/*! 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 110 This exercise illustrates that p... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 110

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 100 customer orders to fill. Each order requires one component part that is purchased from a supplier. However, typically, \(2 \%\) of the components are identified as defective, and the components can be assumed to be independent. (a) If the manufacturer stocks 100 components, what is the probability that the 100 orders can be filled without reordering components? (b) If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components? (c) If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

Short Answer

Expert verified
(a) Approximately 0.1326; (b) Approximately 0.9165; (c) Approximately 0.9933.

Step by step solution

01

Understand the Problem Context

We need to calculate the probability that the manufacturer can fill 100 orders without needing to reorder extra components. Given that 2% of the components are defective, we need to consider how stocking different quantities of components affects this probability.
02

Define Defective Probability

The probability that a single component is defective is 2%, or \( p = 0.02 \). The probability that it is not defective is \( 1 - p = 0.98 \).
03

Setting Up the Binomial Distribution

We can model this scenario using a binomial distribution, where the number of defective components (successes) follows a binomial distribution. Let \( n \) be the stock and \( k \) be the total number of defective components that can be tolerated while still fulfilling the orders.
04

Calculate Probability for 100 Components

For \( n = 100 \), we need exactly 0 defective components for all 100 to function (since there's no excess to replace any defective pieces). The probability is calculated as \( P(X = 0) = \binom{100}{0} (0.02)^0 (0.98)^{100} = (0.98)^{100} \).
05

Calculate Probability for 102 Components

For \( n = 102 \), we can tolerate up to \( k = 2 \) defective components. Calculate \( P(X \leq 2) \) as \[ P(X \leq 2) = \sum_{x=0}^{2} \binom{102}{x} (0.02)^x (0.98)^{102-x} \].
06

Calculate Probability for 105 Components

For \( n = 105 \), we can tolerate up to \( k = 5 \) defective components. Calculate \( P(X \leq 5) \) as \[ P(X \leq 5) = \sum_{x=0}^{5} \binom{105}{x} (0.02)^x (0.98)^{105-x} \].
07

Final Calculation and Conclusion

Carry out the calculations for steps 4 to 6 using a calculator or statistical software to obtain the respective probabilities for each situation, ensuring all necessary terms in the summation are calculated.

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Key Concepts

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

Binomial Distribution
When dealing with quality control in manufacturing, understanding the binomial distribution is essential. This distribution models scenarios where there are two possible outcomes for each trial鈥攕uch as defective or non-defective components.
In our context, each component can either be defective or not. The probability of a defective component is 2%, which serves as the success probability in our binomial experiment. This distribution helps us predict the likelihood of obtaining a certain number, say "k," of defective components out of several trials, denoted by "n".
Key aspects of the binomial distribution are:
  • A fixed number of trials (components inspected).
  • Only two outcomes: defective or not defective.
  • Constant probability of defectiveness for each trial.
    • Understanding this helps companies anticipate the extent of potential defects, which aids in stock planning for order fulfillment.
Defective Components Probability
Calculating the defective components' probability allows manufacturers to assess how many defective parts they might expect. For a specific stock size, figuring out the probability of encountering no more than a certain number of defective units is crucial.
We use the probability of a non-defective component, calculated as 1 minus the defect chance, in conjunction with binomial probability formulas.
You may encounter expressions like \[ P(X \leq k) = \sum_{x=0}^{k} \binom{n}{x} (p)^x (1-p)^{n-x} \], where \( n \) is the total stock, \( k \) is the maximum tolerable defects, and \( p = 0.02 \).
These calculations enable predicting if stocked parts suffice without encountering shortages, hence crucial for preventing unexpected stalling of production due to part defects.
Order Fulfillment
Order fulfillment in manufacturing involves ensuring all customer orders are completed accurately and promptly. A significant challenge arises with defective components that can disrupt the supply chain.
Maintaining adequate stock to buffer against defects ensures that orders are unmet. In our exercise, different stocking levels (100, 102, and 105 items) correspond to varying defect tolerances.
  • With 100 components, zero defects are permitted.
  • With 102, up to two defective parts can be accommodated.
  • 105 components allow for up to five defects.
This proactive planning of component reserves based on calculated probabilities sustains order fulfillment even when some parts are below acceptable standards. Thus, integrating statistical tools like binomial distribution enables efficient risk management and enhances customer satisfaction.

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Most popular questions from this chapter

The sample space of a random experiment is {a, b, c, d, e, f}, and each outcome is equally likely. A random variable is defined as follows: $$ \begin{array}{|c|c|c|c|c|c|c|} \hline \text { outcome } & a & b & c & d & e & f \\ \hline x & 0 & 0 & 1.5 & 1.5 & 2 & 3 \\ \hline \end{array} $$ Determine the probability mass function of \(a\). Use the probability mass function to determine the following probabilities: (a) \(P(X=1.5)\) (b) \(P(0.53)\) (d) \(P(0 \leq X<2)\) (e) \(P(X=0\) or \(X=2)\)

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume that the probability that a wafer passes the test is 0.8 and that wafers are independent. Determine the probability mass function of the number of wafers from a lot that pass the test.

The probability that a visitor to a Web site provides contact data for additional information is \(0.01 .\) Assume that 1000 visitors to the site behave independently. Determine the following probabilities: (a) No visitor provides contact data. (b) Exactly 10 visitors provide contact data. (c) More than 3 visitors provide contact data.

Assume that the number of errors along a magnetic recording surface is a Poisson random variable with a mean of one error every \(10^{5}\) bits. A sector of data consists of 4096 eight-bit bytes. (a) What is the probability of more than one error in a sector? (b) What is the mean number of sectors until an error occurs?

Samples of rejuvenated mitochondria are mutated (defective) in \(1 \%\) of cases. Suppose that 15 samples are studied and can be considered to be independent for mutation. Determine the following probabilities. The binomial table in Appendix A can help. (a) No samples are mutated. (b) At most one sample is mutated. (c) More than half the samples are mutated.
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