/*! 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 27 Exercises 27 to 29 refer to the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exercises 27 to 29 refer to the following setting. Choose an American household at random and let the random variable \(X\) be the number of cars (including \(\mathrm{SUVs}\) and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars: \begin{tabular}{lcccccc} \hline Number of cars \(X:\) & 0 & 1 & 2 & 3 & 4 & 5 \\ Probability: & 0.09 & 0.36 & 0.35 & 0.13 & 0.05 & 0.02 \\ \hline \end{tabular} 27\. What's the expected number of cars in a randomly selected American household? (a) 1.00 (b) 1.75 (c) 1.84 (d) 2.00 (e) 2.50

Short Answer

Expert verified
The expected number of cars is 1.75, choice (b).

Step by step solution

01

Understanding the Probability Model

The probability model shows the number of cars a household can own, denoted by the variable \(X\), and their corresponding probabilities. We need to use this to find the expected number of cars, which is the mean of the distribution.
02

Formula for Expected Value

The expected value \(E(X)\) of a discrete random variable \(X\) is calculated by taking the sum of each value of \(X\) multiplied by its probability. The formula is \(E(X) = \sum (x_i \cdot p_i)\), where \(x_i\) are the values and \(p_i\) are the probabilities.
03

Calculating the Components

Calculate each component: Multiply each number of cars \(x\) by its probability \(p\).- For \(x=0\), \(0 \cdot 0.09 = 0\) - For \(x=1\), \(1 \cdot 0.36 = 0.36\) - For \(x=2\), \(2 \cdot 0.35 = 0.70\) - For \(x=3\), \(3 \cdot 0.13 = 0.39\) - For \(x=4\), \(4 \cdot 0.05 = 0.20\) - For \(x=5\), \(5 \cdot 0.02 = 0.10\)
04

Summing the Components to Estimate Expected Value

Add the results of the calculated components from Step 3 to determine the expected value:\(E(X) = 0 + 0.36 + 0.70 + 0.39 + 0.20 + 0.10 = 1.75\)
05

Conclusion

The expected number of cars owned by an American household is 1.75, corresponding to choice (b).

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.

Probability Distribution
Probability distribution is a fundamental concept in statistics, especially when dealing with random variables. It provides a list of all the possible values a random variable can take, along with the probability associated with each value. In our exercise, the random variable is the number of cars owned by an American household. The probability distribution is demonstrated in a table, presenting various numbers of cars from 0 to 5, along with the likelihood of each occurrence. Probability distributions are crucial because they help us understand how a random variable behaves. When examining them, ensure that all probabilities add up to 1, as they represent the total certainty of all possible outcomes. This trait is evident in the distribution outlined, where each probability, when summed, equals 1. Additionally, probability distributions can be of different types, such as discrete or continuous, depending on whether the random variables are countable or not.
Discrete Random Variable
A discrete random variable is a type of variable that can take on a countable number of specific values. For instance, in our current scenario, the random variable is the number of cars a household owns, which can specifically be 0, 1, 2, 3, 4, or 5. This contrasts with continuous random variables, which can assume an infinite number of values within a range. Understanding discrete random variables is important as they often relate to real-world integers. To fully comprehend them, visualize scenarios where only a limited set of outcomes is possible. Discrete random variables help in establishing structured and countable outcomes, making calculations involving probability distributions more manageable.
Mean of Distribution
The mean of distribution, often referred to as the expected value, offers insight into the central tendency of a random variable's probability distribution. It is essentially the average or the expected outcome you can anticipate if you were to repeat a random experiment multiple times.To compute the mean of a probability distribution for a discrete random variable, multiply each possible value the random variable can take by its corresponding probability, and then sum all these products. This approach reflects the formula: \[ \text{Mean (Expected Value)} = E(X) = \sum (x_i \cdot p_i) \]where \(x_i\) is a value that the random variable can assume, and \(p_i\) is the probability of \(x_i\).Using the exercise example: - The calculation involves multiplying each number of cars by its probability and then adding the results: \[ E(X) = 0 \cdot 0.09 + 1 \cdot 0.36 + 2 \cdot 0.35 + 3 \cdot 0.13 + 4 \cdot 0.05 + 5 \cdot 0.02 = 1.75 \] The result, 1.75, represents the expected number of cars owned in an American household, showcasing how the mean of a distribution summarizes the variable’s trend over time.

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

Running a mile A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean 7.11 minutes and standard deviation 0.74 minute. \({ }^{10}\) Choose a student at random from this group and call his time for the mile \(Y\). Find \(P(Y<6)\) and interpret the result.

Rainy days Imagine that we randomly select a day from the past 10 years. Let \(X\) be the recorded rainfall on this date at the airport in Orlando, Florida, and \(Y\) be the recorded rainfall on this date at Disney World just outside Orlando. Suppose that you know the means \(\mu_{X}\) and \(\mu_{Y}\) and the variances \(\sigma_{X}^{2}\) and \(\sigma_{Y}^{2}\) of both variables. (a) Is it reasonable to take the mean of the total rainfall \(X+Y\) to be \(\mu_{X}+\mu_{Y} ?\) Explain your answer. (b) Is it reasonable to take the variance of the total rainfall to be \(\sigma_{\mathrm{X}}^{2}+\sigma_{\mathrm{Y}}^{2}\) ? Explain your answer.

Random numbers Let \(Y\) be a number between 0 and 1 produced by a random number generator. Assuming that the random variable \(Y\) has a uniform distribution, find the following probabilities: \(\begin{array}{ll}\text { (a) } & P(Y \leq 0.4)\end{array}\) (b) \(P(Y<0.4)\) \(\begin{array}{ll}\text { (c) } & P(0.1

Loser buys the pizza Leona and Fred are friendiy competitors in high school. Both are about to take the ACT college entrance examination. They agree that if one of them scores 5 or more points better than the other, the loser will buy the winner a pizza. Suppose that in fact Fred and Leona have equal ability, so that each score varies Normally with mean 24 and standard deviation 2 . (The variation is due to luck in guessing and the accident of the specific questions being familiar to the student.) The two scores are independent. What is the probability that the scores differ by 5 or more points in either direction?

Buying stock (5.3,6.1) You purchase a hot stock for \(\$ 1000 .\) The stock either gains \(30 \%\) or loses \(25 \%\) 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? (Comment: You see that these two criteria give different answers to the question "Should I invest?")

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.