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The number of flaws in bolts of cloth in textile manufacturing is assumed to be Poisson distributed with a mean of 0.1 flaw per square meter. (a) What is the probability that there are two flaws in one square meter of cloth? (b) What is the probability that there is one flaw in 10 square meters of cloth? (c) What is the probability that there are no flaws in 20 square meters of cloth? (d) What is the probability that there are at least two flaws in 10 square meters of cloth?

Short Answer

Expert verified
(a) 0.0045, (b) 0.3679, (c) 0.1353, (d) 0.2642.

Step by step solution

01

Understand the Poisson Distribution Formula

The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space. The formula for the probability of observing \( x \) events is given by \( P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!} \), where \( \lambda \) is the average number of events in the interval, \( x \) is the number of events we want to find the probability for, and \( e \) is the base of the natural logarithm.
02

Solve for Part (a)

First, we need to determine \( \lambda \), which is the average number of flaws per square meter. Here, \( \lambda = 0.1 \). We want to find the probability of 2 flaws, so \( x = 2 \). Substitute these into the formula:\[ P(X = 2) = \frac{0.1^2 \cdot e^{-0.1}}{2!} = \frac{0.01 \cdot e^{-0.1}}{2} \approx 0.0045. \]
03

Calculate lambda for Other Intervals

For different lengths of cloth, the average number of flaws \( \lambda \) changes accordingly. For 10 square meters, \( \lambda = 10 \times 0.1 = 1 \). For 20 square meters, \( \lambda = 20 \times 0.1 = 2 \). We'll use these in the subsequent calculations.
04

Solve for Part (b)

For 10 square meters, \( \lambda = 1 \). We need the probability of having 1 flaw, so \( x = 1 \). Substitute into the formula:\[ P(X = 1) = \frac{1^1 \cdot e^{-1}}{1!} = 1 \cdot e^{-1} \approx 0.3679. \]
05

Solve for Part (c)

When dealing with 20 square meters, \( \lambda = 2 \). We need the probability there are no flaws, so \( x = 0 \). The formula becomes:\[ P(X = 0) = \frac{2^0 \cdot e^{-2}}{0!} = e^{-2} \approx 0.1353. \]
06

Solve for Part (d)

For at least 2 flaws in 10 square meters (\( \lambda = 1 \)), this is equivalent to 1 minus the probability of having fewer than 2 flaws (0 or 1 flaw). Calculate:\[ P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) \]\[ P(X = 0) = \frac{1^0 \cdot e^{-1}}{0!} = e^{-1} \approx 0.3679 \]\[ P(X = 1) = \frac{1^1 \cdot e^{-1}}{1!} = e^{-1} \approx 0.3679 \]Adding these gives \( 0.3679 + 0.3679 \approx 0.7358 \). The probability of at least 2 flaws is \( 1 - 0.7358 = 0.2642 \).

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental concept in statistics that helps predict the likelihood of various outcomes. When it comes to the Poisson distribution, we're dealing with calculating the probability of a number of events—such as defects occurring in fabric—within a specified space or time. It's essential to understand that the Poisson formula is given by \( P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!} \), where \( \lambda \) represents the average rate of occurrence, \( x \) is the number of events, and \( e \) is Euler's number. Here, for example, if the mean value of flaws is \( 0.1 \) per square meter, and we need to find the probability of observing exactly 2 flaws, the formula adjusts these values accordingly. This systematic method allows calculating probabilities for different potential events, imparting valuable insights into expected outcomes.
Textile Manufacturing
Textile manufacturing involves a variety of processes, from weaving to finishing, with a primary goal of producing quality fabric. However, despite meticulous craftsmanship, the occurrence of flaws like misweaves or small holes is inevitable. These flaws can considerably influence the fabric’s final quality, making flaw detection and probability analysis essential components of the manufacturing process.
When analyzing these flaws through statistical tools like the Poisson distribution, manufacturers gain foresight into the frequency of flawed areas. Understanding such probabilities can guide quality control processes, assisting in improving manufacturing standards, adjusting machinery settings, and training personnel effectively.
  • Improving Quality: Enhanced flaw prediction leads to higher quality textiles.
  • Cost-Efficiency: Reduces waste by identifying and addressing common problem areas early.
Thus, incorporating statistical analysis directly into textile production yields significant operational advantages, elevating the overall standard of the goods produced.
Flaw Detection
Flaw detection in textiles is a critical task that integrates both human oversight and technology. The main objective here is to identify any deviations or imperfections in the fabric immediately. This proactive approach minimizes product defects reaching the end-user.
A Poisson distribution can model these detection processes by examining past data on flaw frequency.
  • Automated Systems: Utilize sensors and AI to continuously monitor fabric quality.
  • Manual Inspection: Committed inspectors evaluate textiles visually or with magnification tools.
Using statistical measures such as Poisson distribution helps predict and manage potential flaws by providing probabilities of their occurrences. It's influential in setting quality thresholds, designing corrective action, and ensuring that production lines remain efficient and problems are quickly addressed.
Mean Value Analysis
Mean value analysis is a significant statistical concept especially useful in operations like textile manufacturing. Here, the mean represents the average rate of flaw occurrence over a unit area, such as per square meter. Calculating mean values allows textile manufacturers to assess whether their production processes consistently meet quality expectations.
In the context of the Poisson distribution, the mean (represented as \( \lambda \)) plays a vital role in estimating probabilities and informing manufacturing strategies. This analysis can be extended across different scales—for instance, finding the mean over larger fabric areas or sequential time periods.
  • Quality Benchmarks: Establishes expected standards for flaw frequency.
  • Operations Enhancement: Data-driven insights improve both short-term adjustments and long-term planning.
Mean value analysis, thus, equips manufacturing managers with robust data to monitor, control, and optimize the quality of their textile products.

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Most popular questions from this chapter

Orders arrive at a Web site according to a Poisson process with a mean of 12 per hour. Determine the following: (a) Probability of no orders in five minutes. (b) Probability of 3 or more orders in five minutes. (c) Length of a time interval such that the probability of no orders in an interval of this length is 0.001 .

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