/*! 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 23 The owners of a motel in Florida... [FREE SOLUTION] | 91Ó°ÊÓ

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The owners of a motel in Florida have noticed that in the long run, about \(40 \%\) of the people who stop and inquire about a room for the night actually rent a room. (a) Quota Problem How many inquiries must the owner answer to be \(99 \%\) sure of renting at least one room? (b) If 25 separate inquiries are made about rooms, what is the expected number of inquiries that will result in room rentals?

Short Answer

Expert verified
(a) 10 inquiries; (b) 10 rentals.

Step by step solution

01

Understand the Problem for Part (a)

We need to determine the number of inquiries required to ensure a 99% probability of renting at least one room. This is a typical "at least one" probability problem.
02

Use Complement Rule for Probability

The probability that one individual inquiry will not result in a room rental is \(1 - 0.4 = 0.6\). The probability that none of the inquiries result in a room rental after \(n\) inquiries is \(0.6^n\). We want at least one rental, which means \(1 - 0.6^n = 0.99\).
03

Solve the Inequality

To solve \(1 - 0.6^n = 0.99\), we rearrange it to \(0.6^n = 0.01\). We can solve for \(n\) by taking the logarithm of both sides: \(n \cdot \log(0.6) = \log(0.01)\). Thus, \(n = \frac{\log(0.01)}{\log(0.6)}\).
04

Calculate the Number of Inquiries

Insert the values into the logarithmic equation to find \(n\): \(n = \frac{-2}{-0.2218} \approx 9.03\). Since \(n\) must be an integer, we round up to 10 inquiries to ensure 99% certainty.
05

Answer Part (a)

The owners need to answer at least 10 inquiries to be 99% sure of renting at least one room.
06

Understand the Problem for Part (b)

We need to find the expected number of rentals when there are 25 inquiries.
07

Calculate Expected Value

The expected number of rentals is calculated using the expected value formula \(E = np\), where \(n = 25\) and \(p = 0.4\).
08

Compute the Expected Number of Rentals

Substitute the values into the formula: \(E = 25 \times 0.4 = 10\).
09

Answer Part (b)

The expected number of inquiries that will result in room rentals is 10.

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Key Concepts

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

Probability
Probability measures the likelihood of an event occurring. It's a fundamental concept in statistics used to predict outcomes. For instance, if you have a 40% chance of renting a room to an inquiring customer, in probability terms, this would be written as a probability of 0.4. In exercises, you'll often see scenarios asking "what are the chances" of an event, such as renting a room.
  • Probability of an event = Number of favorable outcomes / Total number of possible outcomes
  • The probability is always a number between 0 and 1
  • A probability of 0 means an event will not happen, and a probability of 1 means an event will certainly happen
Using probability helps us make informed predictions based on quantitative data, especially when dealing with uncertainty, like deciding how many inquiries are necessary to ensure a high likelihood of room rentals. Breaking down real-world problems into simple probability terms makes them easier to solve.
Expected Value
Expected value gives us the average outcome from a set of possible outcomes, each with a specific probability. It's like the long-term average you'd expect from a random process if you repeated it many times. To calculate the expected value, multiply each possible outcome by its probability and then sum these results. In our motel scenario, each room inquiry has a 40% chance of resulting in a booking. If 25 people ask about rooms, we compute the expected number of bookings as follows:
  • Formula: \(E = n \times p\)
  • For 25 inquiries and a 40% rental rate: \(E = 25 \times 0.4\)
  • This results in an expected value of 10 bookings
The expected value helps businesses anticipate outcomes and manage resources effectively, since they can predict how many rooms they might rent on average given historical inquiry data.
Complement Rule
The complement rule is a handy tool in probability, helping us find the probability of an event happening by looking at the probability of its complement (the event not happening). It's based on the principle that either an event occurs or it doesn't, encompassing all possible outcomes. The complement rule is crucial when calculating probabilities like needing a 99% chance to rent at least one room.
  • The probability of an event = 1 - Probability of its complement
  • In probability terms, complement of an event \(A\) is represented as \(A^c\)
  • For example, if the chance of not renting a room from one inquiry is 0.6, then the chance of renting is 1 - 0.6
In practice, the complement rule simplifies problems like how many inquiries you need to ensure high booking probabilities by considering what would happen if none resulted in bookings.
Logarithms
Logarithms are mathematical operations that answer the question: "To what power must a certain base be raised to obtain a certain number?" They are the inverses of exponentiation, playing a crucial role in solving equations involving exponential growth or decay, such as probability problems involving repeated events.
  • A logarithm \( ext{log}_b(a)\) answers: "What power must \(b\) be raised to get \(a\)?"
  • Common logs use base 10, while natural logs use base \(e\) (approximately 2.718)
  • If you face an equation like \(b^n = a\), taking the log of both sides rephrases it to \(n \cdot \text{log}_b(b) = \text{log}_b(a)\)
In our problem, we use logarithms to solve \(0.6^n = 0.01\), translating to \(n = \frac{\text{log}(0.01)}{\text{log}(0.6)}\). This lets us find \(n\), the minimum inquiries needed to ensure a 99% chance of renting a room. Using logs in probability helps tackle tricky exponentials with ease.

