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Condé Nast Traveler publishes a Gold List of the top hotels all over the world. The Broadmoor Hotel in Colorado Springs contains 700 rooms and is on the 2004 Gold List (Condé Nast Traveler; January 2004 ). Suppose Broadmoor's marketing group forecasts a mean demand of 670 rooms for the coming weekend. Assume that demand for the upcoming weekend is normally distributed with a standard deviation of 30 a. What is the probability that all the hotel's rooms will be rented? b. What is the probability that 50 or more rooms will not be rented? c. Would you recommend the hotel consider offering a promotion to increase demand? What considerations would be important?

Short Answer

Expert verified
P(all rented) = 0.1587; P(50+ not rented) = 0.2514; Consider promotion due to unoccupied rooms.

Step by step solution

01

Understand the Given Parameters

We're given that the demand for rooms is normally distributed with a mean (\( \mu \)) of 670 and a standard deviation (\( \sigma \)) of 30. The hotel has 700 rooms total, and we want to calculate probabilities related to the demand.
02

Calculate the Z-Score for Probability a

To find the probability that all rooms are rented, calculate the probability that demand is 700 or more. Calculate the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where \( X = 700 \). Thus, \[ Z = \frac{700 - 670}{30} = 1.00 \].
03

Find the Probability from the Z-Score for Part a

Use standard normal distribution tables or calculators to find the probability that \( Z \geq 1.00 \). This gives the probability that demand is 700 or more. Look up \( Z = 1.00 \) in the table, which gives a probability of 0.8413. Thus, the probability that demand is 700 or more is \( 1 - 0.8413 = 0.1587 \).
04

Calculate the Z-Score for Probability b

For 50 or more rooms not being rented, put it in terms of demand: demand is less than or equal to 650 rooms (since 700 - 650 = 50). Use the Z-score formula:\[ Z = \frac{650 - 670}{30} = -0.67 \].
05

Find the Probability from the Z-Score for Part b

Use the normal distribution table to find the probability that \( Z \leq -0.67 \). The table indicates a probability of approximately 0.2514. This is the probability that 650 or fewer rooms are rented, meaning 50 or more are not rented.
06

Considerations for Promotion c

For part c, the probability that all rooms will be rented is relatively low (15.87%), and the probability that a significant number of rooms (50 or more) remain vacant is notable (25.14%). Hence, offering a promotion could help increase demand. Important considerations include balancing additional revenue from filling more rooms against the cost of promotions, and potential impacts on branding and customer perception.

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Key Concepts

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

Probability Calculation
To grasp probability calculation in a normal distribution, it's crucial to understand how we determine the likelihood of certain outcomes within the data's constraints. In the hotel scenario, we calculated the probability of all 700 rooms being rented. This requires finding the probability that demand will meet or exceed the available rooms.
This is done using the properties of the normal distribution, which is symmetrical and characterized by its mean and standard deviation. To calculate a probability like this, we start by determining a Z-Score. This Z-Score converts raw data (in this case, room demand) into a standardized format that can be referenced through a standard normal distribution table. This table provides probabilities for any Z-Score you might compute. By converting the demand level (700 rooms) into a Z-Score, we get a statistical measure that allows us to retrieve the probability of the demand meeting those levels from standard normal distribution tables or calculators.
Z-Score
To understand the Z-Score, think of it as a conversion tool. It changes a specific data point from a dataset into a form that expresses how far away it is from the average (mean), measured in terms of standard deviation. This transformation is essential for using the standard normal distribution. For example, in the problem at hand, we used the Z-Score formula: \[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the specific value of interest (e.g., 700 rooms), \(\mu\) is the mean (average demand, 670 rooms), and \(\sigma\) is the standard deviation (30). Thus, the Z-Score helps us determine how unusual or usual a demand value is within the context of our normal distribution. Once calculated, the Z-Score translates our original problem into a universal language of probability. This can be read directly from standard normal distribution tables, which tell us about the overarching chances of various outcomes – whether they be common or rare.
Statistical Analysis
Statistical analysis unveils patterns and insights from data. In analyzing the demand for hotel rooms, we used statistical methods to predict future demand. This insight allows the hotel to make informed decisions, such as whether to run promotions. The calculated probabilities indicate how often certain demand levels occur. Here, the likelihood that all rooms are rented is low, while a significant number of rooms potentially remaining vacant is notably higher. Such insights are pivotal for making strategic business decisions. For a business, understanding statistical outcomes could mean balancing the potential extra revenue from increased occupancy against the promotional costs. Additionally, statistical analysis also considers the impact on brand image and customer loyalty when deciding if promotions should be used. This analysis effectively combines quantitative data (like probability and Z-Scores) with business acumen to craft strategies that align with organizational goals.

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Most popular questions from this chapter

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