/*! 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 58 A very large batch of components... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2 . a. What is the probability that the batch will be accepted when the actual proportion of defectives is .01? .05?.10? 20 ?.25? b. Let \(p\) denote the actual proportion of defectives in the batch. A graph of \(P\) (batch is accepted) as a function of \(p\), with \(p\) on the horizontal axis and \(P\) (batch is accepted) on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan. Use the results of part (a) to sketch this curve for \(0 \leq p \leq 1\). c. Repeat parts (a) and (b) with " 1 " replacing " 2 " in the acceptance sampling plan. d. Repeat parts (a) and (b) with " 15 " replacing " \(10 "\) in the acceptance sampling plan. e. Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?

Short Answer

Expert verified
The sampling plan in part (c) is most satisfactory as it has the lowest risk of accepting a high proportion of defective batches.

Step by step solution

01

Define the Binomial Random Variable

The problem involves determining probabilities for a sample of size 10 from a large population. A binomial random variable is a suitable model because there is a fixed number of trials (n = 10), each trial has two possible outcomes (defective or not defective), the trials are independent, and the probability of a defective component remains constant across trials. Let the random variable X denote the number of defective components found in the 10-component sample.
02

Express the Acceptance Criteria

The batch is accepted if the number of defective components in the sample is at most 2. Hence, we need to find the probability that X is less than or equal to 2.
03

Use the Binomial Probability Formula

The probability of finding k defective components in a sample of 10 is given by the binomial formula: \(P(X = k) = \binom{10}{k} p^k (1-p)^{10-k}\), where \(\binom{10}{k}\) is the number of ways to choose k successes (defective components) out of 10 trials, and \(p\) is the probability of finding a defective component.
04

Calculate Probability for Part (a)

For each given proportion \(p\) (.01, .05, .10, .20, .25), calculate \(P(X \leq 2)\) using: \(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\). Compute this separately for each \(p\).
05

Sketch the Operating Characteristic Curve (Part b)

Using the results from Step 4, plot the calculated probabilities \(P(X \leq 2)\) against the proportions (\(p\)) on a graph, with \(p\) on the horizontal axis and probability on the vertical axis.
06

Modify Acceptance Criteria (Part c and d)

For part (c), change the acceptance criterion to 'at most 1' defective component. For part (d), increase the sample size to 15 and check 'at most 2' defective components. Repeat Steps 3-5 for these new criteria.
07

Compare the Sampling Plans (Part e)

Compare the effectiveness of the three sampling plans by evaluating which plan consistently results in a desirable level of accepting batches across different values of defective proportion \(p\). Consider the risk of accepting a bad batch versus the risk of rejecting a good batch when deciding which sampling plan is most satisfactory.

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.

Binomial Probability
In the context of acceptance sampling, binomial probability refers to the likelihood of observing a certain number of defective items in a sample, based on a set probability. We use this approach because the inspection process is designed to determine if a batch meets specific standards. For example, suppose you receive a batch of components and you decide to sample 10 from this batch. If each component has a fixed probability of being defective, and they are inspected independently, the setup fits a binomial model.Here's what you need to know about binomial probability:
  • Trials are independent: Each sample does not affect the outcome of another.
  • Fixed number of trials: In this exercise, we sample 10 components.
  • Two outcomes: Each component can either be defective or non-defective.
  • Constant probability: The probability of selecting a defective component remains the same for each sample.
To calculate the probability of finding exactly k defective components, use the binomial formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\] where \(n\) is the total number of trials and \(p\) is the probability of defectiveness.
Operating Characteristic Curve
The Operating Characteristic (OC) Curve is a crucial concept in quality control and acceptance sampling. It visually represents the probability of accepting a batch based on different levels of defectiveness, helping you understand and align your sampling plan with quality standards.In this exercise, you plot the OC curve by calculating the probability of acceptance at various defect proportions, \(p\). The graph is plotted with \(p\) on the horizontal axis and the probability of acceptance on the vertical axis. Here is why an OC curve is important:
  • Visual analysis: It allows you to easily assess how effective the sampling plan is at identifying acceptable batches.
  • Decision support: Helps stakeholders make informed decisions regarding quality control procedures and risk assessment.
  • Comparative assessment: Useful for comparing different sampling plans to select the most efficient one.
When you sketch the OC curve using different acceptance criteria, it offers direct insight into how the sampling plan's sensitivity shifts with varying defect levels.
Sampling Plan
A sampling plan stipulates how a group will be inspected to decide whether it meets required standards. It outlines the criteria for inspecting defectiveness, essentially balancing the risks of accepting poor quality and rejecting satisfactory batches. In the given exercise, different sampling plans were considered by altering acceptance criteria and sample size. Consider these elements when creating a satisfactory sampling plan:
  • Sample Size: Larger samples give more insight, but also require more resources to inspect.
  • Acceptance Criteria: Defining what constitutes an acceptable number of defects (e.g., accepting a batch if 2 or fewer are found).
  • Defect Probability: Consider the assumed defect probability to align sampling plans with actual batch quality.
Choosing the most suitable plan involves comparing how each approach performs across different defect probabilities and weighing the trade-offs between false acceptance and rejection. This ensures quality control processes are both effective and resource-efficient.

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

According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Planning and Mgmnt., 2005: 383-393), the drought length \(Y\) is the number of consecutive time intervals in which the water supply remains below a critical value \(y_{0}\) (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). The cited paper proposes a geometric distribution with \(p=.409\) for this random variable. a. What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals? b. What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

Suppose that you read through this year's issues of the New York Times and record each number that appears in a news article-the income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of each number, which could be \(1,2, \ldots, 8\), or 9 . Your first thought might be that the leading digit \(X\) of a randomly selected number would be equally likely to be one of the nine possibilities (a discrete uniform distribution). However, much empirical evidence as well as some theoretical arguments suggest an alternative probability distribution called Benford's law: \(p(x)=P(1\) st digit is \(x)=\log _{10}\left(\frac{x+1}{x}\right) \quad x=1,2, \ldots, 9\) a. Without computing individual probabilities from this formula, show that it specifies a legitimate pmf. b. Now compute the individual probabilities and compare to the corresponding discrete uniform distribution. c. Obtain the cdf of \(X\). d. Using the cdf, what is the probability that the leading digit is at most 3 ? At least 5 ? [Note: Benford's law is the basis for some auditing procedures used to detect fraud in financial reporting-for example, by the Internal Revenue Service.]

Suppose that \(p=P(\) male birth \()=.5\). A couple wishes to have exactly two female children in their family. They will have children until this condition is fulfilled. a. What is the probability that the family has \(x\) male children? b. What is the probability that the family has four children? c. What is the probability that the family has at most four children? d. How many male children would you expect this family to have? How many children would you expect this family to have?

A concrete beam may fail either by shear \((S)\) or flexure \((F)\). Suppose that three failed beams are randomly selected and the type of failure is determined for each one. Let \(X=\) the number of beams among the three selected that failed by shear. List each outcome in the sample space along with the associated value of \(X\).

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, \(25 \%\) of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let \(X\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(X\). [Hint: Let \(S\) denote a homeowner who has insurance and \(F\) one who does not. Then one possible outcome is SFSS, with probability \((.25)(.75)(.25)(.25)\) and associated \(X\) value 3 . There are 15 other outcomes.] b. Draw the corresponding probability histogram. c. What is the most likely value for \(X\) ? d. What is the probability that at least two of the four selected have earthquake insurance?

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.