Homework Help is Here – Start Your Trial Now!
Math
Probability
**Title: Probability of Selecting a Household with 1-3 Cars in a Town** **Introduction:** In this exercise, we will analyze the distribution of the number of cars per household in a town with 1,000 households. We will then calculate the probability of randomly selecting a household that has between one and three cars, inclusive. **Problem Statement:** The random variable \( X \) represents the number of cars per household in a town of 1,000 households. We need to find the probability of randomly selecting a household that has between one and three cars, inclusive. **Data Representation:** The data is presented in the following table: | Cars (X) | Households | |----------|------------| | 0 | 125 | | 1 | 428 | | 2 | 256 | | 3 | 108 | | 4 | 83 | **Options:** You are asked to select the correct probability from the following options: - 0.125 - 0.208 - 0.792 (selected) - 0.428 **Calculation:** To find the probability of selecting a household with between 1 and 3 cars, inclusive, we sum the number of households with 1, 2, and 3 cars, then divide by the total number of households: - Households with 1 car: 428 - Households with 2 cars: 256 - Households with 3 cars: 108 Total households with 1-3 cars: \( 428 + 256 + 108 = 792 \) Total number of households: 1000 Probability \( P(1 \leq X \leq 3) = \frac{792}{1000} = 0.792 \) Thus, the selected answer is **0.792**. **Conclusion:** By using the provided data and calculating the relevant probability, we concluded that the probability of randomly selecting a household from this town that owns between one and three cars is \( \mathbf{0.792} \). This value is derived from the sum of the frequencies of households owning 1, 2, and 3 cars divided by the total number of households.

**Title: Probability of Selecting a Household with 1-3 Cars in a Town** **Introduction:** In this exercise, we will analyze the distribution of the number of cars per household in a town with 1,000 households. We will then calculate the probability of randomly selecting a household that has between one and three cars, inclusive. **Problem Statement:** The random variable \( X \) represents the number of cars per household in a town of 1,000 households. We need to find the probability of randomly selecting a household that has between one and three cars, inclusive. **Data Representation:** The data is presented in the following table: | Cars (X) | Households | |----------|------------| | 0 | 125 | | 1 | 428 | | 2 | 256 | | 3 | 108 | | 4 | 83 | **Options:** You are asked to select the correct probability from the following options: - 0.125 - 0.208 - 0.792 (selected) - 0.428 **Calculation:** To find the probability of selecting a household with between 1 and 3 cars, inclusive, we sum the number of households with 1, 2, and 3 cars, then divide by the total number of households: - Households with 1 car: 428 - Households with 2 cars: 256 - Households with 3 cars: 108 Total households with 1-3 cars: \( 428 + 256 + 108 = 792 \) Total number of households: 1000 Probability \( P(1 \leq X \leq 3) = \frac{792}{1000} = 0.792 \) Thus, the selected answer is **0.792**. **Conclusion:** By using the provided data and calculating the relevant probability, we concluded that the probability of randomly selecting a household from this town that owns between one and three cars is \( \mathbf{0.792} \). This value is derived from the sum of the frequencies of households owning 1, 2, and 3 cars divided by the total number of households.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
See similar textbooks
icon
Related questions
Question
**Title: Probability of Selecting a Household with 1-3 Cars in a Town** **Introduction:** In this exercise, we will analyze the distribution of the number of cars per household in a town with 1,000 households. We will then calculate the probability of randomly selecting a household that has between one and three cars, inclusive. **Problem Statement:** The random variable \( X \) represents the number of cars per household in a town of 1,000 households. We need to find the probability of randomly selecting a household that has between one and three cars, inclusive. **Data Representation:** The data is presented in the following table: | Cars (X) | Households | |----------|------------| | 0 | 125 | | 1 | 428 | | 2 | 256 | | 3 | 108 | | 4 | 83 | **Options:** You are asked to select the correct probability from the following options: - 0.125 - 0.208 - 0.792 (selected) - 0.428 **Calculation:** To find the probability of selecting a household with between 1 and 3 cars, inclusive, we sum the number of households with 1, 2, and 3 cars, then divide by the total number of households: - Households with 1 car: 428 - Households with 2 cars: 256 - Households with 3 cars: 108 Total households with 1-3 cars: \( 428 + 256 + 108 = 792 \) Total number of households: 1000 Probability \( P(1 \leq X \leq 3) = \frac{792}{1000} = 0.792 \) Thus, the selected answer is **0.792**. **Conclusion:** By using the provided data and calculating the relevant probability, we concluded that the probability of randomly selecting a household from this town that owns between one and three cars is \( \mathbf{0.792} \). This value is derived from the sum of the frequencies of households owning 1, 2, and 3 cars divided by the total number of households.
Transcribed Image Text:**Title: Probability of Selecting a Household with 1-3 Cars in a Town** **Introduction:** In this exercise, we will analyze the distribution of the number of cars per household in a town with 1,000 households. We will then calculate the probability of randomly selecting a household that has between one and three cars, inclusive. **Problem Statement:** The random variable \( X \) represents the number of cars per household in a town of 1,000 households. We need to find the probability of randomly selecting a household that has between one and three cars, inclusive. **Data Representation:** The data is presented in the following table: | Cars (X) | Households | |----------|------------| | 0 | 125 | | 1 | 428 | | 2 | 256 | | 3 | 108 | | 4 | 83 | **Options:** You are asked to select the correct probability from the following options: - 0.125 - 0.208 - 0.792 (selected) - 0.428 **Calculation:** To find the probability of selecting a household with between 1 and 3 cars, inclusive, we sum the number of households with 1, 2, and 3 cars, then divide by the total number of households: - Households with 1 car: 428 - Households with 2 cars: 256 - Households with 3 cars: 108 Total households with 1-3 cars: \( 428 + 256 + 108 = 792 \) Total number of households: 1000 Probability \( P(1 \leq X \leq 3) = \frac{792}{1000} = 0.792 \) Thus, the selected answer is **0.792**. **Conclusion:** By using the provided data and calculating the relevant probability, we concluded that the probability of randomly selecting a household from this town that owns between one and three cars is \( \mathbf{0.792} \). This value is derived from the sum of the frequencies of households owning 1, 2, and 3 cars divided by the total number of households.
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON
A First Course in Probability
A First Course in Probability
Probability
ISBN:
9780321794772
Author:
Sheldon Ross
Publisher:
PEARSON
SEE MORE TEXTBOOKS