/*! 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 25 Quota Problem: Motel Rooms The o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Quota Problem: Motel Rooms 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) At least 10 inquiries are needed. (b) The expected number is 10 room rentals.

Step by step solution

01

Understand the Problem Context

We are dealing with probabilities where the probability of renting a room from an inquiry is \( p = 0.4 \). Part (a) involves finding the number of inquiries needed to be 99% certain of success (renting at least one room). Part (b) requires calculating the expected number of successful room rentals from 25 inquiries.
02

Solve Part (a) – Define the Variables and Use the Complement Rule

For part (a), we define the probability of not renting a room from a single inquiry as \( q = 1 - p = 0.6 \). We want the probability of renting at least one room (success) to be 0.99. The probability of zero rentals in \( n \) inquiries is \( q^n \). Thus, the equation is \( 1 - q^n = 0.99 \). Simplifying, we get \( q^n = 0.01 \).
03

Calculate the Number of Inquiries in Part (a)

Converting the equation \( 0.6^n = 0.01 \) into a logarithmic form: \( n = \frac{\log(0.01)}{\log(0.6)} \). Using a calculator, we find: \( n = \frac{-2}{-0.2218} \approx 9.03 \). Rounding up since we can't have a fraction of an inquiry, \( n = 10 \).
04

Solve Part (b) – Calculate Expected Value

For part (b), the expected number of room rentals when 25 inquiries are made is calculated using the formula for expected value in a binomial distribution, \( E(X) = np \). Here, \( n = 25 \) and \( p = 0.4 \). Thus, \( E(X) = 25 \times 0.4 = 10 \).

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.

Understanding Probability in Room Rentals
Probability is a way to measure how likely an event is to occur. In our motel room problem, the probability of renting a room when an inquiry is made is represented by the number 0.4 or 40%. This means that each time someone inquires about a room, there is a 40% chance they will decide to rent one.

In part (a) of the problem, we want to find out how many inquiries are needed to be 99% sure that at least one room is rented out. To solve this, we use the **Complement Rule**. The complement of an event is when that event does not happen. Here, the probability of not renting a room from an inquiry is 0.6, which is simply 1 - 0.4.

We aim to calculate how many inquiries ("n") make it practically impossible to have zero rentals. The formula used resembles this:
  • Choose the complement probability
  • Set your desired probability of the event occurring
  • Transform this into an equation
For our motel situation, we solve for a 0.99 success probability by forming and solving the geometric expression: \[ 1 - 0.6^n = 0.99 \] This formula helps determine the number of inquiries needed to reach a desired level of certainty.
Calculating Expected Value for Room Rentals
Expected value is a concept of finding the average outcome when an experiment is repeated many times. In terms of probability, it gives you the long-term average or mean of the outcomes when you try something multiple times.

In part (b) of our problem, we calculate the expected number of room rentals from 25 inquiries using expected value. The formula used is: \[ E(X) = np \] where "n" is the number of trials (inquiries in this case) and "p" is the probability of success (renting a room).

With 25 inquiries and a 40% rental probability, the expected value of rentals is:
  • Multiply the number of inquiries by the probability of success
  • Get an expected value of 10
This means that across many groups of 25 inquiries, on average, you expect 10 rooms to be rented.
The Role of Binomial Distribution in Room Rentals
A fundamental concept in probability and statistics is the **Binomial Distribution**, which models the number of successes in a series of trials where each trial is independent.

For our problem with the motel, each inquiry about a room can be thought of as a trial with a *success probability* of 0.4 (the room gets rented). The binomial distribution helps us understand and calculate specific outcomes, like how many rooms might get rented.

The probability mass function for a binomial distribution gives the likelihood of achieving "x" successes (i.e., room rentals) across "n" trials (i.e., inquiries). The binomial formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] This formula accounts for:
  • Total number of inquiries (n)
  • Number of rentals (successes, k)
  • Probability of renting from an inquiry (p)
  • Probability of not renting (1-p)
Understanding this distribution allows us to predict how many times a specific event may happen across different trials, such as estimating room occupancy based on past inquiries.

