/*! 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} Problem 21 An insurance company offers four... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An insurance company offers four different deductible levels-none, low, medium, and high-for its homeowner's policyholders and three different levels- low, medium, and high-for its automobile policyholders. The accompanying table gives proportions for the various categories of policyholders who have both types of insurance. For example, the proportion of individuals with both low homeowner's deductible and low auto deductible is .06 (6\% of all such individuals). $$ \begin{array}{lcccc} & {}{}{\text { Homeowner's }} \\ { } \text { Auto } & \mathbf{N} & \mathbf{L} & \mathbf{M} & \mathbf{H} \\ \hline \mathbf{L} & .04 & .06 & .05 & .03 \\ \mathbf{M} & .07 & .10 & .20 & .10 \\ \mathbf{H} & .02 & .03 & .15 & .15 \\ \hline \end{array} $$ Suppose an individual having both types of policies is randomly selected. a. What is the probability that the individual has a medium auto deductible and a high homeowner's deductible? b. What is the probability that the individual has a low auto deductible? A low homeowner's deductible? c. What is the probability that the individual is in the same category for both auto and homeowner's deductibles? d. Based on your answer in part (c), what is the probability that the two categories are different? e. What is the probability that the individual has at least one low deductible level? f. Using the answer in part (e), what is the probability that neither deductible level is low?

Short Answer

Expert verified
a) 0.10, b) 0.18 (auto), 0.19 (homeowner), c) 0.41, d) 0.59, e) 0.31, f) 0.69.

Step by step solution

01

Find Probability for Medium Auto, High Homeowner's Deductible

To find the probability of an individual having a medium auto deductible and a high homeowner's deductible, look at the intersection of the "Medium" row and the "High" column in the table: P (Medium Auto, High Homeowner) = 0.10.
02

Find Probability for Low Auto Deductible

The probability of a low auto deductible is the sum of the probabilities across the "Low" auto deductible row: P(Low Auto) = 0.04 + 0.06 + 0.05 + 0.03 = 0.18.
03

Find Probability for Low Homeowner's Deductible

The probability of a low homeowner's deductible is the sum of the probabilities down the "Low" homeowner's column: P(Low Homeowner) = 0.06 + 0.10 + 0.03 = 0.19.
04

Find Probability for Same Deductible Category

Add the probabilities where both deductible levels (auto and homeowner's) are the same (diagonally from top-left to bottom-right): P(Same Category) = P(Low, Low) + P(Medium, Medium) + P(High, High) = 0.06 + 0.20 + 0.15 = 0.41.
05

Find Probability for Different Deductible Categories

Use the probability of being in the same category from Step 4: P(Different Categories) = 1 - P(Same Category) = 1 - 0.41 = 0.59.
06

Find Probability for At Least One Low Deductible

Add the probabilities of any row or column where "low" is part of the combination: P(At Least One Low) = P(Low Auto) + P(Low Homeowner) - P(Low Auto, Low Homeowner) = 0.18 + 0.19 - 0.06 = 0.31.
07

Find Probability for Neither Deductible Being Low

Use the probability of having at least one low deductible from Step 6: P(Neither Low) = 1 - P(At Least One Low) = 1 - 0.31 = 0.69.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Deductible Levels
Deductible levels in insurance refer to the amount of money a policyholder must pay out-of-pocket before the insurance company pays any expenses. A homeowner's policy might offer deductible levels such as none, low, medium, and high. For auto insurance, deductibles typically range from low to high.
Here’s why deductible levels are important:
  • Financial Planning: Higher deductibles often result in lower premium costs, while lower deductibles mean higher premiums.
  • Risk Assessment: Choosing a deductible level is a decision based on assessing the financial risk you're willing to assume.
  • Policy Customization: Offers flexibility in tailoring the policy to meet individual financial circumstances and comfort levels.
Understanding the relationship between deductible levels and insurance costs can help policyholders make informed decisions, balancing between their short-term financial capacity and long-term risk management.
Exploring Insurance Probabilities
Insurance probabilities provide insight into the likelihood of certain outcomes, such as having a specific deductible level. These probabilities can be calculated using data from various policy categories.
For instance:
  • Calculating Probabilities: To find the probability of a policyholder having a low auto deductible, add all the probabilities in the "Low" auto row.
  • Intersections: Similarly, the joint probability of having a medium auto deductible and a high homeowner's deductible can be derived from the intersecting table cell.
Each probability tells us about the distribution of policyholders across different categories, helping both insurers and policyholders to understand common combinations of coverage options. For example, knowing the probability that a policyholder has a low deductible level might influence an insurance company's pricing strategy.
The Essentials of Joint Probability Distribution
Joint probability distribution is a statistical measure that helps us understand the likelihood of two or more events happening together. In the context of insurance, it lets us see how combinations of deductible levels line up across multiple categories.
Here's how it works:
  • Joint Probability: The intersection of two categories, such as low homeowner’s and low auto deductible, represents the joint probability for that combination.
  • Overall Distribution Analysis: By looking at the entire distribution, it helps predict outcomes and assess risk.
  • Dependency Understanding: It can highlight dependencies between different insurance aspects, which might not be obvious at first glance.
Understanding joint probability distribution is crucial in fields like insurance, where multiple factors play a role in policyholder decisions. It guides not only individual choices but also broader strategic planning for insurance providers.

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

Disregarding the possibility of a February 29 birthday, suppose a randomly selected individual is equally likely to have been born on any one of the other 365 days. a. If ten people are randomly selected, what is the probability that all have different birthdays? That at least two have the same birthday? b. With \(k\) replacing ten in part (a), what is the smallest \(k\) for which there is at least a \(50-50\) chance that two or more people will have the same birthday? c. If ten people are randomly selected, what is the probability that either at least two have the same birthday or at least two have the same last three digits of their Social Security numbers? [Note: The article "Methods for Studying Coincidences" (F. Mosteller and P. Diaconis, J. Amer: Stat. Assoc., 1989: 853–861) discusses problems of this type.]

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2,000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\\{\) the first board is green \(\\}\) and \(B=\\{\) the second board is green \(\\}\). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles. Let \(A\) denote the event that the New York flight is full and define events \(B\) and \(C\) analogously for the other two flights. Suppose \(P(A)=.6, P(B)=.5, P(C)=.4\) and the three events are independent. What is the probability that a. All three flights are full? That at least one flight is not full? b. Only the New York flight is full? That exactly one of the three flights is full?

Components of a certain type are shipped to a supplier in batches of ten. Suppose that \(50 \%\) of all such batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0,1 , and 2 defective components being in the batch under each of the following conditions? a. Neither tested component is defective. b. One of the two tested components is defective.

An academic department with five faculty membersAnderson, Box, Cox, Cramer, and Fisher-must select two of its members to serve on a personnel review committee. Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting the names on identical pieces of paper and then randomly selecting two. a. What is the probability that both Anderson and Box will be selected? [Hint: List the equally likely outcomes.] b. What is the probability that at least one of the two members whose name begins with \(C\) is selected? c. If the five faculty members have taught for \(3,6,7,10\), and 14 years, respectively, at the university, what is the probability that the two chosen representatives have a total of at least 15 years' teaching experience there?

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.