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Chapter 5: Problem 2

An airline reports that for a particular flight operating daily between Phoenix and Atlanta, the probability of an on-time arrival is 0.86. Give a relative frequency interpretation of this probability.

Short Answer

Expert verified
The relative frequency interpretation of the given probability 0.86 is that, on average, for every 50 flights operating daily between Phoenix and Atlanta, there will likely be 43 on-time arrivals.

Step by step solution

01

Define the probability

The probability of an on-time arrival for the flight is given as 0.86. This means that there is an 86% chance that the flight will arrive on time.
02

Convert the probability to a fraction

To convert the probability into a fraction, we can write 0.86 as a fraction over 100. This will give us the fraction \(\frac{86}{100}\).
03

Simplify the fraction (if possible)

We can simplify the fraction \(\frac{86}{100}\) by dividing both the numerator and the denominator by the greatest common divisor (which is 2 in this case): \(\frac{86}{100} = \frac{86\div2}{100\div2} = \frac{43}{50}\)
04

Interpret the relative frequency

The simplified fraction \(\frac{43}{50}\) represents the relative frequency of on-time arrivals for the flight. This interpretation means that, on average, for every 50 flights operating daily between Phoenix and Atlanta, there will likely be 43 on-time arrivals.

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Key Concepts

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

Relative Frequency Interpretation
Understanding the concept of relative frequency is essential when interpreting probabilities. When we say the probability of an event, such as a flight arriving on time, is 0.86, we are using the relative frequency interpretation of probability.

The relative frequency interpretation relates to the number of times an event occurs compared to the total number of trials. If we consider a large number of flights, the probability of 0.86 suggests that for every 100 flights we observe, approximately 86 of them are expected to arrive on time. It is important to note that this is an 'average' or an 'expectation' based on the probability value provided, and actual outcomes might vary.

This concept helps in making informed predictions about future events based on past occurrences. Students often find it easier to grasp probabilities when they can visualize them as the count of successes in a series of trials.
Probability as a Fraction
When dealing with probabilities, expressing them as fractions can be incredibly insightful. For example, the probability of the flight arriving on time is initially given as 0.86, a decimal. However, when we express this as a fraction, it becomes clearer how many out of a set number of trials result in the desired outcome.

In this instance, we convert the decimal to a fraction by considering the decimal as a part of 100. Thus, 0.86 becomes \(\frac{86}{100}\). This fraction represents the same idea as the decimal: out of 100 flights, we expect 86 to be on time. Using fractions can make it more tangible, especially when simplifying to the lowest terms to show the smallest possible 'group' that this probability might apply to.
Simplifying Fractions
Simplifying fractions is a valuable skill in mathematics, particularly in probability. It involves reducing a fraction to its simplest form, where the numerator and denominator have no common factors other than 1. To simplify the fraction \(\frac{86}{100}\), we find the greatest common divisor of both numbers, which is 2, and divide the numerator and denominator by this number.

After simplification, we get \(\frac{43}{50}\). This tells us that for every 50 occurrences (flights, in this context), we expect 43 successful outcomes (on-time arrivals). Students should note that simplifying fractions does not change the value of the probability; it only makes it easier to understand and often easier to work with in various probability problems.

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Most popular questions from this chapter

5.65 The authors of the paper "Do Physicians Know When Their Diagnoses Are Correct?" (Journal of General Internal Medicine [2005]: 334-339) presented detailed case studies to medical students and to faculty at medical schools. Each participant was asked to provide a diagnosis in the case and also to indicate whether his or her confidence in the correctness of the diagnosis was high or low. Define the events \(C\), \(I,\) and \(H\) as follows: \(C=\) event that diagnosis is correct \(I=\) event that diagnosis is incorrect \(H=\) event that confidence in the correctness of the diagnosis is high a. Data appearing in the paper were used to estimate the following probabilities for medical students: $$ \begin{array}{r} P(C)=0.261 \\ P(I)=0.739 \\ P(H \mid C)=0.375 \\ P(H \mid I)=0.073 \end{array} $$ Use Bayes' Rule to calculate the probability of a correct diagnosis given that the student's confidence level in the correctness of the diagnosis is high. b. Data from the paper were also used to estimate the following probabilities for medical school faculty: $$ \begin{array}{r} P(C)=0.495 \\ P(I)=0.505 \\ P(H \mid C)=0.537 \\ P(H \mid I)=0.252 \end{array} $$ Calculate \(P(C \mid H)\) for medical school faculty. How does the value of this probability compare to the value of \(P(C \mid H)\) for students calculated in Part (a)?

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.

In a small city, approximately \(15 \%\) of those eligible are called for jury duty in any one calendar year. People are selected for jury duty at random from those eligible, and the same individual cannot be called more than once in the same year. What is the probability that an eligible person in this city is selected in both of the next 2 years? All of the next 3 years?

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.14\) and \(P(F)=0.76\) i. \(\quad\) 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 \(B, P(A)=0.24, P(B)=0.24\) and \(P(A \cup B)=0.48\). Are \(A\) and \(B\) mutually exclusive? How can you tell?

Lyme disease is transmitted by infected ticks. Several tests are available for people with symptoms of Lyme disease. One of these tests is the EIA/IFA test. The paper "Lyme Disease Testing by Large Commercial Laboratories in the United States" (Clinical Infectious Disease [2014]: \(676-681\) ) found that \(11.4 \%\) of those tested actually had Lyme disease. Consider the following events: \(+\) represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities are based on percentages given in the paper: $$ \begin{array}{r} P(L)=0.114 \\ P\left(L^{C}\right)=0.886 \end{array} $$ $$ \begin{array}{c} P(+\mid L)=0.933 \\ P(-\mid L)=0.067 \\ P\left(+\mid L^{C}\right)=0.039 \\ P\left(-\mid L^{C}\right)=0.961 \end{array} $$ a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a hypothetical 1000 table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use information from the table constructed in Part (b) to calculate this probability.
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