/*! 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 34 Magnetron tulses are produced fr... [FREE SOLUTION] | 91Ó°ÊÓ

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Magnetron tulses are produced from an automated assembly line. A sampling plan is used periodically to assess quality on the lengths of the tubes. This measurement is subject to uncertainty. It is thought that the probability that a random tube meets length specification is \(0.99 .\) A sampling plan is used in which the lengths of 5 random tubes are measured. (a) Show that the probability function of \(Y\), the number out of 5 that meet length specification, is given by the following discrete probability function $$ \begin{aligned} f(y) &=\frac{5 !}{y !(5-y) !}(0.99)^{y}(0.01)^{5-y.} \\ \text { for } y=0.1,2,3,4,5. \end{aligned} $$ (b) Suppose random selections are made off the line and 3 are outside specifications. Use \(f(y)\) above either to support or refute the conjecture that the probability is 0.99 that a single tube meets specifications.

Short Answer

Expert verified
The validity of the conjecture that the probability is 0.99 that a single tube meets specifications is determined by evaluating the probability function for \( f(2) \), then comparing it with expected outcomes. If the calculated probability is significantly low, it suggests that the assumption of \( p = 0.99 \) may not be accurate.

Step by step solution

01

Understand the Problem

Notice that the tubes meeting the specification is a 'success', denoted by \( y \), and each tube measurement is independent of the others. The probability of success \( p = 0.99 \) and the probability of failure \( q = 0.01 \), with \( f(y) \) being the number of successes.
02

Confirm the Probability Function

The given function is a standard form of the binomial probability mass function (pmf). It is calculated as \( f(y) = \frac{n!}{y!(n−y)!} * p^y * q^{n-y} \). In this case, \( n = 5 \), \( y \) is the number of tubes that meet the specification. If we substitute these values into the function, we get \( f(y) = \frac{5!}{y!(5−y)!} * (0.99)^y * (0.01)^{5-y} \) where \( 0 \leq y \leq 5 \). This exactly matches with a given probability function. Thus, we have confirmed that the given function is correct.
03

Applying the Probability Function to the Scenario

Here we are asked to check if the calculated probability \( p = 0.99 \) is consistent with the situation where 3 out of five tubes are outside specification. For this, let's evaluate \( f(2) \), as \( 2 \) is the number of tubes that meet the specification, while \( 3 \) of them do not: \( f(2) = \frac{5!}{2!(5−2)!} * (0.99)^2 * (0.01)^{5-2} \).
04

Interpret the Results

The value yielded by \( f(2) \) is the probability of exactly \( 2 \) tubes meeting specification given that the true probability of meeting specification is \( 0.99 \). If this calculated probability, \( f(2) \), is significantly low, it may suggest that our initial assumption of \( p = 0.99 \) may not be accurate or representative of the real situation.

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Key Concepts

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

Probability Function
The probability function is a mathematical representation of how likely different outcomes are in an experiment or process. Think of it as a recipe which tells us the likelihood of each outcome in a scenario where there's randomness involved. In the context of an automated tube assembly line, the discrete probability function we discuss is specifically known as a binomial probability mass function (pmf).

Why binomial? Since each tube can either meet the specification (success) or not (failure) and there's a fixed number of trials (five tubes), we have a classic case of a binomial experiment. The probability of success is denoted by p, which is given as 0.99, meaning there's a 99% chance a tube is within specifications. Conversely, the probability of failure is q (1 - p), which is 0.01. The pmf formula: \[ f(y) = \frac{n!}{y!(n−y)!} * p^y * q^{n-y} \] is used where n is the number of trials (5), and y is the number of successes (tubes meeting specifications).

This formula gives us the probabilities associated with 0, 1, 2, 3, 4, or 5 tubes meeting the length specification, allowing us to predict quality and consistency in production.
Sampling Plan
In quality control, a sampling plan is a strategy that defines how samples should be selected and tested from a production lot to make decisions about the whole lot. In our exercise, a sampling plan where 5 random tubes are measured is in action. This strategy seeks to capture a 'snapshot' of the production's adherence to length specifications while minimizing inspection time and costs.

In creating a sampling plan, several factors must be considered:
  • Sample size (in the example, it's 5 tubes).
  • Frequency of sampling (how often tubes are tested).
  • Acceptance criteria (the rules to decide if a lot meets quality standards).


Using the binomial probability function within the sampling plan provides us with a statistical basis for evaluating the potential risk of non-conformance and helps to make objective decisions about the entire production process.
Specification Compliance
Specification compliance is a key component in manufacturing quality control. It refers to how well the manufactured products adhere to defined standards and criteria – in this case, the length of the tubes. Maintaining high compliance means that the products are consistently within specifications, which, in turn, indicates a high-quality production process.

Our example focuses on the probability that each tube meets the length specification set forth by the manufacturer. By applying the binomial probability mass function, we can quantitatively assess the compliance level of a small sample which represents the larger lot. An unexpected result (like finding that three out of five tubes are out of specification) calls for a deeper inspection. If the compliance rate suggested by the pmf is much higher than what is observed in a sample, it raises questions about process stability or the accuracy of the probability function used, and further investigation would be necessary to maintain specification compliance.

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Most popular questions from this chapter

The time to failure in hours of an important piece of electronic equipment used in a manufactured DVD player las the density function $$ I(x)=\left\\{\begin{array}{ll} \frac{1}{2000} \exp (-a: / 2000), & x \geq 0, \\ 0, & x<0. \end{array}\right. $$ (a) Find \(F(x)\). (b) Determine the probability that the component (and thus the DVD player) lasts more than 1000 hours before the component needs to be replaced. (c) Determine the probability that the component fails before 2000 hours.

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