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

Criminal Justice: Jury Duty Have you ever tried to get out of jury duty? About \(25 \%\) of those called will find an excuse (work, poor health, travel out of town, etc.) to avoid jury duty (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York). If 12 people are called for jury duty, (a) what is the probability that all 12 will be available to serve on the jury? (b) what is the probability that 6 or more will not be available to serve on the jury? (c) Find the expected number of those available to serve on the jury. What is the standard deviation? (d) Quota Problem How many people \(n\) must the jury commissioner contact to be \(95.9 \%\) sure of finding at least 12 people who are available to serve?

Combination of Random Variables: Repair Service A computer repair shop has two work centers. The first center examines the computer to see what is wrong, and the second center repairs the computer. Let \(x_{1}\) and \(x_{2}\) be random variables representing the lengths of time in minutes to examine a computer \(\left(x_{1}\right)\) and to repair a computer \(\left(x_{2}\right) .\) Assume \(x_{1}\) and \(x_{2}\) are independent random variables. Long-term history has shown the following times: Examine computer, \(x_{1}: \mu_{1}=28.1\) minutes; \(\sigma_{1}=8.2\) minutes Repair computer, \(x_{2}: \mu_{2}=90.5\) minutes; \(\sigma_{2}=15.2\) minutes (a) Let \(W=x_{1}+x_{2}\) be a random variable representing the total time to examine and repair the computer. Compute the mean, variance, and standard deviation of \(W\). (b) Suppose it costs \(\$ 1.50\) per minute to examine the computer and \(\$ 2.75\) per minute to repair the computer. Then \(W=1.50 x_{1}+2.75 x_{2}\) is a random variable representing the service charges (without parts). Compute the mean, variance, and standard deviation of \(W\). (c) The shop charges a flat rate of \(\$ 1.50\) per minute to examine the computer, and if no repairs are ordered, there is also an additional \(\$ 50\) service charge. Let \(L=1.5 x_{1}+50\). Compute the mean, variance, and standard deviation of \(L\).

Expected Value: Life Insurance Sara is a 60 -year-old Anglo female in reasonably good health. She wants to take out a \(\$ 50,000\) term (that is, straight death benefit) life insurance policy until she is \(65 .\) The policy will expire on her 65 th birthday. The probability of death in a given year is provided by the Vital Statistics Section of the Statistical Abstract of the United States (116th Edition). \begin{tabular}{l|ccccc} \hline\(x=\) age & 60 & 61 & 62 & 63 & 64 \\ \hline\(P\) (death at this age) & \(0.00756\) & \(0.00825\) & \(0.00896\) & \(0.00965\) & \(0.01035\) \\ \hline \end{tabular} Sara is applying to Big Rock Insurance Company for her term insurance policy. (a) What is the probability that Sara will die in her 60 th year? Using this probability and the \(\$ 50,000\) death benefit, what is the expected cost to Big Rock Insurance? (b) Repeat part (a) for years \(61,62,63\), and 64 . What would be the total expected cost to Big Rock Insurance over the years 60 through \(64 ?\) (c) Interpretation If Big Rock Insurance wants to make a profit of \(\$ 700\) above the expected total cost paid out for Sara's death, how much should it charge for the policy? (d) Interpretation If Big Rock Insurance Company charges \(\$ 5000\) for the policy, how much profit does the company expect to make?

Basic Computation: Binomial Distribution Consider a binomial experiment with \(n=7\) trials where the probability of success on a single trial is \(p=0.30\). (a) Find \(P(r=0)\). (b) Find \(P(r \geq 1)\) by using the complement rule.

Psychology: Deceit Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detector tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and counting backward by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies! Lies!! Lies!!! The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), \(85 \%\) of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of nine students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What is the probability that (a) all the students are able to pass the polygraph examination? (b) more than half the students are able to pass the polygraph examination? (c) no more than four of the students are able to pass the polygraph examination? (d) all the students fail the polygraph examination?

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.