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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

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 4)=0.8 $$ \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 3)=0.7 $$ \(P(\) Seed 2 defeats Seed 3\()=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats \(\operatorname{Seed} 4)=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.

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