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More than 50 million guests stay at bed and breakfasts (B\&Bs) each year. The website for the Bed and Breakfast Inns of North America, which averages seven visitors per minute, enables many B\&Bs to attract guests (Time, September 2001). a. Compute the probability of no website visitors in a one-minute period. b. Compute the probability of two or more website visitors in a one-minute period. c. Compute the probability of one or more website visitors in a 30 -second period. d. Compute the probability of five or more website visitors in a one-minute period.

Short Answer

Expert verified
a. 0.00091; b. 0.99273; c. 0.9698; d. 0.825

Step by step solution

01

Define the Parameters

We will model this problem using a Poisson distribution, where the average rate (\(\lambda\)) is the average number of visitors per minute to the website. Given that the website averages seven visitors per minute, we have \(\lambda = 7\).
02

Calculate Probability of No Visitors in One Minute (Part a)

The probability of \(x\) events (visitors) in a Poisson distribution is given by \( P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \). We are calculating the probability of 0 visitors (\(x = 0\)) in one minute:\[ P(X = 0) = \frac{e^{-7} \times 7^0}{0!} = e^{-7} \approx 0.00091 \]
03

Calculate Probability of Two or More Visitors in One Minute (Part b)

First, calculate the probability of fewer than two visitors, i.e., \( P(X = 0) \) and \( P(X = 1) \), and then find the complementary probability:\[ P(X = 1) = \frac{e^{-7} \times 7^1}{1!} = 7e^{-7} \approx 0.00636 \]\[ P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) \approx 1 - (0.00091 + 0.00636) = 0.99273 \]
04

Calculate Probability of One or More Visitors in 30 Seconds (Part c)

For a 30-second period, \(\lambda\) becomes 3.5 (half of 7). Find the probability of one or more visitors:\[ P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{e^{-3.5} \times 3.5^0}{0!} = 1 - e^{-3.5} \approx 1 - 0.030197 \approx 0.9698 \]
05

Calculate Probability of Five or More Visitors in One Minute (Part d)

Calculate the probability of fewer than five visitors (\( P(X < 5) \)) and then the complement:\[ P(X = 0) = 0.00091, \, P(X = 1) = 0.00636, \, P(X = 2) = \frac{e^{-7} \times 7^2}{2!} \approx 0.0221, \, P(X = 3) = \frac{e^{-7} \times 7^3}{3!} \approx 0.0516, \, P(X = 4) = \frac{e^{-7} \times 7^4}{4!} \approx 0.0902 \]Sum these probabilities for fewer than five visitors:\[ P(X < 5) \approx 0.175 \]The probability of five or more visitors:\[ P(X \geq 5) = 1 - P(X < 5) \approx 1 - 0.175 = 0.825 \]

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental aspect of statistics, which helps us determine the likelihood of different outcomes. In the context of the Poisson distribution, probability calculations can be used to predict the number of events—like visitors to a website—happening within a certain timeframe.

The Poisson distribution is particularly helpful when events occur independently, and the average rate of occurrence is known. To compute probabilities, we use the formula: \[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \]
  • \(P(X = x)\) is the probability of \(x\) events happening within a given period.
  • \(\lambda\) is the average number of events in that period.
  • \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
  • \(x!\) is the factorial of \(x\), which is the product of all positive integers up to \(x\).

Calculate the probability for specific outcomes by plugging in the right values into the formula. For instance, to find the probability of no visitors in one minute when the average is seven, set \(x = 0\), giving the result of \(0.00091\), a very low probability.

This estimation helps understand how rare or common an event could be given the average rate.
Statistical Models
Statistical models are essential tools for data analysis, helping to describe patterns and relationships within data. The Poisson distribution is a type of statistical model that is both discrete and handy for modeling count data.

When we employ a Poisson model, we assume that the number of events in disjoint time intervals is independent and identically distributed as Poisson observations. It is a great choice for modeling situations where we count occurrences, such as:
  • Number of visitors to a website.
  • Call arrivals at a call center.
  • Number of emails received per hour.

This model is powerful because it provides a mathematical framework to predict the probability of a certain number of events happening, given a known average rate. In our case, with an average rate of seven visitors per minute, the model offers probabilities for various numbers of visitors such as no visitors, which is calculated using the earlier explained formula.

Using the Poisson model, complex real-world problems become manageable and interpretable, allowing us to use mathematical insights for predictions and decisions.
Event Probability
Event probability focuses on the chance of one or more occurrences within a specific time or space. In scenarios modeled with a Poisson distribution, such computations become particularly straightforward, as they offer a way to determine how likely or unlikely an event is to happen.

In practical terms, calculating the probability of two or more visitors usually involves finding out the complement of fewer than two. The complement rule—where the probabilities of all possible outcomes sum up to 1—helps in determining probabilities for complex events such as:
  • \( P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) \)
  • In the exercise, this was computed and approximated as 0.99273, meaning there's a very high likelihood of two or more visitors.

Remembering the event probability can guide planning and expectations in real-world applications, from ensuring adequate web server resources based on visitor spikes to anticipating customer flow in businesses.

With a structured approach using the Poisson distribution, one can efficiently navigate the uncertainties and variabilities of count-based events, making probability an essential component in decision-making.

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Most popular questions from this chapter

Twelve of the top 20 finishers in the 2009 PGA Championship at Hazeltine National Golf Club in Chaska, Minnesota, used a Titleist brand golf ball (GolfBallTest website, November 12,2009 . Suppose these results are representative of the probability that a randomly selected PGA Tour player uses a Titleist brand golf ball. For a sample of 15 PGA Tour players, make the following calculations. a. Compute the probability that exactly 10 of the 15 PGA Tour players use a Titleist brand golf ball. b. Compute the probability that more than 10 of the 15 PGA Tour players use a Titleist brand golf ball. c. For a sample of 15 PGA Tour players, compute the expected number of players who use a Titleist brand golf ball. d. For a sample of 15 PGA Tour players, compute the variance and standard deviation of the number of players who use a Titleist brand golf ball.

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A poll conducted by Zogby International showed that of those Americans who said music plays a "very important" role in their lives, \(30 \%\) said their local radio stations "always" play the kind of music they like (Zogby website, January 12,2004 ). Suppose a sample of 800 people who say music plays an important role in their lives is taken. a. How many would you expect to say that their local radio stations always play the kind of music they like? b. What is the standard deviation of the number of respondents who think their local radio stations always play the kind of music they like? c. What is the standard deviation of the number of respondents who do not think their local radio stations always play the kind of music they like?

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