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The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes" (J. of Pipeline Systems Engr. and Practice, May 2012: 36-46) proposed using the Poisson distribution to model the number of failures in pipelines of various types. Suppose that for cast-iron pipe of a particular length, the expected number of failures is 1 (very close to one of the cases considered in the article). Then \(X\), the number of failures, has a Poisson distribution with \(\mu=1\). a. Obtain \(P(X \leq 5)\) by using Appendix Table A.2. b. Determine \(P(X=2)\) first from the pmf formula and then from Appendix Table A.2. c. Determine \(P(2 \leq X \leq 4)\). d. What is the probability that \(X\) exceeds its mean value by more than one standard deviation?

Short Answer

Expert verified
a: Use CDF to find \(P(X \leq 5)\); b: Use PMF and table to confirm \(P(X = 2)\); c: Sum of \(P(X = 2, 3, 4)\); d: Calculate \(P(X \geq 3)\).

Step by step solution

01

Understanding the Poisson Distribution

The Poisson distribution is defined for the number of events in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. Here, the expected number of failures, \(\mu = 1\). Thus, the random variable \(X\) follows a Poisson distribution with \(\lambda = 1\).
02

Obtain \(P(X \leq 5)\) using the Poisson CDF

Use the cumulative distribution function (CDF) for a Poisson random variable with the parameter \(\lambda = 1\) from statistical tables (Appendix Table A.2).\[P(X \leq 5) = \sum_{x=0}^{5} e^{-1} \frac{1^x}{x!}\]Calculate terms individually and sum them up to obtain the probability.
03

Calculate \(P(X=2)\) using the PMF

The probability mass function (PMF) for a Poisson distribution is given by:\[P(X = k) = e^{-\lambda} \frac{\lambda^k}{k!}\]Substituting \(\lambda = 1\) and \(k = 2\):\[P(X = 2) = e^{-1} \frac{1^2}{2!} = \frac{e^{-1}}{2}\]Perform the calculation to find the probability.
04

Validate \(P(X=2)\) using Appendix Table A.2

Check Appendix Table A.2 for the Poisson distribution with \(\lambda = 1\) and \(X = 2\) to confirm the result from Step 2.
05

Calculate \(P(2 \leq X \leq 4)\)

To find \(P(2 \leq X \leq 4)\), calculate the probabilities of each value in the range and sum them:\[P(2 \leq X \leq 4) = P(X=2) + P(X=3) + P(X=4)\]Calculate each term using the PMF formula and sum the results.
06

Determine \(P(X > 2)\) for the Standard Deviation Check

Calculate the standard deviation \(\sigma\) for the Poisson distribution:\[\sigma = \sqrt{\lambda} = \sqrt{1} = 1\]Find \(P(X > 2)\), meaning \(X\) exceeds the mean by more than one standard deviation. This is equivalent to calculating:\[P(X \geq 3)\]Use the complement rule \(1 - P(X \leq 2)\) to calculate it.

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (PMF) is a key concept when dealing with discrete probability distributions, such as the Poisson distribution. The PMF provides the probability that a discrete random variable is exactly equal to a specific value. In the context of the Poisson distribution, the PMF can be written as:\[P(X = k) = e^{-\lambda} \frac{\lambda^k}{k!}\]where:
  • \(e\) is the base of the natural logarithm
  • \(\lambda\) is the average rate at which events occur
  • \(k\) is the specific number of events
  • \(k!\) is the factorial of \(k\)
For instance, if \(\lambda = 1\) (where the mean number of pipeline failures is 1), the probability of exactly 2 failures, \(P(X = 2)\), can be found using this formula. Calculating the PMF involves substituting the values into the formula and computing the result. It helps in determining the likelihood of observing a certain number of failures and is crucial for assessing specific scenarios, such as a given number of pipeline failures over a fixed interval.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) for discrete variables like those in a Poisson distribution represents the probability that a random variable \(X\) is less than or equal to a certain value. The CDF is an aggregation of the PMF values over a range of interest. When you want to know, for example, \(P(X \leq 5)\), you're using the CDF.For the Poisson distribution:\[P(X \leq k) = \sum_{x=0}^{k} e^{-\lambda} \frac{\lambda^x}{x!}\]This formula sums the probabilities of \(X = 0\) up to \(X = k\). By using the CDF, instead of calculating the probability for each individual point and adding them up manually, one can look up these values in statistical tables or compute them from the formula. The CDF thus simplifies finding the probability over a range of values, streamlining calculations for scenarios such as determining pipeline failure rates up to a certain number.
Pipeline Failure Analysis
Pipeline failure analysis involves using statistical models to predict and understand various failure scenarios over time or space. The Poisson distribution is especially suited for events that occur independently with a constant mean rate, making it perfect for modeling pipeline failures.In pipeline management, understanding the expected number of failures can inform maintenance decisions and risk assessments. The Poisson distribution, with its single parameter \(\lambda\), where \(\lambda\) is the expected number of occurrences (failures) within a given period or length, provides a practical approach.Benefits of using the Poisson distribution in pipeline failure analysis include:
  • Predictive insights into potential problem areas
  • Cost-saving strategies by preventive maintenance based on failure probabilities
  • Risk management by preparing for likely failure scenarios
These analyses are crucial for ensuring the reliability and safety of infrastructure like water supply systems.
Standard Deviation in Poisson Distribution
The standard deviation in a Poisson distribution shows the spread or variability of the distribution around the mean. For a Poisson distribution, the standard deviation \(\sigma\) is simply the square root of the mean \(\lambda\). This relationship is expressed as:\[\sigma = \sqrt{\lambda}\]In scenarios like the exercise provided, where \(\lambda = 1\), the standard deviation is also 1. The standard deviation helps determine how much data points are likely to deviate from the mean and is useful for checking probabilities that involve deviations from the mean, such as determining the probability that the number of failures will exceed the mean by more than one standard deviation.Key roles of standard deviation in Poisson analysis include:
  • Assessing the variability compared to the mean, aiding in statistical significance assessments
  • Enabling decision-makers to understand and predict the amount of scatter in failure data
  • Facilitating the establishment of thresholds for anomaly detection in pipeline management

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Most popular questions from this chapter

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 \(.999\) ?

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\mu=20\) (suggested in the article "'Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

The probability that a randomly selected box of a certain type of cereal has a particular prize is .2. Suppose you purchase box after box until you have obtained two of these prizes. a. What is the probability that you purchase \(x\) boxes that do not have the desired prize? b. What is the probability that you purchase four boxes? c. What is the probability that you purchase at most four boxes? d. How many boxes without the desired prize do you expect to purchase? How many boxes do you expect to purchase?

Starting at a fixed time, each car entering an intersection is observed to see whether it turns left \((L)\), right \((R)\), or goes straight ahead \((A)\). The experiment terminates as soon as a car is observed to turn left. Let \(X=\) the number of cars observed. What are possible \(X\) values? List five outcomes and their associated \(X\) values.

According to the article "'Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Planning and Mgmnt.s 2005: 383-393), the drought length \(Y\) is the number of consecutive time intervals in which the water supply remains below a critical value \(y_{0}\) (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). The cited paper proposes a geometric distribution with \(p=.409\) for this random variable. a. What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals? b. What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

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