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Jim is a real estate agent who sells large commercial buildings. Because his commission is so large on a single sale, he does not need to sell many buildings to make a good living. History shows that Jim has a record of selling an average of eight large commercial buildings every 275 days. (a) Explain why a Poisson probability distribution would be a good choice for \(r=\) number of buildings sold in a given time interval. (b) In a 60 -day period, what is the probability that Jim will make no sales? one sale? two or more sales? (c) In a 90 -day period, what is the probability that Jim will make no sales? two sales? three or more sales?

Short Answer

Expert verified
(a) Poisson is suitable for sporadic, independent events over a fixed period. (b) 60 days: no sales ≈ 0.174, one sale ≈ 0.304, two or more ≈ 0.522. (c) 90 days: no sales ≈ 0.078, two sales ≈ 0.261, three or more ≈ 0.462.

Step by step solution

01

Justifying the Use of Poisson Distribution

A Poisson distribution is suitable for modeling the number of events (sales) occurring within a defined time period when these events occur independently. In this case, we're examining sales over a fixed time period, such as 60 or 90 days. Sales are infrequent and sporadic, matching the Poisson model, which describes rare events in a given time frame.
02

Determine Average Sales Rate

Jim sells 8 buildings in 275 days. To find the average rate (λ) for a different period, compute: \[ \lambda = \frac{\text{average sales}}{\text{time period}} = \frac{8}{275} \] For a 60-day or 90-day period, the expected number of sales is: \[\lambda_{60} = \frac{8}{275} \times 60 = 1.745 \]\[\lambda_{90} = \frac{8}{275} \times 90 = 2.545 \]
03

Probability of No Sales in 60 Days

Using Poisson formula: \[ P(r; \lambda) = \frac{e^{-\lambda} \lambda^r}{r!} \] For no sales (r=0) in 60 days: \[ P(0; 1.745) = \frac{e^{-1.745} \times 1.745^0}{0!} = e^{-1.745} \approx 0.174 \]
04

Probability of One Sale in 60 Days

Using the Poisson formula for one sale (r=1) in 60 days: \[ P(1; 1.745) = \frac{e^{-1.745} \times 1.745^1}{1!} \approx 0.304 \]
05

Probability of Two or More Sales in 60 Days

Calculate using complement rule: \[ P(r \geq 2) = 1 - P(0; 1.745) - P(1; 1.745) \approx 1 - 0.174 - 0.304 = 0.522 \]
06

Probability of No Sales in 90 Days

To find the probability of no sales (r=0) in 90 days: \[ P(0; 2.545) = \frac{e^{-2.545} \times 2.545^0}{0!} = e^{-2.545} \approx 0.078 \]
07

Probability of Two Sales in 90 Days

Using Poisson formula for two sales (r=2) in 90 days: \[ P(2; 2.545) = \frac{e^{-2.545} \times 2.545^2}{2!} \approx 0.261 \]
08

Probability of Three or More Sales in 90 Days

Calculate using complement rule: \[ P(r \geq 3) = 1 - P(0; 2.545) - P(1; 2.545) - P(2; 2.545) \] Calculate \( P(1; 2.545) \):\[ P(1; 2.545) = \frac{e^{-2.545} \times 2.545^1}{1!} \approx 0.199 \] Then,\[ P(r \geq 3) = 1 - 0.078 - 0.199 - 0.261 = 0.462 \]

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Key Concepts

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

Probability
Probability refers to the chance or likelihood that a particular event will occur. When we talk about probability in the context of Jim's real estate sales, we're trying to predict how likely it is for him to sell a specific number of buildings within certain time periods, like 60 or 90 days.
Understanding probability helps us model different scenarios, like Jim making no sales, one sale, or multiple sales in these time periods. Since Jim's sales are rare and infrequent, it's important to know that probability can be calculated using several statistical models.
In this exercise, the Poisson Distribution is used to model these probabilities, offering us a clear view of expectations based on past data.
Event Rate
Event Rate, often denoted as \( \lambda \), represents the average rate at which events happen over a given time frame. For Jim, this is the average number of buildings he sells over a period of time.
To find Jim's event rate, you divide the total sales by the number of days over which those sales occurred. For instance, Jim sells 8 buildings over 275 days, giving him an event rate of \( \lambda = \frac{8}{275} \approx 0.029 \).
This calculation helps us determine the expected number of sales in different time spans by multiplying \( \lambda \) by those time periods, such as 60 or 90 days. This average rate is crucial as it drives the calculations for probabilities using the Poisson model.
Rare Events
Rare events occur infrequently, and in Jim's case, sales of large commercial buildings are considered infrequent. When the frequency of these events within a given time frame is low, we say these are rare events.
The Poisson Distribution is particularly suited for modeling such rare events because it can provide the probability of a number of events occurring in a fixed period, even if actual occurrences are independent and sporadic.
This helps Jim to predict and set expectations realistically when planning his sales strategy. He can have a probability-driven expectation about when a sale might happen, helping his business planning and decision-making process.
Probability Distribution
A Probability Distribution is a statistical function that describes the likelihood of various outcomes. For Poisson Distribution, it specifically models the probability of a certain number of events happening in a fixed interval of time or space.
In Jim's exercise, the Poisson Distribution helps us calculate the probability of 0, 1, 2, or more sales within 60 or 90-day periods. It uses Jim's average rate (or event rate, \( \lambda \)) to find these probabilities through a formula:
  • \( P(r; \lambda) = \frac{e^{-\lambda} \lambda^r}{r!} \)
This formula indicates that once you know the event rate and time frame, you can precisely calculate probabilities for any number of sales. This distribution provides a powerful way to model real-world situations where events are happening rarely and independently.

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Most popular questions from this chapter

Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell Poll was reported in USA Today. According to the survey, \(37 \%\) of adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability that (a) out of five adults, none is concerned that employers are monitoring phone calls. (b) out of five adults, all are concerned that employers are monitoring phone calls. (c) out of five adults, exactly three are concerned that employers are monitoring phone calls.

Consider a binomial distribution with \(n=10\) trials and the probability of success on a single trial \(p=0.05\) (a) Is the distribution skewed left, skewed right, or symmetric? (b) Compute the expected number of successes in 10 trials. (c) Given the low probability of success \(p\) on a single trial, would you expect \(P(r \leq 1)\) to be very high or very low? Explain. (d) Given the low probability of success \(p\) on a single trial, would you expect \(P(r \geq 8)\) to be very high or very low? Explain.

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Let \(r\) be a binomial random variable representing the number of successes out of \(n\) trials. (a) Explain why the sample space for \(r\) consists of the set \(\\{0,1,2, \ldots, n\\}\) and why the sum of the probabilities of all the entries in the entire sample space must be 1. (b) Explain why \(P(r \geq 1)=1-P(0)\) (c) Explain why \(P(r \geq 2)=1-P(0)-P(1)\) (d) Explain why \(P(r \geq m)=1-P(0)-P(1)-\cdots-P(m-1)\) for \(1 \leq m \leq n\).

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