/*! 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 101 A charter bus company is adverti... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A charter bus company is advertising a singles outing on a bus that holds 60 passengers. The company has found that, on average, \(10 \%\) of ticket holders do not show up for such trips; hence, the company routinely overbooks such trips. Assume that passengers act independently of one another. a. If the company sells 65 tickets, what is the probability that the bus can hold all the ticket holders who actually show up? In other words, find the probability that 60 or fewer passengers show up. b. What is the largest number of tickets the company can sell and still be at least \(95 \%\) sure that the bus can hold all the ticket holders who actually show up?

Short Answer

Expert verified
To answer the first question, the probability that 60 or fewer passengers show up when 65 tickets are sold needs to be calculated using the binomial probability formula. For the second question, an iterative process needs to be carried out starting from 60 tickets, and the number of tickets should be incremented till cumulative probability exceeds 0.95.

Step by step solution

01

Calculate probability for first question

The probability that 60 or fewer passengers show up can be calculated using the binomial probability formula. Hence, the cumulative probability (P(X ≤ 60) given n=65, p=0.90) needs to be calculated. In a more manual approach, P(X ≤ 60) = P(X=0) + P(X=1) +...+ P(X=60). We can utilize statistical software, calculators or binomial tables to make the computations relatively easier.
02

Calculate tickets required for second question

To find the largest number of tickets that can be sold while still maintaining at least a 95% certainty that all passengers can be accommodated, an iterative process needs to be carried out. Start with 60 tickets and calculate the cumulative probability. Increment the number of tickets until the cumulative probability first exceeds 0.95. Since we are looking for at least 95% confidence, we should take the largest value of n that falls below or at 0.95.

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!

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

A variation of a roulette wheel has slots that are not of equal size. Instead, the width of any slot is proportional to the probability that a standard normal random variable \(z\) takes on a value between \(a\) and \((a+.1)\), where \(a=-3.0,-2.9,-2.8, \ldots, 2.9,3.0 .\) In other words, there are slots for the intervals \((-3.0,-2.9),(-2.9,-2.8),(-2.8,-2.7)\) through \((2.9,3.0)\). There is one more slot that represents the probability that \(z\) falls outside the interval \((-3.0,3.0)\). Find the following probabilities. a. The ball lands in the slot representing \((.3, .4)\). b. The ball lands in any of the slots representing \((-.1, .4)\). c. In at least one out of five games, the ball lands in the slot representing \((-.1, .4)\) d. In at least 100 out of 500 games, the ball lands in the slot representing \((.4, .5)\).

A company that has a large number of supermarket grocery stores claims that customers who pay by personal check spend an average of \(\$ 87\) on groceries at these stores with a standard deviation of \(\$ 22\). Assume that the expenses incurred on groceries by all such customers at these stores are normally distributed. a. Find the probability that a randomly selected customer who pays by check spends more than \(\$ 114\) on groceries. b. What percentage of customers paying by check spend between \(\$ 40\) and \(\$ 60\) on groceries? c. What percentage of customers paying by check spend between \(\$ 70\) and \(\$ 105\) ? d. Is it possible for a customer paying by check to spend more than \(\$ 185\) ? Explain.

quarter), Britons spend an a… # According to a 2004 survey by the telecommunications division of British Gas (Source: http://www. literacytrust.org.uk/Database/texting.html#quarter), Britons spend an average of 225 minutes per day communicating electronically (on a fixed landline phone, on a mobile phone, by emailing, by texting, and so on). Assume that currently such times for all Britons are normally distributed with a mean of 225 minutes per day and a standard deviation of 62 minutes per day. What percentage of Britons communicate electronically for a. less than 60 minutes per day b. more than 360 minutes per day c. between 120 and 180 minutes per day d. between 240 and 300 minutes per day?

A 2007 article states that \(4.8 \%\) of U.S. households are "linguistically isolated," which means that all members of the household aged 14 years and older have difficulty speaking English (Source: http://www.antara.co.id/en/arc/2007/9/12/five-percent-of-us-families-dont- speak-english-report//). Assume that this percentage is true for the current population of U.S. households. Find the probability that in a random sample of 750 U.S. households, more than 45 would be classified as "linguistically isolated""

Find the following areas under a normal distribution curve with \(\mu=12\) and \(\sigma=2\). a. Area between \(x=7.76\) and \(x=12\) b. Area between \(x=14.48\) and \(x=16.54\) c. Area from \(x=8.22\) to \(x=10.06\)

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.