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Many companies use a quality control technique called acceptance sampling to monitor incoming shipments of parts, raw materials, and so on. In the electronics industry, component parts are commonly shipped from suppliers in large lots. Inspection of a sample of \(n\) components can be viewed as the \(n\) trials of a binomial experiment. The outcome for each component tested (trial) will be that the component is classified as good or defective. Reynolds Electronics accepts a lot from a particular supplier if the defective components in the lot do not exceed \(1 \%\). Suppose a random sample of five items from a recent shipment is tested. a. Assume that \(1 \%\) of the shipment is defective. Compute the probability that no items in the sample are defective. b. Assume that \(1 \%\) of the shipment is defective. Compute the probability that exactly one item in the sample is defective. c. What is the probability of observing one or more defective items in the sample if \(1 \%\) of the shipment is defective? d. Would you feel comfortable accepting the shipment if one item was found to be defective? Why or why not?

Short Answer

Expert verified
a. 0.95099; b. 0.04805; c. 0.04901. d. Acceptance depends on quality standard interpretation, might still be within accepted rates.

Step by step solution

01

Determine Parameters

We are dealing with a binomial distribution here. \( n = 5 \) (number of trials) and \( p = 0.01 \) (probability of success for a defective item). So, the probability distribution is binomial: \( B(n=5, p=0.01) \).
02

Probability of No Defective Items (Part a)

For part a, we need the probability that no items in the sample of five are defective. This can be expressed as \( P(X=0) \). Use the binomial formula: \[ P(X = k) = \binom{n}{k} \, p^k \, (1-p)^{n-k} \] Substitute \( n = 5, k = 0, p = 0.01 \): \[ P(X=0) = \binom{5}{0} \, (0.01)^0 \, (0.99)^5 = 1 \times 1 \times 0.99^5 \approx 0.95099 \].
03

Probability of Exactly One Defective Item (Part b)

For part b, we compute the probability of exactly one defective item in the sample, \( P(X=1) \). Again use the binomial formula: \[ P(X=1) = \binom{5}{1} \, (0.01)^1 \, (0.99)^4 \] Substitute the values: \[ P(X=1) = 5 \, (0.01) \, (0.99)^4 = 5 \times 0.01 \times 0.960596 \approx 0.04805 \].
04

Probability of One or More Defective Items (Part c)

For part c, the probability that one or more items in the sample are defective is the complement of the event that no items are defective. So, \[ P(X \geq 1) = 1 - P(X=0) = 1 - 0.95099 \approx 0.04901 \].
05

Decision on Accepting the Shipment (Part d)

For part d, you must decide if you would feel comfortable accepting the shipment if one item is defective. Since the probability of finding one defective item given the entire shipment being 1% defective is about 4.8% (from Step 3), one defective item might not be surprising, suggesting the overall quality may be acceptable. However, further investigation could be considered as shipment conditions vary.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a fundamental concept in statistics, especially useful for quality control processes like acceptance sampling. This distribution describes scenarios where there are two possible outcomes, commonly labeled as "success" and "failure." In the context of quality control, you could label a defective item as a "success" to find how many defective items exist.

To understand it better, let's see its characteristics:
  • Each trial is independent, meaning the outcome of one does not affect another.
  • There are a fixed number of trials, noted as \(n\).
  • Each trial has the same probability of "success" (defective), recorded as \(p\).
  • The interested number of "successes" in \(n\) trials can be denoted as \(k\).
The binomial distribution is often expressed with the formula \[ P(X = k) = \binom{n}{k} \, p^k \, (1-p)^{n-k} \].

Plugging the parameters helps calculate the probability for any scenario, such as no defective items or just one, as seen in this solution. The formula shows us how the likelihood shifts depending on the number of trials or the defect rate. For example, if fewer tests are done, the presence of defects can become more significant. Understanding these principles can be pivotal in maintaining quality in manufacturing processes.
Quality Control
Quality control is essential in many industries to ensure products meet certain standards. In electronics and similar fields, companies often inspect shipments using acceptance sampling methods.

Acceptance sampling involves testing a random sample from a batch. The outcome of the sample determines if the entire batch is accepted or rejected. For Reynolds Electronics, inspecting a sample from a shipment serves to ensure that defective parts do not exceed a pre-set threshold (in this case, 1%).

Key Benefits of Quality Control:
  • Reduced Costs: Identifying defective products early minimizes waste and potential recalls.
  • Customer Satisfaction: Ensures that received products meet customer requirements.
  • Brand Reputation: Consistently delivering quality products helps maintain a positive brand image.
When sampling shows defect rates acceptable by the set standards, companies can confidently accept their shipments. However, finding defects can mean more testing or rejecting the batch. This method allows companies to systematically check product quality without inspecting every single unit.
Probability Calculation
Calculating probabilities is crucial in determining the acceptance of product shipments, especially when considering potential defects. In acceptance sampling, the probability calculation relies on the binomial distribution as shown previously. Each calculation provides insight into the likelihood of certain defect levels.

For example, calculating the probability that no defects exist in a shipment requires substituting the specific parameters \( n \) (number of trials) and \( p \) (defect probability) into the binomial formula. This gives insights into single scenarios, like either no defects or only one defect in the sample. These probability calculations allow companies to determine how likely they are to encounter certain defect levels under their established quality control thresholds.

Steps in Probability Calculation for Quality Control:
  • Define the total number of trials and what constitutes a "success" (defect).
  • Identify the probability of defect, typically given as a percentage of the shipment.
  • Utilize the binomial formula to find probabilities for different outcomes, like zero or one defect.
  • Interpret these probabilities to make informed decisions about the shipment's quality.
These calculations inform decisions, minimizing risk by predicting defect levels and enabling proactive management of product quality. This understanding helps ensure that the shipment quality aligns with predefined standards. Calculating probabilities is, therefore, not just about numbers; it's a key component in strategic decision-making in production settings.

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Most popular questions from this chapter

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