/*! 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 28 Pyramid Lake is located in Nevad... [FREE SOLUTION] | 91Ó°ÊÓ

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Pyramid Lake is located in Nevada on the Paiute Indian Reservation. This lake is famous for large cutthroat trout. The mean number of trout (large and small) caught from a boat is 0.667 fish per hour (Reference: Creel Chronicle, Vol. \(3,\) No. 2 ). Suppose you rent a boat and go fishing for 8 hours. Let \(r\) be a random variable that represents the number of fish you catch in the 8 -hour period. (a) Explain why a Poisson probability distribution is appropriate for \(r\). What is \(\lambda\) for the 8 -hour fishing trip? Round \(\lambda\) to the nearest tenth so that you can use Table 4 of Appendix II for Poisson probabilities. (b) If you have already caught three trout, what is the probability you will catch a total of seven or more trout? Compute \(P(r \geq 7 | r \geq 3) .\) See Hint below. (c) If you have already caught four trout, what is the probability you will catch a total of fewer than nine trout? Compute \(P(r<9 | r \geq 4) .\) See Hint below. (d) List at least three other areas besides fishing to which you think conditional Poisson probabilities can be applied. Hint for solution: Review item \(6,\) conditional probability, in the summary of basic probability rules at the end of Section \(4.2 .\) Note that $$P(A | B)=\frac{P(A \text { and } B)}{P(B)}$$ and show that in part (b), $$P(r \geq 7 | r \geq 3)=\frac{P[(r \geq 7) \text { and }(r \geq 3)]}{P(r \geq 3)}=\frac{P(r \geq 7)}{P(r \geq 3)}$$

Short Answer

Expert verified
(a) Poisson is appropriate; \( \lambda = 5.3. \) (b) Calculate with conditional probability formula. (c) Use the formula with Poisson probabilities. (d) Traffic, call centers, email filtering.

Step by step solution

01

Justify Poisson Distribution

The Poisson probability distribution is suitable for modeling the count of events happening in a fixed interval of time or space when these events occur with a known constant mean rate and independently of each other. Here, the random variable \( r \) describes the number of fish caught, and the events fit the criteria of a Poisson distribution as they describe a number of events (fish caught) happening at a constant rate (mean 0.667 fish/hour) over an 8-hour period.
02

Calculate \( \lambda \) for the 8-hour Period

To find the mean number \( \lambda \) used in the Poisson distribution, multiply the rate of catching fish per hour by the number of hours. From the problem statement, the mean rate is 0.667 fish/hour. For an 8-hour trip: \[ \lambda = 0.667 \times 8 = 5.336. \] Round this to the nearest tenth to simplify computations: \( \lambda \approx 5.3 \).
03

Calculate \( P(r \geq 7 | r \geq 3) \)

We use conditional probability. First, calculate \( P(r \geq 7) \) using the Poisson distribution: this involves summing probabilities for all \( r \geq 7 \). Then, find \( P(r \geq 3) \) by summing probabilities for all \( r \geq 3 \). As per the hint, \( P(r \geq 7 | r \geq 3) = \frac{P(r \geq 7)}{P(r \geq 3)}. \) Utilize Poisson probability tables to obtain these values.
04

Calculate \( P(r

To compute \( P(r < 9 | r \geq 4) \), first determine \( P(r < 9) \) by summing probabilities for \( r \, \) values from 0 to 8. Next, find \( P(r \geq 4) \) by summing probabilities starting from \( r = 4 \). Use the conditional probability formula: \( P(r<9 | r \geq 4) = \frac{P((r<9) \ and \ (r \geq 4))}{P(r \geq 4)} = \frac{P(4 \leq r < 9)}{P(r \geq 4)}. \) The probability tables will give the needed values.
05

Applications of Conditional Poisson Probabilities

Besides fishing, conditional Poisson probabilities can be applied in areas such as traffic flow (e.g., the number of cars passing a checkpoint), call center management (e.g., incoming calls per minute), and email or spam filtering (e.g., receiving emails in a given timeframe). These situations involve counting events occurring at a constant average rate.

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Key Concepts

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

Conditional Probability
Conditional probability is a measure of the likelihood of an event occurring given that another event has already taken place. It's an essential concept in probability theory and particularly useful in situations where additional information is known.

In our fishing scenario, for example, you want to know the probability of catching a total of seven or more fish, given that you've already caught three. This is symbolized as \( P(r \geq 7 | r \geq 3) \). To calculate this, you would use the formula for conditional probability:
  • \( P(A | B) = \frac{P(A \cap B)}{P(B)} \)
In this instance, the event \( A \) is catching seven or more fish, and \( B \) is having caught at least three fish.

