarddeviation. If a probability distribution is not given, identify the requirements that are notsatisfied.

Ted is not particularly creative. He uses the pickupline “If I could rearrange the alphabet, I’d put U and I together.”The random variable xis the number of women Ted approachesbefore encountering one who reacts positively.

x

P(x)

1

0.001

2

0.009

3

0.030

4

0.060

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free