/*! 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 86 Organisms are present in ballast... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Organisms are present in ballast water discharged from a ship according to a Poisson process with a concentration of 10 organisms/ \(\mathrm{m}^{3}\) [the article "4 Counting at Low Concentrations: The Statistical Challenges of Verifying Ballast Water Discharge Standards" (Ecological Applications, 2013: 339-351) considers using the Poisson process for this purpose]. a. What is the probability that one cubic meter of discharge contains at least 8 organisms? b. What is the probability that the number of organisms in \(1.5 \mathrm{~m}^{3}\) of discharge exceeds its mean value by more than one standard deviation? c. For what amount of discharge would the probability of containing at least 1 organism be \(.999\) ?

Short Answer

Expert verified
a. 0.932, b. 0.158, c. 0.69 m³

Step by step solution

01

Define the Poisson process for part (a)

For part (a), we are given that the concentration of organisms follows a Poisson process with a mean rate, \(\lambda = 10\) organisms per cubic meter. For the Poisson distribution, the probability mass function is given by:\[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]where \(X\) is the number of organisms, \(k\) is a specific count (in this case \( \geq 8\)), and \(\lambda = 10\).
02

Calculate the cumulative probability for part (a)

To find the probability that there are at least 8 organisms in one cubic meter, calculate the cumulative probability for \(X < 8\) and subtract from 1:\[P(X \geq 8) = 1 - \sum_{k = 0}^{7} \frac{e^{-10} \cdot 10^k}{k!}\]Compute this using a calculator or statistical software.
03

Define the Poisson process for part (b)

For part (b), we consider a volume of \(1.5 \, \text{m}^3\). The new mean \(\lambda\) for this volume is \(1.5 \times 10 = 15\). The standard deviation for a Poisson distribution is the square root of the mean, so \(\sigma = \sqrt{15}\).
04

Set up inequality for part (b)

We need the probability that the number of organisms exceeds its mean plus one standard deviation:\[P(X > 15 + \sqrt{15})\]Calculate \(15 + \sqrt{15}\), and compute the cumulative probability \(P(X \leq \lfloor 15 + \sqrt{15} \rfloor)\), where \(\lfloor \cdot \rfloor\) denotes the floor function, then subtract from 1.
05

Solve the cumulative probability for part (b)

Calculate:\[P(X > 15 + \sqrt{15}) = 1 - \sum_{k=0}^{\lfloor 15 + \sqrt{15} \rfloor} \frac{e^{-15} \cdot 15^k}{k!}\]Use a calculator or statistical software to find this value.
06

Determine volume for part (c)

For part (c), we want to find the volume \(V\) such that \(P(X \geq 1) = 0.999\). The mean becomes \(\lambda = 10 \cdot V\). We have:\[P(X = 0) = e^{-10V}\]Ensure that:\[1 - e^{-10V} = 0.999\]Solving for \(V\) gives:\[e^{-10V} = 0.001\]Take the natural logarithm of both sides to solve for \(V\).
07

Solve the equation for part (c)

Compute \(V\):\[V = -\frac{\ln(0.001)}{10}\]Using a calculator, find the numerical value for \(V\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Calculations
Probability calculations are essential in determining how likely certain events are to occur. In the context of the Poisson distribution, the focus is on finding the probability of a certain number of events happening within a fixed interval of time or space. Using the formula for the probability mass function (PMF),
  • \[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]
  • "\(X\)" is the random variable representing the number of events,
  • "\(k\)" is the number of events of interest,
  • "\(\lambda\)" is the average number of events in the interval,
the probability of a specific number of occurrences can be calculated. For instance, to find the probability of at least 8 organisms in a cubic meter, calculate the sum of probabilities for having fewer than 8 organisms and subtract from 1. This requires calculating each probability for \(k = 0\) to \(k = 7\), summing them up, and using:
  • \[P(X \geq 8) = 1 - \sum_{k=0}^{7} \frac{e^{-10} \cdot 10^k}{k!}\]
This cumulative approach helps in computing probabilities over a range of outcomes.
Mean and Standard Deviation
In a Poisson distribution, the mean and standard deviation are key parameters that describe the data. For this exercise, the mean (\(\lambda\)) is particularly vital as it represents the average rate of occurrences over a fixed interval. Given the problem's scenario:
  • Mean, \(\lambda = 10\) organisms per cubic meter for 1 \(\text{m}^3\).
When dealing with a different volume, such as 1.5 \(\text{m}^3\), the mean changes to:
  • \(1.5 \times 10 = 15\).
The standard deviation in a Poisson distribution is the square root of the mean, thus:
  • \(\sigma = \sqrt{15}\).
Understanding that the mean gives you the expected number of events, while the standard deviation tells you how much the number of events can vary, is crucial. To see if the count exceeded its mean by one standard deviation, compute:
  • \(P(X > 15 + \sqrt{15})\).
This step requires summing up probabilities for all values untill this threshold, enhancing your understanding of fluctuating data around expected values.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) is a powerful tool that gives the probability that a random variable is less than or equal to a certain value. In Poisson distribution calculations, it helps to determine probabilities for ranges of values rather than just a single outcome. For example, if you want to find the probability that there are zero organisms present, the CDF provides:
  • \[P(X = 0) = e^{-10V}\]
Calculating CDF involves summing up probabilities from the starting point \(k = 0\) up to the desired value. This is used in calculating the probability of "at least" scenarios by subtracting from 1. In part (c) of the example problem, we want the probability of having at least one organism to be 0.999. The steps are:
  • Ensure \(P(X = 0) = e^{-10V}\) simplifies to \[1 - e^{-10V} = 0.999\]
  • Solve for \(V\) as \[V = -\frac{\ln(0.001)}{10}\]
The CDF not only simplifies calculations for these probability ranges but also provides a clearer picture of the likelihood of events occurring below or above a certain threshold. This understanding is crucial for real-world applications like determining the adequacy of ballast water discharge in avoiding ecological harm.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, \(60 \%\) can be repaired, whereas the other \(40 \%\) must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

The Centers for Disease Control and Prevention reported in 2012 that 1 in 88 American children had been diagnosed with an autism spectrum disorder (ASD). a. If a random sample of 200 American children is selected, what are the expected value and standard deviation of the number who have been diagnosed with ASD? b. Referring back to (a), calculate the approximate probability that at least 2 children in the sample have been diagnosed with ASD? c. If the sample size is 352 , what is the approximate probability that fewer than 5 of the selected children have been diagnosed with ASD?

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is \begin{tabular}{l|cccc} \(y\) & 0 & 1 & 2 & 3 \\ \hline\(p(y)\) & \(.60\) & \(.25\) & \(.10\) & \(.05\) \end{tabular} a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

Possible values of \(X\), the number of components in a system submitted for repair that must be replaced, are 1 , 2,3 , and 4 with corresponding probabilities .15, .35,.35, and \(.15\), respectively. a. Calculate \(E(X)\) and then \(E(5-X)\). b. Would the repair facility be better off charging a flat fee of \(\$ 75\) or else the amount \(\$[150 /(5-X)]\) ? [Note: It is not generally true that \(E(c / Y)=c / E(Y)\).]

Three brothers and their wives decide to have children until each family has two female children. What is the pmf of \(X=\) the total number of male children born to the brothers? What is \(E(X)\), and how does it compare to the expected number of male children born to each brother?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.