/*! 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 96 According to the article "Optimi... [FREE SOLUTION] | 91Ó°ÊÓ

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According to the article "Optimization of Distribution Parameters for Estimating Probability of Crack Detection" (J. of Aircraft, 2009: 2090-2097), the following "Palmberg" equation is commonly used to determine the probability \(P_{d}(c)\) of detecting a crack of size \(c\) in an aircraft structure: $$ P_{d}(c)=\frac{\left(c / c^{*}\right)^{\beta}}{1+\left(c / c^{*}\right)^{\beta}} $$ where \(c^{*}\) is the crack size that corresponds to a \(.5\) detection probability (and thus is an assessment of the quality of the inspection process). a. Verify that \(P_{d}\left(c^{*}\right)=.5\) b. What is \(P_{d}\left(2 c^{*}\right)\) when \(\beta=4\) ? c. Suppose an inspector inspects two different panels, one with a crack size of \(c^{*}\) and the other with a crack size of \(2 c^{*}\). Again assuming \(\beta=4\) and also that the results of the two inspections are independent of one another, what is the probability that exactly one of the two cracks will be detected? d. What happens to \(P_{d}(c)\) as \(\beta \rightarrow \infty\) ?

Short Answer

Expert verified
a) Verified, b) \(0.9412\), c) \(0.5\), d) Step function: 0 or 1.

Step by step solution

01

Verify P_d(c^*) = 0.5

To verify that \( P_{d}(c^{*}) = 0.5 \) we simply substitute \( c = c^{*} \) into the Palmberg equation:\[P_{d}(c^{*}) = \frac{\left(c^{*} / c^{*}\right)^{\beta}}{1 + \left(c^{*} / c^{*}\right)^{\beta}} = \frac{1^{\beta}}{1 + 1^{\beta}} = \frac{1}{2}.\]Thus, our verification is complete.
02

Calculate P_d(2c^*) for β=4

We need to determine \( P_{d}(2c^{*}) \) when \( \beta = 4 \). Plugging into the equation we get:\[P_{d}(2c^{*}) = \frac{\left(2c^{*} / c^{*}\right)^{4}}{1 + \left(2c^{*} / c^{*}\right)^{4}} = \frac{2^{4}}{1 + 2^{4}} = \frac{16}{17}.\]Thus, \( P_{d}(2c^{*}) = \frac{16}{17} \approx 0.9412 \).
03

Probability of Exactly One Detection

To find the probability of exactly one detection, we first find the probabilities for detecting each crack. From Step 1, \( P_{d}(c^{*}) = 0.5 \). For the second crack, from Step 2, \( P_{d}(2c^{*}) = \frac{16}{17} \). Since the inspections are independent, the probability of exactly one detection is:\[ P(\text{exactly one detected}) = P_{d}(c^{*}) \cdot (1 - P_{d}(2c^{*})) + (1 - P_{d}(c^{*})) \cdot P_{d}(2c^{*}). \]Substituting in values:\[= 0.5 \times \left(1 - \frac{16}{17}\right) + (1 - 0.5) \times \frac{16}{17} = 0.5 \times \frac{1}{17} + 0.5 \times \frac{16}{17} = \frac{1}{34} + \frac{8}{17} = \frac{1 + 16}{34} = \frac{17}{34} = 0.5.\]Therefore, the probability of exactly one crack being detected is 0.5.
04

Limit Behavior as β → ∞

We need to evaluate how \( P_{d}(c) \) behaves as \( \beta \rightarrow \infty \). Consider:\[P_{d}(c) = \frac{\left(c / c^{*}\right)^{\beta}}{1 + \left(c / c^{*}\right)^{\beta}}. \]If \( c < c^{*} \), then \( \left(c / c^{*}\right)^{\beta} \rightarrow 0 \) as \( \beta \rightarrow \infty \), so \( P_{d}(c) \rightarrow 0 \).If \( c > c^{*} \), then \( \left(c / c^{*}\right)^{\beta} \rightarrow \infty \), so \( P_{d}(c) \rightarrow 1 \).If \( c = c^{*} \), \( P_{d}(c) \) stays exactly \( 0.5 \) regardless of \( \beta \).Thus, as \( \beta \rightarrow \infty \), \( P_{d}(c) \) approaches a step function, being mostly 0 or 1.

