/*! 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 10 Suppose a homeowner spends \(\$ ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose a homeowner spends \(\$ 300\) for a home insurance policy that will pay out \(\$ 200,000\) if the home is destroyed by fire. Let \(Y=\) the profit made by the company on a single policy. From previous data, the probability that a home in this area will be destroyed by fire is 0.0002 . (a) Make a table that shows the probability distribution of Y. (b) Compute the expected value of Y. Explain what this result means for the insurance company.

Short Answer

Expert verified
The expected profit per policy is \$260, indicating profitability for the company.

Step by step solution

01

Define Random Variable

Let's define the random variable \( Y \) which represents the profit made by the company on a single policy. There are two possible outcomes:1. The home is destroyed by fire: the company suffers a loss and pays out \( \\(200,000 \), receiving \( \\)300 \) as premium. Hence, \( Y = 300 - 200,000 = -199,700 \).2. The home is not destroyed by fire: the company gains the premium of \( \$300 \) without any payout. Hence, \( Y = 300 \).
02

Probability Distribution Table

Create a table for the probability distribution of \( Y \):| \( Y \) (Profit) | Probability ||-------------------|-------------|| \( 300 \) | \( 1 - 0.0002 = 0.9998 \) || \( -199,700 \) | \( 0.0002 \) |
03

Calculate Expected Value

The expected value \( E(Y) \) is calculated using the formula for expectation: \[ E(Y) = \sum (Y_i \times P(Y_i)) \] Plugging the values from the table:\[ E(Y) = (300 \times 0.9998) + (-199,700 \times 0.0002) \] \[ E(Y) = 299.94 - 39.94 = 260 \]
04

Interpret Expected Value

The expected value \( E(Y) = 260 \) means that on average, the insurance company expects to make \( \$260 \) profit per policy sold. This represents long-term profitability, indicating that selling each policy is statistically beneficial for the company.

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.

Expected Value
The expected value is a central concept in probability and statistics, providing insights into the average or mean outcome of a random event. In this exercise, calculating the expected value helps the insurance company predict its average profit per policy. The expected value, often denoted as \( E(Y) \), is essentially a weighted average of all possible outcomes, weighted by their probabilities. For the insurance policy, there are two main outcomes determining the company's profit:
  • If the home isn't destroyed by fire, the profit is \( \\(300 \) since the company keeps the premium paid.
  • If the home is destroyed, the company pays \( \\)200,000 \) but retains the \( \\(300 \) premium, leading to a loss of \( -\\)199,700 \).
To compute the expected value, the formula \( E(Y) = \sum (Y_i \times P(Y_i)) \) is used. You multiply each possible profit (or loss) by its probability of occurrence, then add these results. In this case:
  • Probability of making \( \\(300 \): 0.9998
  • Probability of a \( -\\)199,700 \) loss: 0.0002
So, \( E(Y) = (300 \times 0.9998) + (-199,700 \times 0.0002) = 260 \). This calculation suggests that, despite rare large losses, the policy is expected to generate profit when large numbers of policies are considered over time.
Random Variable
Random variables are fundamental to understanding probability distributions and expected values. They can take on different values depending on the outcome of a random event.In our example, the random variable \( Y \) represents the insurance company's profit from a single policy. As with many random variables, it is tied to a probability distribution, which maps each possible outcome (profit in this case) to its probability.With our insurance scenario, there are two potential values for \( Y \):
  • Profit of \( \\(300 \) when no fire occurs.
  • Loss of \( -\\)199,700 \) when there is a fire.
These outcomes are quantified using probabilities:
  • The probability the home won't catch fire, leading to a profit of \( \\(300 \), is 0.9998.
  • The probability of a fire occurring, resulting in a \( -\\)199,700 \) loss, is 0.0002.
Understanding random variables and their distributions is crucial to see how individual possibilities contribute to the overall picture, such as the company's expected profits from sold policies. It makes probability more tangible and reveals the potential variance around average results.
Insurance Profit Calculation
Insurance profit calculations leverage probability and expected values to examine the dynamic nature of risk. In the context of this problem, it involves assessing the potential profit or loss from selling insurance policies over time.Insurance companies operate by estimating risks via historical data. Here, the calculation uses probabilities of various outcomes:
  • The premium payment received: \( \\(300 \), occurs with a high probability of not having to pay out \( \\)200,000 \), which is 0.9998.
  • The rare, yet financially impactful event of a house fire results in a significant payout of \( \\(200,000 \), only occurring with a low probability of 0.0002.
The calculation of expected value gives insurers an overview of long-term profitability. With each policy, on average, they expect to receive \( \\)260 \) after considering potential payouts for fire-related claims. This insight, while abstract, governs practical business decisions, informing premiums to charge and spotlighting the importance of spreading the risk by selling many policies. An effective insurance profit calculation helps ensure a company remains viable, balancing client security with fiscal prudence.

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

Binomial setting? A binomial distribution will be approximately correct as a model for one of these two sports settings and not for the other. Explain why by briefly discussing both settings. (a) A National Football League kicker has made 80\(\%\) of his field goal attempts in the past. This season he attempts 20 field goals. The attempts differ widely in distance, angle, wind, and so on. (b) A National Basketball Association player has made 80\(\%\) of his free-throw attempts in the past. This season he takes 150 free throws. Basketball free throws are always attempted from 15 feet away with no interference from other players.

Statistics for investing \((3.2)\) Joe's retirement plan invests in stocks through an "index fund" that follows the behavior of the stock market as a whole, as measured by the Standard \& Poor's (S\&P) 500 stock index. Joe wants to buy a mutual fund that does not track the index closely. He reads that monthly returns from Fidelity Technology Fund have correlation \(r=\) 0.77 with the S\&P 500 index and that Fidelity Real Estate Fund has correlation \(r=0.37\) with the index. (a) Which of these funds has the closer relationship to returns from the stock market as a whole? How do you know? (b) Does the information given tell Joe anything about which fund had higher returns?

Buying stock \((5.3,6.1)\) You purchase a hot stock for \(\$ 1000\) . The stock either gains 30\(\%\) or loses 5\(\%\) each day, each with probability 0.5 . Its returns on consecutive days are independent of each other. You plan to sell the stock after two days. (a) What are the possible values of the stock after two days, and what is the probability for each value? What is the probability that the stock is worth more after two days than the \(\$ 1000\) you paid for it? (b) What is the mean value of the stock after two days?

Rhubarb Suppose you purchase a bundle of 10 bare-root rhubarb plants. The sales clerk tells you that on average you can expect 5\(\%\) of the plants to die before producing any rhubarb. Assume that the bundle is a random sample of plants. Let Y = the number of plants that die before producing any rhubarb. Use the binomial probability formula to find \(P(Y=1)\) . Interpret this result in context.

ITBS scores The Normal distribution with mean \(\mu=6.8\) and standard deviation \(\sigma=1.6\) is a good description of the lowa Test of Basic Skills (ITBS) vocabulary scores of seventh-grade students in Gary, Indiana. Call the score of a randomly chosen student \(X\) for short. Find \(P(X \geq 9)\) and interpret the result. Follow the four-step process.

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.