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Most popular questions from this chapter

Innocent until proven guilty? In Japanese criminal trials, about \(95 \%\) of the defendants are found guilty. In the United States, about \(60 \%\) of the defendants are found guilty in criminal trials (Source: The Book of Risks, by Larry Laudan, John Wiley and Sons). Suppose you are a news reporter following seven criminal trials. (a) If the trials were in Japan, what is the probability that all the defendants would be found guilty? What is this probability if the trials were in the United States? (b) Of the seven trials, what is the expected number of guilty verdicts in Japan? What is the expected number in the United States? What is the standard deviation in each case? (c) Quota Problem As a U.S. news reporter, how many trials \(n\) would you need to cover to be at least \(99 \%\) sure of two or more convictions? How many trials \(n\) would you need if you covered trials in Japan?

The Wall Street Journal reported that approximately \(25 \%\) of the people who are told a product is improved will believe that it is, in fact, improved. The remaining \(75 \%\) believe that this is just hype (the same old thing with no real improvement). Suppose a marketing study consists of a random sample of eight people who are given a sales talk about a new, improved product. (a) Make a histogram showing the probability that \(r=0\) to 8 people believe the product is, in fact, improved. (b) Compute the mean and standard deviation of this probability distribution. (c) Quota Problem How many people are needed in the marketing study to be \(99 \%\) sure that at least one person believes the product to be improved? (Hint: Note that \(P(r \geq 1)=0.99\) is equivalent to \(1-P(0)=0.99\), or \(P(0)=0.01 .)\)

Lost Bags USA Today reported that for all airlines, the number of lost bags was May: \(6.02\) per 1000 passengers \(\quad\) December: \(12.78\) per 1000 passengers Note: A passenger could lose more than one bag. (a) Let \(r=\) number of bags lost per 1000 passengers in May. Explain why the Poisson distribution would be a good choice for the random variable \(r\). What is \(\lambda\) to the nearest tenth? (b) In the month of May, what is the probability that out of 1000 passengers, no bags are lost? that 3 or more bags are lost? that 6 or more bags are lost? (c) In the month of December, what is the probability that out of 1000 passengers, no bags are lost? that 6 or more bags are lost? that 12 or more bags are lost? (Round \(\lambda\) to the nearest whole number.)

Consider two binomial distributions, with \(n\) trials each. The first distribution has a higher probability of success on each trial than the second. How does the expected value of the first distribution compare to that of the second?

The Denver Post reported that a recent audit of Los Angeles 911 calls showed that \(85 \%\) were not emergencies. Suppose the 911 operators in Los Angeles have just received four calls. (a) What is the probability that all four calls are, in fact, emergencies? (b) What is the probability that three or more calls are not emergencies? (c) Quota Problem How many calls \(n\) would the 911 operators need to answer to be \(96 \%\) (or more) sure that at least one call is, in fact, an emergency?

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