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The number of imperfections in the weave of a certain textile has a Poisson distribution with a mean of 4 per square yard. Find the probability that a a. 1-square-yard sample will contain at least one imperfection. b. 3-square-yard sample will contain at least one imperfection.

Short Answer

Expert verified
a. 0.9817 b. 0.99999386

Step by step solution

01

Understanding the Problem

The problem involves the Poisson distribution which models the number of times an event occurs within a fixed interval. We have a mean of 4 imperfections per square yard. We need to find the probability of having at least one imperfection for 1-square-yard and 3-square-yard samples.
02

Finding Probability for 1-Square-Yard

To find the probability that a 1-square-yard sample contains at least one imperfection, we first find the probability of having zero imperfections and then subtract this from 1. The probability of having zero imperfections is given by the Poisson probability formula: \( P(X=k) = \frac{{e^{- ext{mean}} \cdot ext{mean}^k}}{{k!}} \). For \( k = 0 \), \( P(X=0) = \frac{{e^{-4} \cdot 4^0}}{{0!}} = e^{-4} \). Therefore, the probability of at least one imperfection is \( 1 - e^{-4} \).
03

Calculating Probability for 1-Square-Yard

Using the value \( e^{-4} \approx 0.0183 \), the probability of having at least one imperfection is \( 1 - 0.0183 = 0.9817 \).
04

Finding Mean for 3-Square-Yard

For a 3-square-yard sample, the average number of imperfections is 3 times the mean for 1 square yard. Therefore, the mean \( \lambda \) becomes \( 4 imes 3 = 12 \).
05

Finding Probability for 3-Square-Yard

Similarly, find the probability of having zero imperfections for a mean of 12 and subtract from 1. \( P(X=0) = \frac{{e^{-12} \, 12^0}}{{0!}} = e^{-12} \). Hence, the probability of having at least one imperfection is \( 1 - e^{-12} \).
06

Calculating Probability for 3-Square-Yard

Using \( e^{-12} \approx 0.00000614 \), the probability for at least one imperfection is \( 1 - 0.00000614 = 0.99999386 \).

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Key Concepts

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

Mathematical Statistics
Mathematical statistics serves as the foundation for many statistical methods, including the Poisson distribution. This branch of mathematics helps us understand how data behaves and provides tools to model various types of probability distributions. In the context of the exercise given, the Poisson distribution is used due to its suitability for modeling the number of events (imperfections, in this case) over a specific interval (or area). By using mathematical statistics, we can harness complex formulas to simplify the process of calculating probabilities and make predictions based on data.
  • Understand that mathematical statistics allows us to use theoretical models to describe real-world phenomena.
  • It involves the development and application of methods to collect, analyze, and interpret data.
  • In mathematical statistics, we use the Poisson formula: \( P(X=k) = \frac{e^{- ext{mean}} \cdot \ ext{mean}^k}{k!} \) to calculate specific outcomes.
In our exercise, this formula lets us compute the probability of zero imperfections and thus find the outcome of having at least one imperfection in a given area.
Probability Calculation
Probability calculation is essential for understanding events' likelihoods under different conditions. When you use probability in a practical sense, like calculating imperfections in textile samples, it gives insight into expected outcomes given historical data.For probability calculations using the Poisson distribution:
1. Determine the average rate (mean) of occurrence within a fixed interval. Here it’s 4 imperfections per square yard.
2. Apply the Poisson formula to determine the probability of zero events (imperfections).
The formula for zero imperfections reverts to an exponential function, such as \( e^{-4} \) for a 1-square-yard example.Remember, probability calculation isn't just about plugging numbers into a formula:
  • It requires understanding the context, such as knowing why a Poisson model fits.
  • Check the time or area interval the mean applies to, ensuring accuracy in results.
  • Finally, derivate the conditional probability (e.g. at least one imperfection) by manipulating the base probabilities, like subtracting zero probability from one.
Effective probability calculation allows for better predictive insights, crucial for quality control in manufacturing.
Statistical Modelling
Statistical modeling involves creating models that represent or approximate complex real-world behaviors, such as the frequency of imperfections in textiles. Through statistical modeling, we make data actionable by predicting, simulating, or explaining phenomena using sound mathematical principles. In the case of the Poisson distribution, we model the likelihood of a certain number of imperfections appearing within a fixed area (like a square yard of fabric). This approach is essential for:
  • Predicting future outcomes based on known probabilities, such as how often a fabric might have flaws.
  • Improving quality control processes by setting expectations for what is considered 'normal' variability.
  • Helping businesses allocate resources more effectively, focusing efforts on monitoring and reducing imperfections.
When applying statistical modelling, check assumptions, choose appropriate models (like Poisson for count-based data), and validate your results against observed data. Understanding and managing model limitations are also key to achieving more accurate and reliable predictions. This process helps assess risks and variabilities inherent in processes or products, facilitating better decision-making.

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Most popular questions from this chapter

A group of six software packages available to solve a linear programming problem has been ranked from 1 to 6 (best to worst). An engineering firm, unaware of the rankings, randomly selected and then purchased two of the packages. Let \(Y\) denote the number of packages purchased by the firm that are ranked \(3,4,5,\) or \(6 .\) Give the probability distribution for \(Y\).

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