/*! 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 10 Inclusions are defects in poured... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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) 0.9179, b) Calculate using \( \lambda = 12.5 \), c) 1.842 mm³, d) 3 inclusions/mm³.

Step by step solution

01

Understanding the Poisson Distribution

In this exercise, we are dealing with a Poisson distribution, which models the number of times an event occurs within a fixed interval or space. The parameter for the Poisson distribution is the mean number of occurrences, denoted as \( \lambda \). In this problem, \( \lambda = 2.5 \) for a volume of 1 cubic millimeter.
02

Calculating Probability of At Least One Inclusion (Part a)

For part a, we need to find the probability of at least one inclusion in a cubic millimeter. The probability of having zero inclusions is \( P(X = 0) = e^{-\lambda} \frac{\lambda^0}{0!} = e^{-2.5} \). Therefore, the probability of having at least one inclusion is \( 1 - P(X = 0) = 1 - e^{-2.5} \).
03

Evaluating Probability for Part a

By calculating \( e^{-2.5} \approx 0.0821 \), the probability of at least one inclusion is \( 1 - 0.0821 = 0.9179 \). Therefore, the probability of at least one inclusion per cubic millimeter is about 0.9179.
04

Calculating Probability for At Least Five Inclusions in 5 Cubic Millimeters (Part b)

Here, the mean number of inclusions in 5 cubic millimeters is \( 5 \times 2.5 = 12.5 \). We want \( P(X \geq 5) \). First, calculate \( P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \), using the Poisson probability mass function \( P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!} \). Subtract the cumulative probability from 1 to get \( P(X \geq 5) = 1 - P(X < 5) \).
05

Calculating Individual Probabilities for Part b

- \( P(X = 0) = e^{-12.5}\frac{12.5^0}{0!} = e^{-12.5}\)- \( P(X = 1) = e^{-12.5}\frac{12.5^1}{1!} \)- Repeat for \( k = 2, 3, 4 \), calculate each term, and sum the probabilities for \( P(X < 5) \).
06

Evaluating Probability for Part b

Using cumulative calculations or a statistical tool, calculate \( P(X < 5) \) and then find \( P(X \geq 5) \). The numerical evaluation should result in this cumulative sum for \( P(X < 5) \) and the subsequent subtraction from 1 will give \( P(X \geq 5) \).
07

Finding Volume for Probability of 0.99 (Part c)

Set the equation for the probability of at least one inclusion as \( 1 - e^{-\lambda V} = 0.99 \), where \( V \) is the volume in cubic millimeters. Solving for \( V \), you obtain \( e^{-\lambda V} = 0.01 \), so \( \lambda V = -\ln(0.01) \). Hence \( V = \frac{-\ln(0.01)}{\lambda} \). Use \( \lambda = 2.5 \) to calculate \( V \).
08

Calculating Volume for Part c

\( V = \frac{-\ln(0.01)}{2.5} \approx \frac{4.605}{2.5} \approx 1.842 \). This implies a volume of 1.842 cubic millimeters is needed to ensure a probability of 0.99 for at least one inclusion.
09

Finding Mean for Desired Probability (Part d)

Given the probability of at least one inclusion is 0.95, use \( 1 - e^{-\mu} = 0.95 \) to solve for \( \mu \). Resulting in \( e^{-\mu} = 0.05 \), therefore \( \mu = -\ln(0.05) \). Calculate \( \mu \) for the desired rate of inclusions per cubic millimeter.
10

Calculating Mean for Part d

\( \mu = -\ln(0.05) \approx 2.996 \). Hence, a mean of approximately 3 inclusions per cubic millimeter is needed to achieve a probability of 0.95 for at least one inclusion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Calculation
The probability calculation using the Poisson distribution is essential for understanding such scenarios as casting defects. When dealing with the probability of events, like occurrences of inclusions, the Poisson distribution gives us a helpful tool. It helps to predict the likelihood of a given number of events happening in a fixed interval of time or space.
For instance, the exercise asks to compute the probability of having at least one inclusion in a cubic millimeter. Here, the key is to calculate the probability of having no inclusions (zero events) and then subtract this from 1. This approach is typical for questions involving 'at least one' event because it's often simpler to first find the complement (zero occurrences).
In our case, given a mean \( \lambda = 2.5 \) for each cubic millimeter, the Poisson formula is used to calculate this complement:
  • First, find \( P(X = 0) = e^{-2.5} \frac{2.5^0}{0!} = e^{-2.5} \).
  • Then, the probability of at least one inclusion is \( 1 - P(X = 0) \).
By evaluating \( e^{-2.5} \), we find that this probability is approximately 0.9179, meaning there is a high chance of at least one inclusion occurring in a given cubic millimeter of cast iron.
Mean Number of Events
The mean number of events, denoted as \( \lambda \), plays a significant role in the Poisson distribution. It represents the expected number of occurrences in a given interval. Understanding the mean is crucial when you want to extrapolate the probability over a different volume or to a different event rate.
In this context, the problem illustrates using a mean of 2.5 inclusions per cubic millimeter. This value is adjusted based on the volume or event conditions in questions b and d.
For part b, where inclusions are given in 5 cubic millimeters, the mean adjusts accordingly: \( \lambda = 5 imes 2.5 = 12.5 \). This is important because the probability of a certain number of events changes with the volume being considered. The foundation here is that probabilities must reflect the mean over the correct interval or area being examined.
The variations in mean, therefore, allow for flexibility in predicting the probability of different scenarios, like ensuring inclusions are kept below a certain count in larger volumes.
Inclusions in Cast Iron
Inclusions are imperfections within cast iron, often caused by impurities or foreign elements. These defects can influence the strength and durability of metal products.
When analyzing cast iron's quality, predicting the presence and frequency of inclusions is critical. By modeling these occurrences with a Poisson distribution, manufacturers can anticipate and mitigate quality issues.
The exercise uses this model to evaluate probabilities of inclusion in specific volumes of cast iron. This involves analyzing different scenarios to determine how often these defects surpass a threshold, affecting the material's quality. For example, part c sought to find a material volume that guarantees a high probability (0.99) of capturing at least one inclusion, highlighting the importance of precision in inspection processes.
In the real world, understanding the likelihood of inclusions not only aids in quality control but also in adjusting manufacturing methods to minimize such defects. This attentiveness ensures the robustness and longevity of the final cast iron products.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An array of 30 LED bulbs is used in an automotive light. The probability that a bulb is defective is 0.001 and defective bulbs occur independently. Determine the following a. Probability that an automotive light has two or more defective bulbs. b. Expected number of automotive lights to check to obtain one with two or more defective bulbs.

Samples of rejuvenated mitochondria are mutated (defective) in \(1 \%\) of cases. Suppose that 15 samples are studied and can be considered to be independent for mutation. Determine the following probabilities. a. No samples are mutated. b. At most one sample is mutated. c. More than half the samples are mutated.

A Web site randomly selects among 10 products to discount each day. The color printer of interest to you is discounted today. a. What is the expected number of days until this product is again discounted? b. What is the probability that this product is first discounted again exactly 10 days from now? c. If the product is not discounted for the next five days, what is the probability that it is first discounted again 15 days from now? d. What is the probability that this product is first discounted again within three or fewer days?

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. Determine the following probabilities: a. exactly 5 calls in one hour b. 3 or fewer calls in one hour c. exactly 15 calls in two hours d. exactly 5 calls in 30 minutes

An electronic product contains 40 integrated circuits. The probability that any integrated circuit is defective is \(0.01,\) and the integrated circuits are independent. The product operates only if there are no defective integrated circuits. What is the probability that the product operates?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.