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Chapter 3: Problem 126

Grasshoppers are distributed at random in a large field according to a Poisson distribution with parameter \(\alpha=2\) per square yard. How large should the radius \(R\) of a circular sampling region be taken so that the probability of finding at least one in the region equals \(.99\) ?

Short Answer

Expert verified
The radius \( R \) should be approximately 1.084 yards.

Step by step solution

01

Understanding Poisson Distribution

The scenario describes using a Poisson distribution where grasshoppers are spread randomly across a field. We know the parameter \(\alpha = 2\) represents the average number of grasshoppers per square yard.
02

Defining the Region

Since we're examining a circular region, the area \( A \) of the region can be calculated using the formula for the area of a circle: \( A = \pi R^2 \), where \( R \) is the radius of the circle. This is the area in which we are considering the probability of finding grasshoppers.
03

Setting Up the Poisson Probabilities

The number of grasshoppers in the circular area is still a Poisson random variable, now with parameter \(\lambda = \alpha \times A = 2 \times \pi R^2\). We need the probability of finding at least one grasshopper in this area, which is \( P(X \geq 1) = 1 - P(X = 0) \).
04

Calculating Probability of Zero Grasshoppers

The probability of finding zero grasshoppers in the area is given by the Poisson probability formula: \( P(X = 0) = e^{-\lambda} \). Hence, \( P(X = 0) = e^{-2\pi R^2} \).
05

Setting the Desired Probability

We want \( 1 - P(X = 0) = 0.99 \), which implies \( P(X = 0) = 0.01 \). Thus, \( e^{-2\pi R^2} = 0.01 \).
06

Solving for Radius \(R\)

We take the natural logarithm of both sides to solve for \( R \): \( -2\pi R^2 = \ln(0.01) \). Thus, \( R^2 = -\frac{\ln(0.01)}{2\pi} \). Calculating this gives \( R \approx 1.084 \).
07

Conclusion

The radius \( R \) should be about \( 1.084 \) yards to ensure a 0.99 probability of finding at least one grasshopper.

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Key Concepts

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

Probability
Probability is a measure that quantifies the likelihood of an event occurring. It ranges from 0 to 1, where 0 means the event will not occur, and 1 means it will certainly happen. In the context of Poisson distribution, we are often interested in the probability of a certain number of events occurring in a fixed region or interval.
For this exercise involving grasshoppers, probability helps us determine how likely it is to find at least one grasshopper within a circular area. We use complement probability to tackle this, finding the probability of zero occurrences first and then subtracting from 1. This process is often used in Poisson probability scenarios.
Random Variable
A random variable is a variable that takes on different numerical values based on the outcomes of a random phenomenon. In this problem, the random variable is the number of grasshoppers found in the specified circular area.
In a Poisson distribution, the random variable is usually characterized by a parameter that indicates the expected number of events (in this case, grasshoppers) per unit measurement. The random variable can assume any non-negative integer value, indicating the number of occurrences in that area.
Area of a Circle
The area of a circle is a fundamental concept in geometry, used to describe the space enclosed by the circle. It is calculated using the formula \( A = \pi R^2 \), where \( R \) is the radius. In this exercise, the area is crucial because it defines the size of the sampling region where we are assessing the probability of grasshoppers.
The Poisson parameter for the problem is dependent on the area of this circle, influencing the likelihood calculations. A larger area increases the probability of encountering more grasshoppers, whereas a smaller area reduces it.
Sampling Region
The sampling region is the specific area in which we are interested in studying and calculating probabilities. In this exercise, it is a circular region of a field where we want to know the chance of finding at least one grasshopper.
The choice of the sampling region's size and shape directly affects the Poisson parameters and hence the probability outcome. By adjusting the size of this sampling region, we can determine the probability of detecting grasshoppers. For this problem, an ideal radius is calculated to achieve a certain level of probability, systematically determining the sampling region that meets the desired threshold of detection probability.

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Most popular questions from this chapter

Customers at a gas station pay with a credit card (A), debit card (B), or cash (C). Assume that successive customers make independent choices, with \(P(A)=.5, P(B)=.2\), and \(P(C)=.3 .\) a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning. b. Answer part (a) for the number among the 100 who don't pay with cash.

An article in the Los Angeles Times (Dec. 3 , 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one \(P(\) Wed. \()=.3\), \(P(\) Thurs. \()=.4, P(\) Fri. \()=.2\), and \(P(\) Sat. \()=.1\). Let \(Y=\) the number of days beyond Wednesday that it takes for both magazines to arrive (so possible \(Y\) values are \(0,1,2\), or 3 ). Compute the pmf of \(Y\). [Hint: There are 16 possible outcomes; \(Y(W, W)=0, Y(F, T h)=2\), and so on. \(]\)

Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate \(\alpha=10 / \mathrm{h}\). Suppose that with probability .5 an arriving vehicle will have no equipment violations. a. What is the probability that exactly ten arrive during the hour and all ten have no violations? b. For any fixed \(y \geq 10\), what is the probability that \(y\) arrive during the hour, of which ten have no violations? c. What is the probability that ten "no-violation" cars arrive during the next hour? [Hint: Sum the probabilities in part (b) from \(y=10\) to \(\infty\).]

If the sample space \(\delta\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(s\) will have an infinite set of possible values? If yes, say why. If no, give an example.
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