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Key Concepts

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

Palmberg Equation
The Palmberg Equation is a mathematical formula used in the context of engineering and quality testing, especially for estimating the probability of detecting structural flaws like cracks in aircraft components. It is expressed as: \[ P_{d}(c) = \frac{\left(c / c^{*}\right)^{\beta}}{1+\left(c / c^{*}\right)^{\beta}} \]Here,
  • \( c \) is the size of the crack.
  • \( c^{*} \) represents the characteristic crack size that equates to a 50% probability of being detected. This threshold is crucial as it benchmarks the efficacy of the inspection process.
  • \( \beta \) is a parameter that governs the steepness of the detection curve. It usually captures the sensitivity of the detection process.
This formula indicates the likelihood that a given defect—characterized by its size—will be discovered during an inspection. Understanding this relationship helps in evaluating and enhancing inspection techniques, making them more reliable and effective.
Crack Size Detection
Crack size detection is a critical process in fields such as aerospace engineering, where maintaining structural integrity is paramount. The concept revolves around determining the probability of detecting a crack of specific dimensions. Key factors influencing crack detection include:
  • The size of the crack \( c \), with larger cracks usually easier to detect.
  • The inspection method and its sensitivity, often encapsulated in the parameter \( \beta \) of the Palmberg Equation.
  • Environmental and operational conditions during inspection.
To put this into perspective, in a typical scenario described by the Palmberg Equation, for a crack of size \( c = c^{*} \), detection probability stands at 50%. Identifying how the probability scales with different sizes, such as \( 2c^{*} \), aids in optimizing inspection protocols and ensuring the safety and reliability of structural components.
Independent Events
Understanding the concept of independent events is critical in probability, particularly in situations where multiple inspections or tests are conducted. Independent events imply that the outcome of one event does not influence the outcome of another. For example, when inspecting two separate panels, one with a crack size equal to \( c^{*} \) and another with \( 2c^{*} \), assuming independence means:
  • The detection of one crack does not impact the probability of detecting the other.
  • The combined probability of outcomes, such as detecting exactly one crack, is calculated using the individual probabilities from each panel inspection.
In our specific case, independence allows us to determine the probability of detecting exactly one crack using basic probability rules: \[ P(\text{exactly one detected}) = P_{d}(c^{*}) \cdot (1 - P_{d}(2c^{*})) + (1 - P_{d}(c^{*})) \cdot P_{d}(2c^{*}) \]This rule emphasizes that the dependencies between inspections do not contribute, simplifying complex calculations and scenarios.
Limit Behavior
The limit behavior of a probability model is crucial for understanding how it behaves under extreme conditions. In the context of the Palmberg Equation, examining the limit as \( \beta \rightarrow \infty \) reveals insightful thresholds in detection sensitivity. Here's what happens:
  • If the crack size \( c \) is less than \( c^{*} \), the probability of detection \( P_{d}(c) \) diminishes towards zero as \( \beta \) increases.
  • If \( c \) is greater than \( c^{*} \), the probability surges towards one, suggesting near-certain detection.
  • For exactly \( c = c^{*} \), the probability remains 0.5, showing robustness at this benchmark no matter the value of \( \beta \).
In essence, as \( \beta \) rises, the equation morphs into a step function, distinctly separating detectable from undetectable sizes. This insight is critical for designing detection systems that are finely tuned to capture flaws only when they truly matter, thus balancing false positives and undetected defects.

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Most popular questions from this chapter

One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. a. What is the probability that a red ball is selected from the first box and a red ball is selected from the second box? b. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

A certain system can experience three different types of defects. Let \(A_{i}(i=1,2,3)\) denote the event that the system has a defect of type \(i\). Suppose that $$ \begin{aligned} &P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05 \\ &P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14 \\\ &P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 \end{aligned} $$ a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

Three molecules of type \(A\), three of type \(B\), three of type \(C\), and three of type \(D\) are to be linked together to form a chain molecule. One such chain molecule is \(A B C D A B C D A B C D\), and another is \(B C D D A A A B D B C C\). a. How many such chain molecules are there? [Hint: If the three \(A\) 's were distinguishable from one another- \(A_{1}, A_{2}\), \(A_{3}\)-and the \(B\) 's, C's, and D's were also, how many molecules would there be? How is this number reduced when the subscripts are removed from the \(A\) 's?] b. Suppose a chain molecule of the type described is randomly selected. What is the probability that all three molecules of each type end up next to one another (such as in \(B B B A A A D D D C C C\) )?

Let \(A\) denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let \(B\) be the event that the next request is for help with SAS. Suppose that \(P(A)=.30\) and \(P(B)=.50\). a. Why is it not the case that \(P(A)+P(B)=1\) ? b. Calculate \(P\left(A^{\prime}\right)\). c. Calculate \(P(A \cup B)\). d. Calculate \(P\left(A^{\prime} \cap B^{\prime}\right)\).

An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let \(A\) be the event that the Asian project is successful and \(B\) be the event that the European project is successful. Suppose that \(A\) and \(B\) are independent events with \(P(A)=.4\) and \(P(B)=.7\). a. If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning. b. What is the probability that at least one of the two projects will be successful? c. Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

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