The calculation simplifies to \( \frac{P(r \geq 7)}{P(r \geq 3)} \), using the Poisson probabilities for \( r \geq 7 \) and \( r \geq 3 \). This helps quantify your chances given your prior success.
Mean Rate
The mean rate in a Poisson distribution reflects how frequently an event occurs on average within a given time frame or spatial area. This concept is crucial because it sets the stage for calculating probabilities related to event counts.

In our exercise, the mean rate of catching fish is given as 0.667 fish per hour. To apply this to an 8-hour fishing trip, you multiply 0.667 by 8, yielding a mean of 5.336 fish for the 8-hour period. It is often rounded to a more manageable number for use in Poisson tables, hence, 5.3.

Understanding the mean rate allows you to predict the expected number of events (in this case, fish caught) over your interval of interest. This is pivotal in setting up and interpreting the parameters of your Poisson distribution calculations.
Event Occurrence
In the context of Poisson distribution, event occurrence refers to the actual number of times an event (like catching a fish) happens during a given interval. Each event is assumed to happen independently and with a constant probability over time.

This independence of events is one of the hallmarks of Poisson processes. For instance, the likelihood of catching a fish in one hour may not depend on the catch rate of previous hours, provided the conditions remain the same. This ensures that the distribution can reliably model situations like our fishing trip, where we count fish caught over a span of hours.

With different outcomes ranging from fewer to more catches, Poisson probabilities give a structured way to understand these occurrences using the calculated mean rate.
Count Data
Count data are numerical values that represent the number of occurrences of an event within a fixed criterion of time or space. In our discussion, the random variable \( r \) is the count data representing the number of fish caught.

This kind of data is central to Poisson distribution because it models events counted in discrete units. For example, the count data would be valid if measuring something like the number of emails received in an hour or fish caught in a day.
  • This data type is particularly handy in probabilistic modeling because it often follows predictable probability patterns when provided a mean rate.
Exploring count data through Poisson distribution sheds light on expected outcomes and the variability or randomness that could naturally occur, allowing for more informed predictions and decisions in real-world scenarios.

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Most popular questions from this chapter

Expand Your Knowledge: Multinomial Probability Distribution Consider a multinomial experiment. This means the following: 1\. The trials are independent and repeated under identical conditions. 2\. The outcomes of each trial falls into exactly one of \(k \geq 2\) categories. 3\. The probability that the outcomes of a single trial will fall into ith category is \(p_{i}\) (where \(i=1,2 \ldots, k\) ) and remains the same for each trial. Furthermore, \(p_{1}+p_{2}+\ldots+p_{k}=1\) 4\. Let \(r_{i}\) be a random variable that represents the number of trials in which the outcomes falls into category \(i\). If you have \(n\) trials, then \(r_{1}+r_{2}+\ldots\) \(+r_{k}=n .\) The multinational probability distribution is then $$P\left(r_{1}, r_{2}, \cdots r_{k}\right)=\frac{n !}{r_{1} ! r_{2} ! \cdots r_{2} !} p_{1}^{r_{1}} p_{2}^{(2)} \cdots p_{k}^{r_{2}}$$ How are the multinomial distribution and the binomial distribution related? For the special case \(k=2,\) we use the notation \(r_{1}=r, r_{2}=n-r, p_{1}=p\) and \(p_{2}=q .\) In this special case, the multinomial distribution becomes the binomial distribution. The city of Boulder, Colorado is having an election to determine the establishment of a new municipal electrical power plant. The new plant would emphasize renewable energy (e.g., wind, solar, geothermal). A recent large survey of Boulder voters showed \(50 \%\) favor the new plant, \(30 \%\) oppose it, and \(20 \%\) are undecided. Let \(p_{1}=0.5, p_{2}=0.3,\) and \(p_{3}=0.2 .\) Suppose a random sample of \(n=6\) Boulder voters is taken. What is the probability that (a) \(r_{1}=3\) favor, \(r_{2}=2\) oppose, and \(r_{3}=1\) are undecided regarding the new power plant? (b) \(r_{1}=4\) favor, \(r_{2}=2\) oppose, and \(r_{3}=0\) are undecided regarding the new power plant?

