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Chapter 3: Problem 16

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, \(25 \%\) 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 \(X\). [Hint: Let \(S\) denote a homeowner who has insurance and \(F\) one who does not. Then one possible outcome is SFSS, with probability \((.25)(.75)(.25)(.25)\) and associated \(X\) value 3 . There are 15 other outcomes.] 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. Probability distribution: {0: 0.3164, 1: 0.4219, 2: 0.2109, 3: 0.0469, 4: 0.0039}; b. Histogram drawn; c. Most likely value: 1; d. At least two insured: 0.2617.

Step by step solution

01

Understand the Problem

We are given that 25% of homeowners have earthquake insurance, and we need to determine the probability distribution for the number of insured homeowners out of four randomly selected ones. This involves using the binomial distribution where success (having insurance) probability is 0.25, the failure (not having insurance) probability is 0.75, and we are selecting 4 homeowners.
02

Identify the Binomial Distribution Components

For this problem, define success as a homeowner having insurance (denoted by S). The problem mentions using a binomial distribution model because we have a fixed number of trials, each with the same probability of success. The parameters are: number of trials \(n = 4\) and probability of success \(p = 0.25\). We need to find the probability distribution for \(X\), which represents the number of successes (insured homeowners).
03

Calculate Each Probability Using the Binomial Formula

The binomial probability formula is \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(\binom{n}{k}\) is the binomial coefficient. Calculate for \(X = 0\) to \(X = 4\):- For \(X=0\): \(P(X = 0) = \binom{4}{0} (0.25)^0 (0.75)^4 = 1 \times 0.3164 = 0.3164\)- For \(X=1\): \(P(X = 1) = \binom{4}{1} (0.25)^1 (0.75)^3 = 4 \times 0.25 \times 0.4219 = 0.4219\)- For \(X=2\): \(P(X = 2) = \binom{4}{2} (0.25)^2 (0.75)^2 = 6 \times 0.0625 \times 0.5625 = 0.2109\)- For \(X=3\): \(P(X = 3) = \binom{4}{3} (0.25)^3 (0.75)^1 = 4 \times 0.0156 \times 0.75 = 0.0469\)- For \(X=4\): \(P(X = 4) = \binom{4}{4} (0.25)^4 (0.75)^0 = 1 \times 0.0039 = 0.0039\)The probability distribution of \(X\) is \{0: 0.3164, 1: 0.4219, 2: 0.2109, 3: 0.0469, 4: 0.0039\}.
04

Draw the Probability Histogram

Construct a histogram where the x-axis represents the possible values of \(X\) (0 through 4) and the y-axis represents the probability of each outcome. Each bar corresponds to one possible value of \(X\) with a height equal to its probability from the distribution calculated in Step 3.
05

Identify the Most Likely Value of X

The most likely value of \(X\) is the one with the highest probability. From the distribution, \(X = 1\) has a probability of 0.4219, which is the highest. Therefore, the most likely value for \(X\) is 1.
06

Calculate the Probability of at Least Two Insured

To find the probability that at least two out of four homeowners are insured, sum the probabilities for \(X = 2, 3,\) and \(4\):\[P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.2109 + 0.0469 + 0.0039 = 0.2617\]

