/*! 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 30 When a new machine is functionin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When a new machine is functioning properly, only \(3 \%\) of the items produced are defective. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found. a. Describe the conditions under which this situation would be a binomial experiment. b. Draw a tree diagram similar to Figure 5.3 showing this problem as a two- trial experiment. c. How many experimental outcomes result in exactly one defect being found? d. Compute the probabilities associated with finding no defects, exactly one defect, and two defects.

Short Answer

Expert verified
a) This is a binomial experiment. b) Use the steps in the tree diagram. c) There are 2 outcomes. d) No defect: 0.9409, one defect: 0.0582, two defects: 0.0009.

Step by step solution

01

Identify the Conditions for a Binomial Experiment

A binomial experiment must meet the following criteria: 1. There are a fixed number of trials, in this case, 2 trials (selecting two parts). 2. Each trial has only two possible outcomes: defective or not defective. 3. The probability of success (finding a defective part) is the same for each trial, which is 0.03. 4. The trials are independent, meaning the outcome of one trial doesn't affect the other. Therefore, this scenario can indeed be described as a binomial experiment.
02

Construct a Tree Diagram

To create a tree diagram for this problem, follow these steps: - Begin with a starting point and create two branches: one for defective (D) and one for non-defective (ND) for the first part. - The probability of a part being defective is 0.03 and non-defective is 0.97. - From each of these branches, create two more branches for the second part: again, defective (D) and non-defective (ND). - Label all probabilities accordingly: - First branch: D (0.03), ND (0.97). - From D: D (0.03), ND (0.97); - From ND: D (0.03), ND (0.97).
03

Count Outcomes for Exactly One Defect

To find the number of outcomes with exactly one defective part, follow the tree diagram: 1. First part is defective: path is (D, ND). 2. Second part is defective: path is (ND, D). Each of these paths results in exactly one defective part. Therefore, there are 2 experimental outcomes.
04

Calculate Probabilities for Each Outcome

Using the binomial probability formula:1. Probability of no defects (both non-defective): \[ P(0) = (1 - 0.03)^2 = 0.97^2 = 0.9409. \]2. Probability of exactly one defect: \[ P(1) = \binom{2}{1} \times 0.03 \times 0.97 = 2 \times 0.03 \times 0.97 = 0.0582. \]3. Probability of two defects: \[ P(2) = (0.03)^2 = 0.0009. \]These probabilities add up to 1, which is a good sanity check.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Tree Diagram
A tree diagram is a helpful visual tool used in probability to outline all the possible outcomes of an experiment. In this case, we are interested in analyzing the probability of finding defective parts from a machine.

To construct a tree diagram, begin with a single node representing the start of our experiment. From this node, draw two branches representing the possible outcomes for the first trial: defective (D) and non-defective (ND). The probability of choosing a defective part is 0.03, and the probability of a non-defective part is 0.97.

Continue by branching out from the endpoints of the first branches for the second trial. Each branch from the first defective branch will again split into D and ND, with the same probabilities. Similarly, each branch from the first non-defective branch will also split into D and ND.

This tree diagram helps visually map out the pathways and calculate the probabilities of different scenarios, such as finding exactly one defective part. By assigning probabilities to each branch, you'll see all the potential outcomes and their likelihoods.
Probability Calculation
In probability calculations, identifying the likelihood of different outcomes is crucial. Using the tree diagram, we can easily calculate the probability of each scenario happening.

To find the probability of all outcomes where no defects are found, we look at the path where both parts are non-defective: (ND, ND). Calculate this using the formula:
  • \( P(0) = 0.97 \times 0.97 = 0.9409 \)
This result shows there's a high chance of not finding any defect.

Next, to find the probability of exactly one defect, consider two paths: the first path (D, ND) and the second path (ND, D). Use the binomial formula to compute this probability:
  • \( P(1) = \binom{2}{1} \times 0.03 \times 0.97 = 0.0582 \)
This indicates a moderate chance of finding exactly one defect.

For the probability of finding two defective parts (D, D), consider:
  • \( P(2) = 0.03 \times 0.03 = 0.0009 \)
This is a very slim chance, reflecting the rarity of this outcome. Adding these probabilities should give us a total of 1, confirming the accuracy of the calculations.
Defective Parts Analysis
Analyzing defective parts in a manufacturing process is essential for quality control. By understanding the likelihood of defects, manufacturers can improve processes and reduce waste.

In this exercise, the defective rate is given as 3%. Such information is vital to create strategies for reducing defects. Notably, this problem exemplifies a binomial experiment: evaluating two items where each can be defective or not. Each trial is independent, and the probability of defects is constant.

By examining the experimental outcomes, you can focus on pathways that lead to one defect. The outcomes (D, ND) and (ND, D) both result in exactly one defective part. With visual aids like tree diagrams, you can effortlessly identify these crucial pathways.

This method of analysis helps anticipate the proportion of defective parts over many trials, allowing fine-tuning of the machine or process for optimal efficiency. Keeping an eye on the defect probabilities enables companies to maintain high-quality production standards.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Twelve of the top 20 finishers in the 2009 PGA Championship at Hazeltine National Golf Club in Chaska, Minnesota, used a Titleist brand golf ball (GolfBallTest website, November 12,2009 . Suppose these results are representative of the probability that a randomly selected PGA Tour player uses a Titleist brand golf ball. For a sample of 15 PGA Tour players, make the following calculations. a. Compute the probability that exactly 10 of the 15 PGA Tour players use a Titleist brand golf ball. b. Compute the probability that more than 10 of the 15 PGA Tour players use a Titleist brand golf ball. c. For a sample of 15 PGA Tour players, compute the expected number of players who use a Titleist brand golf ball. d. For a sample of 15 PGA Tour players, compute the variance and standard deviation of the number of players who use a Titleist brand golf ball.

In a survey conducted by the Gallup Organization, respondents were asked, "What is your favorite sport to watch?" Football and basketball ranked number one and two in terms of preference (Gallup website, January 3,2004 ). Assume that in a group of 10 individuals, 7 prefer football and 3 prefer basketball. A random sample of 3 of these individuals is selected. a. What is the probability that exactly 2 prefer football? b. What is the probability that the majority (either 2 or 3) prefer football?

According to a survey conducted by TD Ameritrade, one out of four investors have exchange-traded funds in their portfolios (USA Today, January 11,2007 ). Consider a sample of 20 investors. a. Compute the probability that exactly four investors have exchange-traded funds in their portfolios. b. Compute the probability that at least two of the investors have exchange- traded funds in their portfolios. c. If you found that exactly 12 of the investors have exchange-traded funds in their portfolios, would you doubt the accuracy of the survey results? d. Compute the expected number of investors who have exchange-traded funds in their portfolios.

Three students scheduled interviews for summer employment at the Brookwood Institute. In each case the interview results in either an offer for a position or no offer. Experimental outcomes are defined in terms of the results of the three interviews. a. \(\quad\) List the experimental outcomes. b. Define a random variable that represents the number of offers made. Is the random variable continuous? c. Show the value of the random variable for each of the experimental outcomes.

The budgeting process for a midwestern college resulted in expense forecasts for the coming year (in \(\$$ millions) of \)\$ 9, \$ 10, \$ 11, \$ 12,\( and \)\$ 13 .\( Because the actual expenses are unknown, the following respective probabilities are assigned: \).3, .2, . .25, .05,\( and \).2 .\( a. Show the probability distribution for the expense forecast. b. What is the expected value of the expense forecast for the coming year? c. What is the variance of the expense forecast for the coming year? d. If income projections for the year are estimated at \)\$ 12$ million, comment on the financial position of the college.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.