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Chapter 3: Problem 195

In a manufacturing process that laminates several ceramic layers, \(1 \%\) of the assemblies are defective. Assume that the assemblies is independent. (a) What is the mean number of assemblies that need to be checked to obtain five defective assemblies? (b) What is the standard deviation of the number of assemblies that need to be checked to obtain five defective assemblies? (c) Determine the minimum number of assemblies that need to be checked so that the probability that at least one defective assembly is obtained exceeds \(0.95 .\)

Short Answer

Expert verified
(a) Mean is 500. (b) Standard deviation is approximately 222.36. (c) Minimum 299 assemblies are needed.

Step by step solution

01

Understanding the Problem

We are dealing with a manufacturing process where 1% of assemblies are defective. We need to determine the mean and standard deviation for checking to find five defective ones and the minimum checks to ensure at least one defective with a 0.95 probability.
02

Identifying the Distribution Type

The problem can be modeled using the Negative Binomial distribution, as we are counting the number of trials needed to get a fixed number of successes (defective assemblies). Specifically, we're looking to find 5 defective assemblies.
03

Calculating the Mean for Part (a)

For a Negative Binomial distribution where we want to find 5 defects, the mean number of trials is calculated using the formula: \[ E[T] = \frac{r}{p} \] where \(r = 5\) (number of defects) and \(p = 0.01\) (probability of a defective). The mean is \(\frac{5}{0.01} = 500\).
04

Calculating the Standard Deviation for Part (b)

The standard deviation for the Negative Binomial distribution is calculated using the formula: \[ SD[T] = \sqrt{\frac{r(1-p)}{p^2}} \] With \(r = 5\) and \(p = 0.01\), we get \( \sqrt{\frac{5(0.99)}{0.01^2}} = \sqrt{49500} \approx 222.36 \).
05

Solving Part (c): Exceeding a Probability of 0.95

To find the minimum number \(n\) of assemblies where the probability of finding at least one defect exceeds 0.95, use the complementary probability. \(1 - P(\text{no defect})^n = 0.95\). So \(P(\text{no defect}) = 0.99\) and \((0.99)^n < 0.05\). Solving gives \(n > \frac{\log(0.05)}{\log(0.99)} \approx 299\). Therefore, a minimum of 299 assemblies need to be checked.

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Key Concepts

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

Mean and Standard Deviation
The concepts of mean and standard deviation are fundamental in statistics as they summarize important characteristics of a data distribution.
  • Mean: In a manufacturing context, the mean, or expected value, represents the average outcome of numerous trials. In our case, with the negative binomial distribution, the mean helps predict how many assemblies need to be checked to find five defective items. For example, if 1% of assemblies are defective, then on average, you would expect to check 500 assemblies to find five defective ones, since \( E[T] = \frac{r}{p} = \frac{5}{0.01} = 500 \).
  • Standard Deviation: This metric measures the variability or spread of a set of data points. In our exercise, the standard deviation tells us how much the number of assemblies checked might differ from the mean. A larger standard deviation would suggest that in some samples, you might need to check much more than 500 assemblies, while in others, perhaps far fewer. The calculation for this, given the negative binomial distribution, is \( SD[T] = \sqrt{\frac{r(1-p)}{p^2}} = \sqrt{49500} \approx 222.36 \). This means the number of checks could, in reality, be around 222 more or less than the mean, indicating variability in the process.
Manufacturing Process Quality Control
Quality control in a manufacturing process is crucial to ensure the end product meets predefined quality standards. In our scenario, a key goal is to monitor the assembly line and identify defective items, which might indicate a possible need for corrective measures.
  • Application of Statistics: By understanding patterns in defect occurrence, manufacturers can control and improve quality. Using statistical tools like the negative binomial distribution helps identify the frequency and distribution of defects over a large number of products. This statistical insight is used to predict and manage quality issues proactively.
  • Practical Aspects: Regular quality checks using statistical criteria can help detect anomalies early, reducing waste and rework costs. In this exercise, checking assemblies means identifying a significant number of defects before they impact the final products adversely. It ensures that the manufacturing process maintains its integrity over time.
Implementing effective quality control techniques leads to better resource management, enhancing overall operational efficiency.
Probability of Defective Assemblies
Understanding probabilities allows manufacturing engineers to make informed decisions about factory processes. In our exercise, the probability of finding defective assemblies is essential to determine appropriate quality control levels.
  • Calculating Defective Chances: If each assembly has a 1% defect rate, understanding the probability of uncovering at least one defect is crucial. In our described process, with a 0.95 probability target to find at least one defect, using complementary probability reveals how many assemblies must be checked. The calculation leads to checking 299 or more assemblies to confidently find at least one defect, which safeguards process consistency.
  • Implications for Manufacturing: Probability insights guide how rigorous testing needs to be, depending on acceptable confidence levels regarding defect discovery. A higher probability requirement demands more extensive checks; this ensures high standards are maintained consistently. This statistical understanding helps align testing strategies with quality objectives, ensuring defects are minimized while maximizing resource efficiency.
Recognizing the role of probability can greatly improve decision-making on the factory floor, decreasing the likelihood of defective products reaching consumers.

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Most popular questions from this chapter

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 one 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?

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A An installation technician for a specialized communication system is dispatched to a city only when three or more orders have been placed. Suppose that orders follow a Poisson distribution with a mean of 0.25 per week for a city with a population of 100,000 , and suppose that your city contains a population of 800,000 . (a) What is the probability that a technician is required after a one-week period? (b) If you are the first one in the city to place an order, what is the probability that you have to wait more than two weeks from the time you place your order until a technician is dispatched?

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