/*! 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 43 An insurance company estimates t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An insurance company estimates that it should make an annual profit of $$\$ 150$$ on each homeowner's policy written, with a standard deviation of $$\$ 6000$$. a. Why is the standard deviation so large? b. If it writes only two of these policies, what are the mean and standard deviation of the annual profit? c. If it writes 10,000 of these policies, what are the mean and standard deviation of the annual profit? d. Is the company likely to be profitable? Explain. e. What assumptions underlie your analysis? Can you think of circumstances under which those assumptions might be violated? Explain.

Short Answer

Expert verified
The standard deviation can be high due to the unpredictable nature of claims in insurance, leading to large variations in profits. Two policies would result in a mean profit of \$300 and standard deviation of \$8485.28. For 10,000 policies, the mean profit will be \$1,500,000 and the standard deviation will be \$60,000,000. The company's likelihood of profit is high owing to the large number of policies. The underlying assumptions are that all policies are independent and have the same profit distribution, which might be violated under certain circumstances like widespread natural disasters and changes in the risk environment.

Step by step solution

01

Understanding the Problem

An annual profit of \$150 is expected on each homeowner's policy written, with a variation (standard deviation) of \$6000. The question asks why the standard deviation is so large, and the mean and standard deviation when writing two and 10,000 policies respectively, whether the company is likely to be profitable and the assumptions made.
02

Answering Part A

The standard deviation is large because the profit from each policy can greatly vary due to insurance risks like fires, floods, etc. Some policies will have no claims and some will have large claims. So, despite an average profit of \$150, the profit for individual policy can greatly deviate from the mean.
03

Answering Part B

The mean profit for two policies would be \$300 (\$150 x 2). For standard deviation, if events are independent, variances should be added together instead of standard deviations. Hence, new standard deviation will be \(\sqrt{2} \times \$6000\) or approximately \$8485.28.
04

Answering Part C

The mean profit for 10,000 policies is \$1,500,000 (\$150 x 10,000). The standard deviation will be \$60,000,000 (\$6000 x \( \sqrt{10000} \))
05

Answering Part D

The company is likely to be profitable as the mean profit for 10,000 policies is a large positive number (\$1,500,000). With that many policies, the large losses on some will be offset by the small gains on most, making the distribution of outcomes more predictable. However, there is still risk due to the large standard deviation.
06

Answering Part E

The assumption here is that profits from each policy are independent of each other and have the same distribution. That is, the occurrence of a large claim on one policy does not affect the claims on other policies. Also, it’s assumed that the average profit and standard deviation based on past data will hold true for future policies. These assumptions may be violated if policies are not independent, eg. widespread natural disasters may cause claims on many policies at once, or if the risk environment changes, causing the average profit and standard deviation to change as well.

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.

standard deviation
Standard deviation is an important concept in statistics that helps measure the amount of variation or dispersion in a set of values. In the context of the insurance company's problem, a large standard deviation of \( \$6000 \) signifies that the profits from individual homeowner's policies can vary significantly.

This variation stems from the unpredictable nature of insurance risks, such as fires or floods, that can cause large fluctuations in the payouts. A high standard deviation indicates that while some policies may incur little to no claims, others might experience substantial claims, leading to a wide range of potential profits or losses.
mean
The mean, also known as the average, is a simple yet powerful statistical measure that represents the central value of a dataset. For the insurance company, the mean annual profit per policy is \( \\(150 \). It indicates that, on average, the company expects to earn \( \\)150 \) from each policy over a year.

Calculating the mean for multiple policies involves multiplying the mean of one policy by the total number of policies, thereby providing an overall expectation of profit. For instance, writing two policies would mean a total expected profit of \( \\(300 \) (\( \\)150 \times 2 \)), while writing 10,000 policies translates to \( \$1,500,000 \).
variance
Variance is a statistical metric that provides insight into the degree of spread in a dataset. It is the square of the standard deviation and thus shares a direct relationship with it. In the given insurance scenario, variance helps in understanding how much the individual profits from each policy deviate from the mean expected profit.

For two policies, the calculation involves summing the variances rather than the standard deviations. This results in a variance of \( 2 \times (\$6000^2) \), highlighting the combined variability. For 10,000 policies, the variance is dramatically larger, helping in assessing the risk and spread of outcomes on a massive scale, allowing for better preparedness in profit predictions.
independence in probability
Independence in probability plays a crucial role in the company's analysis of profits. It assumes that the result of one policy does not affect the outcome of another policy. In simpler terms, each policyholder's risk of making a claim is independent of others.

This assumption makes the statistical analysis more straightforward by allowing variances to be added and simplifying the calculation of expected outcomes. However, real-world scenarios may challenge this assumption, such as natural disasters, which can impact many policyholders simultaneously, leading to correlated risks rather than independent ones.
risk assessment
Risk assessment involves evaluating the uncertainties and potential financial impacts that the company may face. For the insurance company, the standard deviation and mean are crucial in understanding potential risks and predicting profitability.

By assessing the large standard deviation alongside the mean profit, the company can gauge the extent of financial variability and potential losses. This helps in formulating strategies to mitigate risks, such as diversifying the types of insurance offered or creating reserves to cover unforeseen events. Understanding risk is key to ensuring financial stability and profitable operations in the insurance industry.

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

A grocery supplier believes that in a dozen eggs, the mean number of broken ones is 0.6 with a standard deviation of 0.5 eggs. You buy 3 dozen eggs without checking them. a. How many broken eggs do you expect to get? b. What's the standard deviation? c. What assumptions did you have to make about the eggs in order to answer this question?

You bet! You roll a die. If it comes up a \(6,\) you win $$\$ 100$$. If not, you get to roll again. If you get a 6 the second time, you win $$\$ 50$$. If not, you lose. a. Create a probability model for the amount you win. b. Find the expected amount you'll win. c. What would you be willing to pay to play this game?

The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and pour one large and one small bowl. a. How much more cereal do you expect to be in the large bowl? b. What's the standard deviation of this difference? c. If the difference follows a Normal model, what's the probability the small bowl contains more cereal than the large one? d. What are the mean and standard deviation of the total amount of cereal in the two bowls? e. If the total follows a Normal model, what's the probability you poured out more than 4.5 ounces of cereal in the two bowls together? f. The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of 16.3 ounces and a standard deviation of 0.2 ounces. Find the expected amount of cereal left in the box and the standard deviation.

An employer pays a mean salary for a 5-day workweek of \(\$ 1250\) with a standard deviation of \(\$ 129 .\) On the weekends, his salary expenses have a mean of \(\$ 450\) with a standard deviation of \(\$ 57 .\) What is the mean and standard deviation of his total weekly salaries?

An automatic filling machine in a factory fills bottles of ketchup with a mean of 16.1 oz and a standard deviation of 0.05 oz with a distribution that can be well modeled by a Normal model. What is the probability that your bottle of ketchup contains less than 16 oz?

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.