Chapter 4: Problem 126
A quality-control plan calls for accepting a large lot of crankshaft bearings if a sample of seven is drawn and none are defective. What is the probability of accepting the lot if none in the lot are defective? If \(1 / 10\) are defective? If \(1 / 2\) are defective?
Short Answer
Expert verified
Answer: The probabilities of accepting the lot under the three scenarios are:
1. When none are defective: 1
2. When 1/10 are defective: \(\left(\frac{9}{10}\right)^7\)
3. When 1/2 are defective: \(\left(\frac{1}{2}\right)^7\)
Step by step solution
01
Identify the given information
We are given that a sample of 7 crankshaft bearings is drawn, and the lot is accepted if none of these 7 are defective. We need to find the probability of accepting the lot under three different scenarios:
1. When none are defective.
2. When 1/10 are defective.
3. When 1/2 are defective.
02
Formula for binomial probability
To solve this problem, we will be using the formula for binomial probability, which is:
\[P(X = k) = {n \choose k} p^k (1-p)^{(n-k)}\],
where \(P(X = k)\) is the probability of getting \(k\) successes in \(n\) trials, \({n \choose k}\) represents the number of combinations of choosing \(k\) items from a set of \(n\) items, \(p\) is the probability of success, and \((1-p)\) is the probability of failure.
03
Calculate the probability when none are defective
In this scenario, all the bearings in the lot are good, so the probability of selecting a non-defective bearing (success) is 1.
We want to find the probability of obtaining 0 defective bearings in all 7 trials, which means all trials result in selecting non-defective bearings.
With \(n = 7\) and \(p = 1\), the probability can be calculated as:
\[P(X = 0) = {7 \choose 0} (1)^0 (1-1)^{(7-0)} = 1 \times 1 \times 0^7 = 1\]
The probability of accepting the lot when none are defective is 1.
04
Calculate the probability when 1/10 are defective
In this scenario, 1/10 of the bearings are defective, so the probability of selecting a non-defective bearing (success) is 9/10.
We want to find the probability of obtaining 0 defective bearings in all 7 trials, which means all trials result in selecting non-defective bearings.
With \(n = 7\) and \(p = \frac{9}{10}\), the probability can be calculated as:
\[P(X = 0) = {7 \choose 0} \left(\frac{9}{10}\right)^7 \left(1-\frac{9}{10}\right)^{(7-0)} = 1 \times \left(\frac{9}{10}\right)^7 \times 0^7 = \left(\frac{9}{10}\right)^7\]
The probability of accepting the lot when 1/10 are defective is \(\left(\frac{9}{10}\right)^7\).
05
Calculate the probability when 1/2 are defective
In this scenario, 1/2 of the bearings are defective, so the probability of selecting a non-defective bearing (success) is 1/2.
We want to find the probability of obtaining 0 defective bearings in all 7 trials, which means all trials result in selecting non-defective bearings.
With \(n = 7\) and \(p = \frac{1}{2}\), the probability can be calculated as:
\[P(X = 0) = {7 \choose 0} \left(\frac{1}{2}\right)^7 \left(1-\frac{1}{2}\right)^{(7-0)} = 1 \times \left(\frac{1}{2}\right)^7 \times 0^7 = \left(\frac{1}{2}\right)^7\]
The probability of accepting the lot when 1/2 are defective is \(\left(\frac{1}{2}\right)^7\).
Thus, the probabilities of accepting the lot under the three scenarios are:
1. When none are defective: 1
2. When 1/10 are defective: \(\left(\frac{9}{10}\right)^7\)
3. When 1/2 are defective: \(\left(\frac{1}{2}\right)^7\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quality Control
Quality control is a vital process in manufacturing and production industries that ensures products meet the required standards. It involves systematic measures to detect and eliminate defects in products to ensure consistency, safety, and efficacy. Quality control is crucial in maintaining the credibility and trust of a brand.
- Testing: Regular inspection and testing of products are conducted to catch defects early and prevent faulty items from reaching customers.
- Inspection Standards: Standards are established, specifying allowed imperfections and defining what a 'defective' product is.
- Sampling: Instead of examining every item, a representative sample is tested to infer the quality level of an entire lot.
Acceptable Quality Level
The Acceptable Quality Level (AQL) is a predetermined limit of defective items allowed in any given batch before it is rejected. It is a crucial parameter in any quality-control process, helping manufacturers decide whether a lot meets quality standards without testing every product individually.
- AQL is expressed as a percentage or ratio that indicates the maximum acceptable number of defective items.
- The concept of AQL allows companies to define what is tolerable in terms of defects while balancing production costs and quality assurance.
- Different products and industries may have varying AQLs depending on the criticality of the product's performance and safety.
Defective Rate
The defective rate is the ratio or percentage of defective items identified in a batch during quality control processes. It is essential for understanding the quality and reliability of the manufacturing process.
- A low defective rate indicates a high-quality process, leading to fewer returns, customer complaints, and higher customer satisfaction.
- Frequent monitoring of the defective rate allows manufacturers to identify problematic patterns and implement corrective actions.
- The defective rate is a direct measure of process performance and an indicator of potential areas for improvement.
Combinatorial Mathematics
Combinatorial mathematics involves the study of counting, arranging, and selecting objects. In quality control, it is often used to determine the probability of specific outcomes based on combinations and permutations of samples.
- The binomial probability formula, used in the given exercise, is a combinatorial approach that evaluates the likelihood of a specific number of successes in a series of independent experiments.
- The formula involves calculating combinations \( {n \choose k} \) which determine how many ways "k" successes can occur in "n" trials.
- This mathematical approach supports decision-making in quality control by providing precise probabilistic assessments.