/*! 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 11 Airlines sometimes overbook flig... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the variable \(x\) as the number of people who actually show up for a sold-out flight. From past experience, the probability distribution of \(x\) is given in the following table: \(\begin{array}{ccccccccc}x & 95 & 96 & 97 & 98 & 99 & 100 & 101 & 102 \\ p(x) & .05 & .10 & .12 & .14 & .24 & .17 & .06 & .04 \\ x & 103 & 104 & 105 & 106 & 107 & 108 & 109 & 110 \\ p(x) & .03 & .02 & .01 & .005 & .005 & .005 & .0037 & .0013\end{array}\) a. What is the probability that the airline can accommodate everyone who shows up for the flight? b. What is the probability that not all passengers can be accommodated? c. If you are trying to get a seat on such a flight and you are number 1 on the standby list, what is the probability that you will be able to take the flight? What if you are number 3 ?

Short Answer

Expert verified
a) The probability that the airline can accommodate everyone who shows up for the flight is 0.82. b) The probability that not all passengers can be accommodated is 0.18. c) If you are number 1 on the standby list, the probability that you will be able to take the flight is 0.88. If you are number 3 on the standby list, the probability is 0.91.

Step by step solution

01

Calculate the probability that everyone gets a seat

This is the summation of the probabilities of 100 or fewer people actually showing up for the flight. Using the table, add up the probabilities for \(x = 95\) to \(x = 100\). The calculation is: \(0.05 + 0.10 + 0.12 + 0.14 + 0.24 + 0.17 = 0.82\). Therefore, the probability that everyone gets accommodated is 0.82.
02

Calculate the probability that not all passengers can be accommodated

This refers to the scenarios where more than 100 passengers show up. To find this, sum up the probabilities for all \(x\) values greater than 100. This is equivalent to subtracting the result from Step 1 from 1, as the total probability must sum up to 1. The calculation here is: \(1 - 0.82 = 0.18\). Therefore, the probability that not all passengers can be accommodated is 0.18.
03

Calculate the probability of getting a seat as a standby passenger

The likelihood of getting on the flight increases with the number of no-shows. If you are number 1 on the standby list, you need 101 or fewer people to show up. Sum the probabilities for \(x\) from 95 to 101, this equals \(0.82 + 0.06 = 0.88\). Therefore, the probability of being able to take the flight as number 1 on the standby list is 0.88. If you are number 3 on the standby list, you need 103 or fewer people to show up, sum the probabilities for \(x\) from 95 to 103, which equals \(0.88 + 0.03 = 0.91\). Therefore, the probability of being able to take the flight as number 3 on the standby list is 0.91.

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.

Probabilistic Analysis
Understanding probabilistic analysis is key to evaluating scenarios where uncertainty is inherent. Take, for example, the airline overbooking issue. Probabilistic analysis helps in predicting passenger behavior and potential outcomes. This analysis is grounded on probability distributions, which in this case, describe the likelihood of different numbers of passengers showing up.

By compiling historical data into a probability distribution table, we can assess risks and make informed decisions. For every flight, there's a different probability for each number of passengers that could potentially show up, and when a company gathers enough data, it can predict future events with greater precision. For instance, we calculated that the airline has a probability of 0.82 of being able to accommodate all passengers, which is a valuable insight for capacity planning and customer service strategies.
Discrete Random Variable
In the context of the airline example, the variable 'x', representing the number of passengers showing up, is a discrete random variable. Discrete random variables can only take on a finite or countably infinite number of values. Here, 'x' can only take values from the list of whole numbers ranging between 95 and 110.

This concept is essential for creating the probability distribution, which is a function that assigns a probability to each possible value of the discrete random variable. In our table, the probability that exactly 98 people show up, for instance, is 0.14. Such clear-cut assignments allow for straightforward calculations and help forecast discrete outcomes like the headcount on a flight.
Statistics in Decision Making
Statistics play a pivotal role in decision making, especially in situations with variability and uncertainty. In the airline scenario, statistical analysis informs decisions such as how many reservations to accept and how to manage the standby list.

Through understanding the probability distribution of passengers showing up, the airline makes strategic decisions to maximize revenue without compromising customer satisfaction. The calculated probabilities, like the 0.18 chance that not all passengers can be accommodated, help airlines to evaluate their overbooking policies. Furthermore, such statistics enable the airline to communicate transparently with standby passengers regarding their chances of securing a seat, which in this case was 0.88 for the first standby passenger and 0.91 for the third.

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

Suppose that \(90 \%\) of all registered California voters favor banning the release of information from exit polls in presidential elections until after the polls in California close. A random sample of 25 registered California voters is to be selected. a. What is the probability that more than 20 favor the ban? b. What is the probability that at least 20 favor the ban? c. What are the mean value and standard deviation of the number of voters in the sample who favor the ban? d. If fewer than 20 voters in the sample favor the ban, is this at odds with the assertion that (at least) \(90 \%\) of California registered voters favors the ban? (Hint: Consider \(P(x<20)\) when \(p=.9 .)\)

Starting at a particular time, each car entering an intersection is observed to see whether it turns left (L) or right (R) or goes straight ahead (S). The experiment terminates as soon as a car is observed to go straight. Let \(y\) denote the number of cars observed. What are possible \(y\) values? List five different outcomes and their associated \(y\) values.

A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well approximated by a normal distribution with mean \(3 \mathrm{~cm}\) and standard deviation \(0.1 \mathrm{~cm}\). The specifications call for corks with diameters between \(2.9\) and \(3.1 \mathrm{~cm}\). A cork not meeting the specifications is considered defective. (A cork that is too small leaks and causes the wine to deteriorate; a cork that is too large doesn't fit in the bottle.) What proportion of corks produced by this machine are defective?

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Suppose that only \(20 \%\) of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the probability that at most 70 mufflers are ultimately replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty, would you question the \(20 \%\) figure? Explain.

Suppose that for a given computer salesperson, the probability distribution of \(x=\) the number of systems sold in 1 month is given by the following table: \(\begin{array}{lllllllll}x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\end{array}\) \(\begin{array}{lllllllll}p(x) & .05 & .10 & .12 & .30 & .30 & .11 & .01 & .01\end{array}\) a. Find the mean value of \(x\) (the mean number of systems sold). b. Find the variance and standard deviation of \(x\). How would you interpret these values? c. What is the probability that the number of systems sold is within 1 standard deviation of its mean value? d. What is the probability that the number of systems sold is more than 2 standard deviations from the mean? $\begin{array}{llll}

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.