What is the age distribution of patients who make office visits to a doctor or nurse? The following table is based on information taken from the Medical Practice Characteristics section of the Statistical Abstract of the United States (116th edition). $$\begin{array}{l|ccccc} \hline \text { Age group, years } & \text { Under 15 } & 15-24 & 25-44 & 45-64 & 65 \text { and older } \\ \hline \text { Percent of office visitors } & 20 \% & 10 \% & 25 \% & 20 \% & 25 \% \\ \hline \end{array}$$ Suppose you are a district manager of a health management organization (HMO) that is monitoring the office of a local doctor or nurse in general family practice. This morning the office you are monitoring has eight office visits on the schedule. What is the probability that (a) at least half the patients are under 15 years old? First, explain how this can be modeled as a binomial distribution with 8 trials, where success is visitor age is under 15 years old and the probability of success is \(20 \%\) (b) from 2 to 5 patients are 65 years old or older (include 2 and 5 )? (c) from 2 to 5 paticnts are 45 years old or older (include 2 and 5 )? Hint: Success is 45 or older. Use the table to compute the probability of success on a single trial. (d) all the patients are under 25 years of age? (e) all the patients are 15 years old or older?

The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of eight syringes taken from the batch. Suppose the batch contains \(1 \%\) defective syringes. (a) Make a histogram showing the probabilities of \(r=0,1,2,3,4,5,6,7,\) and 8 defective syringes in a random sample of eight syringes. (b) Find \(\mu\). What is the expected number of defective syringes the inspector will find? (c) What is the probability that the batch will be accepted? (d) Find \(\sigma\)

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) 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?

This problem shows you how to make a better blend of almost anything. Let \(x_{1}, x_{2}, \ldots x_{n}\) be independent random variables with respective variances \(\sigma_{1}^{2}\) \(\sigma_{2}^{2}, \ldots, \sigma_{n}^{2}.\) Let \(c_{1}, c_{2}, \ldots, c_{n}\) be constant weights such that \(0 \leq c_{i} \leq 1\) and \(c_{1}+c_{2}+\ldots+\) \(c_{n}=1 .\) The linear combination \(w=c_{1} x_{1}+c_{2} x_{2}+\ldots+c_{n} x_{n}\) is a random variable with variance $$\sigma_{w}^{2}=c_{1}^{2} \sigma_{1}^{2}+c_{2}^{2} \sigma_{2}^{2}+\cdots+c_{n}^{2} \sigma_{n}^{2}$$ (a) In your own words write a brief explanation regarding the following statement: The variance of \(w\) is a measure of the consistency or variability of performance or outcomes of the random variable \(w\). To get a more consistent performance out of the blend \(w\), choose weights \(c_{i}\) that make \(\sigma_{w}^{2}\) as small as possible. Now the question is how do we choose weights \(c_{i}\) to make \(\sigma_{w}^{2}\) as small as possible? Glad you asked! A lot of mathematics can be used to show the following choice of weights will minimize \(\sigma_{w}^{2}.\) $$c_{i}=\frac{\frac{1}{\sigma_{i}^{2}}}{\left[\frac{1}{\sigma_{i}^{2}}+\frac{1}{\sigma_{2}^{\frac{1}{2}}}+\cdots+\frac{1}{\sigma_{n}^{2}}\right]} \quad \text { for } i=1,2, \cdots, n$$ (Reference: Introduction to Mathematical Statistics, 4th edition, by Paul Hoel.) (b)Two types of epoxy resin are used to make a new blend of superglue. Both resins have about the same mean breaking strength and act independently. The question is how to blend the resins (with the hardener) to get the most consistent breaking strength. Why is this important, and why would this require minimal \(\sigma_{w}^{2} ?\) Hint: We don't want some bonds to be really strong while others are very weak, resulting in inconsistent bonding. Let \(x_{1}\) and \(x_{2}\) be random variables representing breaking strength (lb) of each resin under uniform testing conditions. If \(\sigma_{1}=8\) lb \(\sigma_{2}=12\) lb, show why a blend of about \(69 \%\) resin 1 and \(31 \%\) resin 2 will result in a superglue with smallest \(\sigma_{w}^{2}\) and most consistent bond strength. (c) Use \(c_{1}=0.69\) and \(c_{2}=0.31\) to compute \(\sigma_{w}\) and show that \(\sigma_{w}\) is less than both \(\sigma_{1}\) and \(\sigma_{2}\) The dictionary meaning of the word synergetic is "working together or cooperating for a better overall effect." Write a brief explanation of how the blend \(w=c_{1} x_{1}+c_{2} x_{2}\) has a synergetic effect for the purpose of reducing variance.

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