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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.

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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.

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Most popular questions from this chapter

A study of the impact of seeking a second opinion about a medical condition is described in the paper "Evaluation of Outcomes from a National Patient- Initiated Second-Opinion Program". Based on a review of 6791 patient-initiated second opinions, the paper states the following: "Second opinions often resulted in changes in diagnosis (14.8\%), treatment \((37.4 \%),\) or changes in both \((10.6 \%)\)." Consider the following two events: \(D=\) event that second opinion results in a change in diagnosis \(T=\) event that second opinion results in a change in treatment a. What are the values of \(P(D), P(T),\) and \(P(D \cap T) ?\) b. Use the given probability information to set up a hypothetical 1000 table with columns corresponding to \(D\) and \(D^{C}\) and rows corresponding to \(T\) and \(T^{C}\). c. What is the probability that a second opinion results in neither a change in diagnosis nor a change in treatment? d. What is the probability that a second opinion results is a change in diagnosis or a change in treatment?

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=0.4\) and \(P\left(E_{2}\right)=0.3\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\operatorname{not} E_{1}\) and not \(E_{2}\) ). c. What is the probability that the firm is successful in at least one of the two bids?

A small college has 2700 students enrolled. Consider the chance experiment of selecting a student at random. For each of the following pairs of events, indicate whether or not you think they are mutually exclusive and explain your reasoning. a. the event that the selected student is a senior and the event that the selected student is majoring in computer science. b. the event that the selected student is female and the event that the selected student is majoring in computer science. c. the event that the selected student's college residence is more than 10 miles from campus and the event that the selected student lives in a college dormitory. d. the event that the selected student is female and the event that the selected student is on the college football team.

The article "A Crash Course in Probability" from The Economist included the following information: The chance of being involved in an airplane crash when flying on an Airbus 330 from London to New York City on Virgin Atlantic Airlines is 1 in \(5,371,369 .\) This was interpreted as meaning that you "would expect to go down if you took this flight every day for 14,716 years." The article also states that a person could "expect to fly on the route for 14,716 years before plummeting into the Atlantic." Comment on why these statements are misleading.

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.41\) and \(P(E)=0.23\). i. What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F) ?\) b. Suppose that for events \(A\) and \(B, P(A)=0.26, P(B)=0.34\), and \(P(A \cup B)=0.47\). Are \(A\) and \(B\) mutually exclusive? How can you tell?

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