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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 1 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

Understanding Poisson Distribution

The Poisson distribution models the count of events occurring within a fixed interval of time or space when these events happen with a known constant mean rate and are independently of each other. The probability mass function for the Poisson distribution is given by: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where \( \lambda \) is the average number of occurrences in the interval and \( k \) is the number of occurrences we want to find the probability for.
02

(a) Calculate Probability for Two Flaws in 1 Square Meter

In this case, \( \lambda = 0.1 \). We want to find the probability \( P(X = 2) \). Substitute these values into the formula:\[ P(X = 2) = \frac{e^{-0.1} (0.1)^2}{2!} = \frac{e^{-0.1} \times 0.01}{2} \approx 0.0045 \]
03

(b) Calculate Probability for One Flaw in 10 Square Meters

Here, the mean number of flaws for 10 square meters is \( \lambda = 10 \times 0.1 = 1 \). We need \( P(X = 1) \):\[ P(X = 1) = \frac{e^{-1} \times 1^1}{1!} = e^{-1} \approx 0.3679 \]
04

(c) Calculate Probability for No Flaws in 20 Square Meters

Now, \( \lambda = 20 \times 0.1 = 2 \). We want \( P(X = 0) \):\[ P(X = 0) = \frac{e^{-2} \times 2^0}{0!} = e^{-2} \approx 0.1353 \]
05

(d) Calculate Probability for At Least Two Flaws in 10 Square Meters

For this, we need \( P(X \geq 2) = 1 - P(X < 2) \), where \( \lambda = 1 \). First, calculate \( P(X = 0) \) and \( P(X = 1) \), then subtract from 1:- \( P(X = 0) = e^{-1} \approx 0.3679 \)- \( P(X = 1) \approx 0.3679 \)Thus, \[ P(X < 2) = P(X = 0) + P(X = 1) = 2 \times e^{-1} \approx 0.7358 \]Therefore,\[ P(X \geq 2) = 1 - 0.7358 = 0.2642 \]

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Key Concepts

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

Probability Calculation
Probability calculation in a Poisson distribution involves a specific formula that helps us predict the likelihood of a certain number of events happening in a fixed interval. For the Poisson distribution, the probability of observing exactly \( k \) events is calculated using the formula:
  • \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Here, \( \lambda \) represents the average rate of occurrence, which is the expected number of events per interval.
For instance, in the context of manufacturing, if you know the average number of flaws per square meter, you can predict the probability of encountering a specific number of flaws. This calculation is crucial for making informed decisions and managing expectations regarding product quality. The poisson distribution is useful when you're dealing with rare events, such as flaws on a production line.
It's essential to substitute the correct values into the formula to find the probability for the given scenario. Understanding this process enables you to handle various situations where events occur independently, helping in probabilistic assessments.
Statistical Modeling
Statistical modeling using the Poisson distribution is a powerful tool that allows us to accurately represent the occurrence of events across a given space or time period. Particularly, it's used when you expect an average rate of events, and those events do not happen at regularly predictable intervals.
This modeling is distinctive because it assumes that each occurrence is independent and happens with a constant mean rate.
  • In our textile example, this involves assessing flaws in the fabric per square meter.
  • The Poisson model helps to predict occurrences over a specified range, even if the range changes, such as considering 1, 10, or 20 square meters in our exercises.
This allows manufacturers to anticipate product defects and better control quality - a significant aspect for operational efficiency.
The model provides more than just numbers; it forms a basis for statistical reasoning, understanding variability in data, and supports decision-making based on empirical evidence.
Event Probability
Understanding event probability within the context of Poisson distribution involves considering how likely certain events are to occur. This distribution is particularly useful when dealing with rare or unexpected events over a fixed space or time.Using our fabric flaws example, event probability tells us how likely it is to find 0, 1, 2, or more flaws as specified in the exercise.
We compute it by focusing on the mean event rate (\( \lambda \)) and using it to evaluate different scenarios.For example:
  • The probability of having two flaws in 10 square meters is calculated by first finding the probability for different smaller counts of flaws and then deriving the combined probability.
  • The calculation \( (P(X \geq 2)) \) requires the complementary probabilities of fewer than 2 events, aligning with the concept that probabilities sum up to 1.
Their inference involves calculating such probabilities gives important insight into the likelihood of certain outcomes, which is crucial for planning, forecasting, and mitigating risks.
Therefore, understanding event probability is integral to interpreting real-world occurrences in varied fields such as quality control, insurance, and risk management.

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Most popular questions from this chapter

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