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Key Concepts

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

Probability Distribution
Probability distribution is a fundamental concept in statistics that describes how probabilities are distributed over the possible values of a random variable. In this context, we have a discrete random variable, meaning it can take only a finite number of distinct values. Here, the variable is the number of insured homeowners out of four, denoted by \(X\). Each outcome has its associated probability calculated using a binomial distribution formula.
The key parameters for our binomial distribution are the number of trials \(n = 4\) and the success probability \(p = 0.25\). The binomial distribution helps us find the probability of having exactly \(k\) successes in \(n\) trials. In mathematical terms, it is represented by:
  • \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)
where \(\binom{n}{k}\) is the binomial coefficient, which signifies how many ways we can choose \(k\) successes out of \(n\) trials. Calculating these probabilities for each potential number of insured homeowners (from 0 to 4) gives us the complete probability distribution of \(X\).
This probability distribution is essential because it gives a complete picture of all possible outcomes and their likelihoods, allowing for informed decisions about risks and expectations.
Probability Histogram
A probability histogram visually represents the probability distribution of a discrete random variable like \(X\) in our problem. It uses bars to show the probability of each outcome on the x-axis, helping to easily interpret and understand the distribution's shape and spread.
To draw a probability histogram for this exercise:
  • List each possible outcome of \(X\) (from 0 to 4) along the horizontal axis.
  • The vertical axis represents the probabilities calculated from the binomial formula.
  • Draw bars at each value of \(X\), with the height corresponding to its probability.
This visualization allows us to quickly identify key characteristics of the distribution, such as its skewness and the most probable outcomes. In our case, the bar for \(X = 1\) will be the highest, indicating it's the most likely outcome, while the bars for \(X = 3\) and \(X = 4\) will be the shortest, reflecting their lower probabilities.
Probability histograms are powerful tools in statistics because they provide a clear and immediate understanding of data that might be less obvious through numbers alone.
Statistical Concepts
Understanding statistical concepts such as binomial distribution, probability, and expectation is crucial when analyzing real-world scenarios like determining the likelihood of homeowners having insurance. This specific example uses the binomial distribution because it fits situations with fixed numbers of independent trials and two possible outcomes (success or failure).
Important terms to know in this context include:
  • Success Probability (\(p\)): The probability of a homeowner having insurance, here \(p = 0.25\).
  • Failure Probability: Complement of success, calculated as \(1 - p = 0.75\).
  • Expected Value: This measures the average result we expect in repeated trials. For a binomial distribution, it is calculated as \(E(X) = n \times p\). Here, \(E(X) = 4 \times 0.25 = 1\), meaning on average, one homeowner out of four will be insured.
  • Variance and Standard Deviation: These measure the spread of the distribution around the expected value. Variance is \(Var(X) = n \times p \times (1-p)\) and standard deviation is the square root of variance.
These concepts allow us to dissect and interpret data beyond just single calculations, providing a more comprehensive understanding of potential outcomes and their implications in any statistical study.

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Most popular questions from this chapter

Starting at a fixed time, each car entering an intersection is observed to see whether it turns left \((L)\), right \((R)\), or goes straight ahead \((A)\). The experiment terminates as soon as a car is observed to turn left. Let \(X=\) the number of cars observed. What are possible \(X\) values? List five outcomes and their associated \(X\) values.

A mail-order computer business has six telephone lines. Let \(X\) denote the number of lines in use at a specified time. Suppose the pmf of \(X\) is as given in the accompanying table. \begin{tabular}{l|ccccccc} \(x\) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(.10\) & \(.15\) & \(.20\) & \(.25\) & \(.20\) & \(.06\) & \(.04\) \end{tabular} Calculate the probability of each of the following events. a. \\{at most three lines are in use\\} b. \\{fewer than three lines are in use\\} c. \\{at least three lines are in use\\} d. \\{between two and five lines, inclusive, are in use\\} e. \\{between two and four lines, inclusive, are not in use\\} f. \\{at least four lines are not in use\\}

A toll bridge charges \(\$ 1.00\) for passenger cars and \(\$ 2.50\) for other vehicles. Suppose that during daytime hours, \(60 \%\) of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? [Hint: Let \(X=\) the number of passenger cars; then the toll revenue \(h(X)\) is a linear function of \(\left.X_{-}\right]\)

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, \(60 \%\) can be repaired, whereas the other \(40 \%\) must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one \(P(\) Wed. \()=.3\), \(P(\) Thurs. \()=.4, \quad P(\) Fri. \()=.2\), and \(P(\) Sat. \()=.1\). Let \(Y=\) the number of days beyond Wednesday that it takes for both magazines to arrive (so possible \(Y\) values are \(0,1,2\), or 3). Compute the pmf of \(Y\). [Hint: There are 16 possible outcomes; \(Y(W, W)=0, Y(F, T h)=2\), and so on.]
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