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Inclusions are defects in poured metal caused by contaminants. The number of (large) inclusions in cast iron follows a Poisson distribution with a mean of 2.5 per cubic millimeter. Determine the following: (a) Probability of at least one inclusion in a cubic millimeter. (b) Probability of at least five inclusions in 5.0 cubic millimeters. (c) Volume of material to inspect such that the probability of at least one inclusion is \(0.99 .\) (d) Instead of a mean of 2.5 per cubic millimeters, the mean inclusions per cubic millimeter such that the probability of at least one inclusion is \(0.95 .\)

Short Answer

Expert verified
(a) 0.918 (b) 0.9921 (c) 1.84 mm³ (d) 2.996 per mm³

Step by step solution

01

Understanding the Poisson Distribution

The Poisson distribution is used to model the number of times an event occurs in a fixed interval of time or space. It is defined by the parameter \( \lambda \), which is the average number of events in the interval. Here, \( \lambda = 2.5 \) for 1 cubic millimeter.
02

Solving Part (a)

We need to find the probability of at least one inclusion in a cubic millimeter. This is the complementary event of having zero inclusions. Thus, \( P(X \geq 1) = 1 - P(X = 0) \). For a Poisson distribution, \( P(X = 0) = \frac{e^{-\lambda} \cdot \lambda^0}{0!} = e^{-2.5} \). Therefore, \( P(X \geq 1) = 1 - e^{-2.5} \approx 0.918 \).
03

Solving Part (b)

For at least five inclusions in 5 cubic millimeters, the mean \( \lambda \) becomes \( 2.5 \times 5 = 12.5 \). We need \( P(Y \geq 5) \), where \( Y \) is a Poisson random variable with \( \lambda = 12.5 \). Using the cumulative distribution function (CDF), calculate \( P(Y < 5) \) and find \( P(Y \geq 5) = 1 - P(Y < 5) \). Based on cumulative tables or software, \( P(Y \geq 5) \approx 0.9921 \).
04

Solving Part (c)

Determine volume when the probability of at least one inclusion is 0.99. Using \( P(X \geq 1) = 1 - e^{-\lambda V} = 0.99 \), solve for \( V \). Thus, \( e^{-2.5V} = 0.01 \). Rearrange to find \( V \): \( V = -\frac{\ln(0.01)}{2.5} \approx 1.84 \) cubic millimeters.
05

Solving Part (d)

Determine new mean \( \lambda' \) per cubic millimeter for \( P(X \geq 1) = 0.95 \). Use \( 1 - e^{-\lambda'} = 0.95 \). Thus, \( e^{-\lambda'} = 0.05 \), and solve for \( \lambda' \). Hence, \( \lambda' = -\ln(0.05) \approx 2.996 \), which is the new mean inclusions per cubic millimeter.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics focused on the analysis of random events. Here, we use probability theory to calculate the chance of finding metal inclusions.
One way to tackle such problems is by using the Poisson distribution. This particular distribution is helpful when we need to figure out the likelihood of a given number of events within a certain space or timeframe. For example, calculating the number of defects in metal.
A Poisson distribution is defined by its parameter, lambda ( \( \lambda \) ), which represents the average event count over the interval. Understanding this concept can easily solve problems like determining how likely an inclusion is present when pouring metal.
  • Events must occur independently.
  • Events occur at a constant average rate.
Cumulative Distribution Function
The cumulative distribution function (CDF) of a probability distribution is a fundamental concept. It's used to calculate the probability of a random variable taking a value less than or equal to a specific point.
In the context of the Poisson distribution, the CDF is vital for understanding an accumulation of events. For instance, if you are checking for at least a certain number of inclusions, the CDF helps you find the sum total of probabilities for having fewer than this number.
To find the probability of at least five inclusions in a set volume, you would subtract from one the probability that there are fewer than five. This makes complex calculations simpler and ensures accurate results. Calculating these probabilities helps ensure precise quality control in manufacturing processes by quantifying defect occurrences.
  • Useful for probabilities leading up to a threshold.
  • Allows for easy calculation by using 1 minus the CDF.
Exponential Function
Exponential functions play a key role in probability distributions, especially Poisson. The exponential function is represented as \( e^x \), where \( e \) is approximately 2.718. This function characterizes continuous growth or decay.
In Poisson distribution calculations, the exponential function helps model the probability of events over a fixed interval.
For example, when assessing the absence of a defect, you often find expressions like \( e^{-\lambda} \). It models this probability accurately because the occurrence rate is constant. Its shape, having a rapid increase or decrease, is ideal in representing stochastic or random processes such as arrival rates or defects in materials.
  • Grows or decays at a rate proportional to its current value.
  • Provides a continuous curve for probabilistic predictions.
Parameter Lambda
The parameter lambda ( \( \lambda \) ) is crucial in using the Poisson distribution. It reflects the average number of events occurring in a set timeframe or area. Understanding lambda aids in accurate calculation of event probabilities.
Lambda helps to predict defects or inclusions when pouring metals. For example, a lambda set to 2.5 indicates that, on average, 2.5 inclusions are expected per cubic millimeter. Adjusting lambda provides flexibility in the Poisson model to match real-world scenarios and manufacturing conditions, making it a versatile prediction tool.
To determine more specific probabilities, manipulating lambda, such as changing the conditions for rarity or frequency of events, can be very insightful.
In practical applications, lambda guides quality control and defect analysis by setting expectations according to empirical or historical data.
  • Average rate of occurrence.
  • Flexibility in model application.

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Most popular questions from this chapter

The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average there are 10 calls per hour. (a) What is the probability that there are exactly 5 calls in one hour? (b) What is the probability that there are 3 or fewer calls in one hour? (c) What is the probability that there are exactly 15 calls in two hours? (d) What is the probability that there are exactly 5 calls in 30 minutes?

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