/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 63 Life insurance The risk of insur... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Q. 63 (page 392)

Life insurance The risk of insuring one person's life is reduced if we insure many people. Suppose that we randomly select two insured 21-year-old males, and that their ages at death are independent. If X1 and X2 are the insurer's income from the two insurance policies, the insurer's average income W on the two policies is

W=X1+X222W=X1+X22

Find the mean and standard deviation of W. (You see that the mean income is the same as for a single policy, but the standard deviation is less.)

Short Answer

Expert verified

The required value for Mean of W,

渭W=$303.35

The required value for Standard deviation of W ,

蟽W=$6864.29

Step by step solution

01

Given Information

X:amount earned by Life Insurance Company on a 5-year term life insurance chosen at random.

Mean,

渭X=$303.35

The standard deviation (SD)

蟽X=$9707.57

The insurer's income from two insurance policies isX1and X2

So that

The average income of the insurer on two policies,

W=X1+X22

02

Find the mean and standard deviation of W

When both X and Y are independent,

Property mean:

渭aX+bY=a渭X+b渭Y

Property variance:

蟽aX+bY2=a2渭X2+b2渭Y2

Since W represents insurer's average income.

Then

W=X1+X22=0.5X1+0.5X2

Thus,

We have

Mean of W,

渭W=渭0.5x1+0.5x2=0.5渭X+0.5渭X=0.5(303.35)+0.5(303.35)=$303.35

Standard deviation of W,

蟽W=蟽0.5x1+0.5x2=0.52蟽x2+0.52蟽x2=0.25(9707.57)2+0.25(9707.57)2鈮�$6864.29

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 91影视!

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

Exercises 21 and 22 examine how Benford鈥檚 law (Exercise 9) can be used to detect fraud.

Benford鈥檚 law and fraud A not-so-clever employee decided to fake his monthly expense report. He believed that the first digits of his expense amounts should be equally likely to be any of the numbers from 1 to 9. In that case, the first digit Yof a randomly selected expense amount would have the probability distribution shown in the histogram.

(a) What鈥檚 P(Y<6)? According to Benford鈥檚 law (see Exercise 9), what proportion of first digits in the employee鈥檚 expense amounts should be greater than 6? How could this information be used to detect a fake expense report?

(b) Explain why the mean of the random variable Yis located at the solid red line in the figure.

(c) According to Benford鈥檚 law, the expected value of the first digit is 渭X=3.441. Explain how this information could be used to detect a fake expense report.

Ana is a dedicated Skee Ball player who always rolls for the 50-point slot.

Ana鈥檚 score Xon a randomly selected roll of the ball has the probability distribution

shown here with mean 渭X=23.8and standard deviation 蟽X=12.63.

A player receives one ticket from the game for every 10points scored. Define T=number of tickets Ana gets on a randomly selected roll.

a. What shape does the probability distribution of Thave?

b. Find the mean of T.

c. Calculate the standard deviation ofT.

Easy-start mower Refer to Exercise 92 .

a. Calculate and interpret the mean of T.

b. Calculate and interpret the standard deviation of T.

How does your web browser get a file from the Internet? Your computer sends a request for the file to a web server, and the web server sends back a response. Let Y=the amount of time (in seconds) after the start of an hour at which a randomly selected request is received by a particular web server. The probability distribution of Ycan be modeled by a uniform density curve on the interval from 0to3600seconds. Define the random variable W=Y/60.

a. Explain what Wrepresents.

b. What probability distribution does Whave?

Auto emissions The amount of nitrogen oxides (NOX) present in the exhaust of a particular model of old car varies from car to car according to a Normal distribution with mean 1.4 grams per mile (g/mi)and standard deviation 0.3g/mi. Two randomly selected cars of this model are tested. One has 1.1g/miof NOX; the other has 1.9g/mi. The test station attendant finds this difference in emissions between two similar cars surprising. If the NOX levels for two randomly chosen cars of this type are independent, find the probability that the difference is greater than 0.8or less than -0.8.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation
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.