/*! 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 5 In a particular state, automobil... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a particular state, automobiles that are more than 10 years old must pass a vehicle inspection in order to be registered. This state reports the probability that a car more than 10 years old will fail the vehicle inspection is 0.09 . Give a relative frequency interpretation of this probability.

Short Answer

Expert verified
The relative frequency interpretation of the given probability (0.09) is that out of 100 vehicles that are more than 10 years old undergoing inspection, we would expect about 9 of them to fail the inspection.

Step by step solution

01

Relative Frequency Interpretation

Relative frequency interpretation refers to the ratio of the number of times an event occurs to the total number of trials. In the given exercise, the event of interest is the car more than 10 years old failing the vehicle inspection.
02

Convert Probability to Relative Frequency

The given probability is 0.09. To interpret this probability in terms of relative frequency, we need to consider that there were 100 trials (a common denominator for understanding probability in terms of percentages). We can express the given probability as a proportion out of 100.
03

Interpret Relative Frequency

So, with a probability of 0.09, we can convert it to a relative frequency of 9 out of 100, or 9%. This means that if we were to observe 100 vehicles that are more than 10 years old undergoing inspection, we would expect about 9 of them to fail the inspection. This provides an insight into the long-term frequency of this event occurring.

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.

Relative Frequency
The concept of relative frequency is a fundamental aspect of understanding probability. It is essentially a way to look at how often something happens compared to the number of opportunities there are for it to happen. Imagine every time an event happens, it adds a tick to a list.

Relative frequency refers to the number of times a particular event occurs divided by the total number of trials or observations. For example, if you flip a coin 100 times and it lands on heads 55 times, the relative frequency of getting heads would be 55/100 or 0.55.

This method is especially useful because it provides an empirical approach to probability, relying on actual data and past occurrences rather than theoretical calculations. It helps us make reasonable predictions about future events based on observed patterns of past events.
Vehicle Inspection
Vehicle inspection is an official process that ensures automobiles meet safety and emissions standards. For vehicles older than 10 years, inspections can identify issues that may compromise the safety of the vehicle or the environment.

Each inspection checks various components of a vehicle, like brakes, lights, and emission controls. If a car fails, it may need repairs before it can be legally registered and driven.

In many regions, regular vehicle inspections help reduce the number of unsafe vehicles on the road, thus contributing to the safety of all drivers and passengers. The probability that a vehicle over 10 years old will fail its inspection gives insight into common aging issues in cars and the effectiveness of the inspection program.
Event Occurrence
In probability, an event occurrence refers to the instance when a particular event actually happens. For example, when discussing the event of a car failing a vehicle inspection, an occurrence is when a specific car doesn't meet the required inspection standards during a check.

Each occurrence is a chance for analyzing data and understanding probabilities more accurately. It helps in understanding how frequently an event happens over a series of trials.

By examining multiple instances and occurrences, one can determine patterns and make predictions, which is crucial for improving processes like vehicle inspections. Being able to predict the likelihood of a car failing based on age and condition can help owners maintain their vehicles better.
Long-term Frequency
Long-term frequency takes the concept of relative frequency and stretches it over a large number of trials or observations, providing a comprehensive way to predict outcomes over time. It is the stable frequency achieved as the number of trials tends to infinity.

In the context of vehicle inspections, if we consistently see that 9% of cars fail, this becomes a "long-term frequency". The more trials we consider, like thousands of car inspections over many years, the more accurate this frequency becomes.

Understanding long-term frequencies is vital because it allows businesses and governments to project trends and plan accordingly. For instance, they can allocate resources to vehicle maintenance programs or adjust inspection criteria to address issues observed through long-term data. This assists in creating a safer road environment and more reliable registration systems.

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

An online store offers two methods of shipping-regular ground service and an expedited 2 -day shipping. Customers may also choose whether or not to have the purchase gift wrapped. Suppose that the events \(E=\) event that the customer chooses expedited shipping \(G=\) event that the customer chooses gift wrap are independent with \(P(E)=0.26\) and \(P(G)=0.12\). a. Construct a hypothetical 1000 table with columns corresponding to whether or not expedited shipping was chosen and rows corresponding to whether or not gift wrap was selected. b. Use the information in the table to calculate \(P(E \cup G)\). Give a long- run relative frequency interpretation of this probability.

The National Center for Health Statistics (www.cdc .gov/nchs/data/nvsr/nvsr64/nvsr64_12.pdf, retrieved April 25,2017 ) gave the following information on births in the United States in 2014 : $$ \begin{array}{|lr|} \hline \text { Type of Birth } & \text { Number of Births } \\ \hline \text { Single birth } & 3,848,214 \\ \text { Twins } & 135,336 \\ \text { Triplets } & 4,233 \\ \text { Quadruplets } & 246 \\ \text { Quintuplets or higher } & 47 \\ \hline \end{array} $$ Use this information to estimate the probability that a randomly selected pregnant woman who gave birth in 2014 a. delivered twins b. delivered quadruplets c. gave birth to more than a single child

The same issue of The Chronicle for Higher Education Almanac referenced in the previous exercise also reported the following information for Ph.D. degrees awarded by U.S. colleges in the 2013-2014 academic year: \- A total of 54,070 Ph.D. degrees were awarded. \- 12,504 of these degrees were in the life sciences. \- 9859 of these degrees were in the physical sciences. \- The remaining degrees were in majors other than life or physical sciences. What is the probability that a randomly selected Ph.D. student who received a degree in \(2013-2014\) a. received an degree in the life sciences? b. received a degree that was not in a life or a physical science? c. did not receive a degree in the physical sciences?

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.64\) and \(P(F)=0.17\) i. What is the value of \(P(E \cap F)\) ? ii. What is the value of \(P(E \cup F)\) ? b. Suppose that \(A\) and \(B\) are events with \(P(A)=0.3, P(B)=0.5\), and \(P(A \cap B)=0.15 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell? c. Suppose that \(A\) and \(B\) are events with \(P(A)=0.65\) and \(P(B)=0.57 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

5.57 There are two traffic lights on Shelly's route from home to work. Let \(E\) denote the event that Shelly must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=0.4, P(F)=0.3\) and \(P(E \cap F)=0.15 .\) In Exercise \(5.25,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that Shelly must stop for at least one light (the probability of the event \(E \cup F\) ). b. The probability that Shelly does not have to stop at either light. c. The probability that Shelly must stop at exactly one of the two lights. d. The probability that Shelly must stop only at the first light.

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.