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Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, \(30 \%\) of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let \(X\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(X\). b. Draw the corresponding probability histogram. c. What is the most likely value for \(X\) ? d. What is the probability that at least two of the four selected have earthquake insurance?

Short Answer

Expert verified
a) See step 3. b) Histogram not drawn. c) Most likely \(X = 1\). d) \(P(X \geq 2) = 0.3483\).

Step by step solution

01

Understanding the Problem

We need to find the probability distribution for the random variable \(X\), which represents the number of insured homeowners among a sample of four. The probability that a single homeowner has insurance is 0.3, while the probability that they do not is 0.7. This scenario follows a binomial distribution with parameters \(n=4\) and \(p=0.3\).
02

Identify the Binomial Distribution Formula

The probability of having exactly \(k\) insured homeowners is given by the binomial formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \(\binom{n}{k}\) is the number of combinations of \(n\) items taken \(k\) at a time.
03

Calculate Probabilities for Each Value of X

For \(n=4\) homeowners, calculate the probabilities for \(k = 0, 1, 2, 3,\) and \(4\):- \(P(X = 0) = \binom{4}{0} (0.3)^0 (0.7)^4 = 0.2401\)- \(P(X = 1) = \binom{4}{1} (0.3)^1 (0.7)^3 = 0.4116\)- \(P(X = 2) = \binom{4}{2} (0.3)^2 (0.7)^2 = 0.2646\)- \(P(X = 3) = \binom{4}{3} (0.3)^3 (0.7)^1 = 0.0756\)- \(P(X = 4) = \binom{4}{4} (0.3)^4 (0.7)^0 = 0.0081\)
04

Draw the Probability Histogram

Plot a histogram with the x-axis representing the number of insured homeowners (\(X=0\) to \(X=4\)) and the y-axis representing the corresponding probabilities. Plot each \(X = k\) with its calculated probability.
05

Determine the Most Likely Value for X

The most likely value of \(X\) is the one with the highest probability. From the calculated probabilities, \(X = 1\) has the highest probability of 0.4116.
06

Calculate the Probability of At Least Two Insured Homeowners

We need \(P(X \geq 2)\) which is the sum of \(P(X = 2)\), \(P(X = 3)\), and \(P(X = 4)\):\[P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.2646 + 0.0756 + 0.0081 = 0.3483\]

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Key Concepts

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

Probability Distribution
In this exercise, we're looking at a probability distribution, which is a fundamental concept in statistics. It describes how probabilities are assigned to possible outcomes of a random experiment. Here, the experiment involves selecting four homeowners to see how many have earthquake insurance.

The distribution of a random variable can be described by listing the possible outcomes (number of insured homeowners) and the probabilities associated with these outcomes. In our case, we've used the binomial distribution, which is suitable when you have a fixed number of independent trials, each with the same probability of success.

Key characteristics of a probability distribution include:
  • It provides a complete description of the range of potential outcomes and their probabilities.
  • The sum of probabilities for all possible outcomes is 1.
In our situation, we've calculated probabilities for each possible number of insured homeowners (0 through 4), given that each has a 30% chance of having insurance.
Earthquake Insurance
Earthquake insurance is a specific type of coverage designed to protect homeowners from financial losses due to earthquake damage. In earthquake-prone areas, like certain parts of California, having this insurance can be crucial. However, not all homeowners have it due to various reasons such as cost, perceived necessity, or other personal factors.

In the context of this exercise, the statistic that 30% of homeowners have earthquake insurance indicates a fraction of the population has chosen this type of protection. This percentage is crucial as it becomes the probability of success when calculating the binomial probability distribution.
  • Understanding the proportion of homeowners with insurance helps in risk assessment.
  • The probability of at least two insured homeowners among four gives insights about insurance penetration in the area.
This data influences not only individual decisions but also how insurance companies and policymakers might strategize in these regions.
Most Likely Value
The 'most likely value' in a probability distribution is the value associated with the highest probability. Sometimes, it's referred to as the mode of the distribution.

In our exercise, the most likely value for the number of insured homeowners, denoted by the variable X, is 1. This is because the probability of exactly one homeowner out of four being insured is the highest at 0.4116.

This concept helps us understand which outcome is most frequent or expected under given circumstances and can be essential when making predictions based on probability. For instance:
  • If the purpose is to allocate resources efficiently, focusing on outcomes with the highest probability can be beneficial.
  • The most likely value provides insight into the common situation among the sample.
While it's the most expected value, it doesn't necessarily reflect the average or "typical" scenario, unless the distribution is perfectly symmetrical.
Probability Histogram
A probability histogram is a graphical representation of a probability distribution. It shows the possible outcomes of a random variable on the x-axis and their corresponding probabilities on the y-axis.

Creating a histogram helps visualize how probabilities are spread over the different outcomes. In this exercise, each bar represents the number of homeowners with insurance (from 0 to 4), with the height of each bar indicating the probability of that number occurring.

Constructing a probability histogram:
  • Helps in visualizing the distribution, making it easier to identify patterns and most likely values.
  • Facilitates comparison between different probabilities and outcomes.
  • Clarifies concepts such as skewness or symmetry in a distribution.
For our scenario, the histogram would clearly demonstrate that having one insured homeowner (X=1) is the most likely, followed by two homeowners. This visual can often simplify understanding of numerical data, which is key in statistical analysis.

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Most popular questions from this chapter

The College Board reports that \(2 \%\) of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test. a. What is the probability that exactly 1 received a special accommodation? b. What is the probability that at least 1 received a special accommodation? c. What is the probability that at least 2 received a special accommodation? d. What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated? e. Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed \(4.5\) hours. What would you expect the average time allowed the 25 selected students to be?

a. Show that \(b(x ; n, 1-p)=b(n-x ; n, p)\). b. Show that \(B(x ; n, 1-p)=1-B(n-x-1 ; n, p)\). [Hint: At most \(x S\) 's is equivalent to at least \((n-x) F\) 's.] c. What do parts (a) and (b) imply about the necessity of including values of \(p\) greater than \(.5\) in Appendix Table A.1?

Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either (1/3.5) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).

A plan for an executive travelers' club has been developed by an airline on the premise that \(10 \%\) of its current customers would qualify for membership. a. Assuming the validity of this premise, among 25 randomly selected current customers, what is the probability that between 2 and 6 (inclusive) qualify for membership? b. Again assuming the validity of the premise, what are the expected number of customers who qualify and the standard deviation of the number who qualify in a random sample of 100 current customers? c. Let \(X\) denote the number in a random sample of 25 current customers who qualify for membership. Consider rejecting the company's premise in favor of the claim that \(p>.10\) if \(x \geq 7\). What is the probability that the company's premise is rejected when it is actually valid? d. Refer to the decision rule introduced in part (c). What is the probability that the company's premise is not rejected even though \(p=.20\) (i.e., \(20 \%\) qualify)?

A newsstand has ordered five copies of a certain issue of a photography magazine. Let \(X=\) the number of individuals who come in to purchase this magazine. If \(X\) has a Poisson distribution with parameter \(\lambda=4\), what is the expected number of copies that are sold?

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