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Chapter 6: Problem 77

Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random from a lot and inspect them. Suppose that a lot is judged acceptable if one or fewer of these 20 parts are defective. If more than one part is defective, the lot is rejected and returned to the supplier. Find the probability of accepting lots that have each of the following (Hint: Identify success with a defective part.): a. \(5 \%\) defective parts b. \(10 \%\) defective parts c. \(20 \%\) defective parts

Short Answer

Expert verified
The probabilities of accepting lots with different percentages of defective parts are: - 5% defective parts: \(P_A\approx 0.3585\) - 10% defective parts: \(P_A\approx 0.1316\) - 20% defective parts: \(P_A\approx 0.0189\)

Step by step solution

01

a. 5% defective parts

We look for the probability that exactly 0 or 1 part out of 20 is defective when the probability of a defective part is 5% (p = 0.05). Using the binomial formula, we get: \[P(x=0) = \binom{20}{0}(0.05)^0(1-0.05)^{20}\] \[P(x=1) = \binom{20}{1}(0.05)^1(1-0.05)^{19}\] The probability of accepting a lot is the sum of these probabilities: \[P_A = P(x=0) + P(x=1)\] Calculating these probabilities, we find the probability of accepting a lot with 5% defective parts is approximately 0.3585.
02

b. 10% defective parts

We now look for the probability that exactly 0 or 1 part out of 20 is defective when the probability of a defective part is 10% (p = 0.10). Using the binomial formula, we get: \[P(x=0) = \binom{20}{0}(0.10)^0(1-0.10)^{20}\] \[P(x=1) = \binom{20}{1}(0.10)^1(1-0.10)^{19}\] The probability of accepting a lot is the sum of these probabilities: \[P_A = P(x=0) + P(x=1)\] Calculating these probabilities, we find the probability of accepting a lot with 10% defective parts is approximately 0.1316.
03

c. 20% defective parts

Lastly, we look for the probability that exactly 0 or 1 part out of 20 is defective when the probability of a defective part is 20% (p = 0.20). Using the binomial formula, we get: \[P(x=0) = \binom{20}{0}(0.20)^0(1-0.20)^{20}\] \[P(x=1) = \binom{20}{1}(0.20)^1(1-0.20)^{19}\] The probability of accepting a lot is the sum of these probabilities: \[P_A = P(x=0) + P(x=1)\] Calculating these probabilities, we find the probability of accepting a lot with 20% defective parts is approximately 0.0189. In summary, the probabilities of accepting lots with different percentages of defective parts are as follows: - 5% defective parts: \(P_A\approx 0.3585\) - 10% defective parts: \(P_A\approx 0.1316\) - 20% defective parts: \(P_A\approx 0.0189\)

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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 key concept in probability theory. It is particularly useful for modeling situations where you have a fixed number of independent trials, each with two possible outcomes: success or failure. In the context of quality control, a 'success' can be thought of as detecting a defective part.

The binomial distribution is defined by two parameters: the number of trials, denoted as \( n \), and the probability of success on a single trial, \( p \). The probability of getting exactly \( k \) successes (defective parts) in \( n \) trials is given by the formula:
  • \( P(x=k) = \binom{n}{k} p^k (1-p)^{n-k} \)
In our exercise, \( n = 20 \) and \( p \) varies based on the defective rate of the supplier's lot. Calculations are done for when 0 or 1 out of the 20 parts are defective.
Quality Control
Quality control is a crucial process in many industries. It ensures that the products or materials meet specific standards and specifications before reaching the market. An essential part of this process is the inspection of incoming materials from suppliers. This involves carefully selecting samples from each batch, examining them for defects, and deciding whether to accept or reject the lot based on the inspection outcomes.

In the exercise provided, quality control involves evaluating lots of parts with a predetermined method for acceptance. Specifically, the acceptance criterion is that no more than one defective part is found in a sample of 20. If more than one part is defective, the lot is considered below standard and rejected.
Defective Rate
The defective rate refers to the proportion of defective items in a batch. It is a critical factor in determining the overall quality of a lot. Companies aim to minimize the defective rate to ensure customer satisfaction and reduce returns. In probability terms, the defective rate corresponds to the probability \( p \) used in the binomial distribution.

In our exercise, we explore different defective rates to calculate the probability of accepting a lot. The different scenarios provided have defective rates of 5%, 10%, and 20%.
  • A 5% defective rate means one part in twenty could be defective.
  • A 10% defective rate implies the likelihood increases to two in twenty.
  • At a 20% defective rate, four parts in twenty could be expected to be defective.
These variations help determine how strict or lenient the quality control acceptance should be.
Acceptance Sampling
Acceptance sampling is a statistical method used in quality control to decide whether a batch of goods should be accepted or rejected. Rather than inspecting every item, a random sample is taken and examined. The results from the sample inspection decide the fate of the entire batch.

This method saves time and cost, as it is often not feasible to check every item. However, acceptance sampling has its risks. There is always the chance of accepting a poor quality lot or rejecting a good one. In the given exercise, acceptance sampling is based on inspecting 20 parts from a lot and accepting the lot if one or no defective parts are found.

The decision criteria used in acceptance sampling heavily rely on statistical probabilities, which can significantly affect the risk levels both consumers and producers are willing to take.

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Most popular questions from this chapter

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