/*! 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 87 In a food processing and packagi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a food processing and packaging plant, there are, on the average, two packaging machine breakdowns per week. Assume the weekly machine breakdowns follow a Poisson distribution. a. What is the probability that there are no machine breakdowns in a given week? b. Calculate the probability that there are no more than two machine breakdowns in a given week.

Short Answer

Expert verified
Answer: The probability of no machine breakdowns in a given week is approximately 13.53%. The probability of no more than 2 machine breakdowns in a given week is approximately 67.67%.

Step by step solution

01

Required Concepts

The given problem follows the Poisson probability distribution. The probability mass function (PMF) of a Poisson distribution is given by: P(x) = (e^(-λ) * λ^x) / x! Where, P(x) = Probability of x machine breakdowns λ = Average number of machine breakdowns per week (λ = 2) e = Base of natural logarithm (e ≈ 2.71828) x = Number of machine breakdowns in a given week #a. We need to find the probability of no machine breakdowns in a given week. So, x = 0.
02

Use the Poisson PMF with x = 0 and λ = 2

The PMF of the Poisson distribution with x = 0 and λ= 2 is given by: P(0) = (e^(-2) * 2^0) / 0!
03

Calculate the probability

Now, we will calculate the probability: P(0) = (2.71828^(-2) * 1) / 1 P(0) ≈ 0.1353 The probability of no machine breakdowns in a given week is approximately 0.1353 or 13.53%. #b. We are asked to find the probability of no more than 2 machine breakdowns in a given week. So, x = 0, 1, and 2.
04

Use the Poisson PMF for x = 0, 1, and 2

For x = 0, 1, and 2, we will use the Poisson PMF: P(0) = (e^(-2) * 2^0) / 0! P(1) = (e^(-2) * 2^1) / 1! P(2) = (e^(-2) * 2^2) / 2!
05

Calculate the probabilities

Calculate the individual probabilities: P(0) ≈ 0.1353 (calculated previously) P(1) = (2.71828^(-2) * 2) / 1 ≈ 0.2707 P(2) = (2.71828^(-2) * 4) / 2 ≈ 0.2707
06

Calculate the probability of no more than 2 breakdowns

Now, we will sum up the probabilities: P(x ≤ 2) = P(0) + P(1) + P(2) P(x ≤ 2) ≈ 0.1353 + 0.2707 + 0.2707 P(x ≤ 2) ≈ 0.6767 The probability of no more than 2 machine breakdowns in a given week is approximately 0.6767 or 67.67%.

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.

Probability Mass Function
The Probability Mass Function (PMF) is a mathematical function used to describe the probability of a specific number of events occurring in a fixed interval of time or space under certain conditions. In the context of a Poisson distribution, the PMF is especially useful to determine the likelihood of a certain number of events occurring when you know the average rate of occurrence.

For the Poisson distribution, the PMF is given by the formula:
  • \[ P(x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \]
In this formula:

  • \( P(x) \) is the probability of observing \( x \) events (e.g., machine breakdowns) in the given interval.
  • \( \lambda \) represents the average number of events (here, 2 machine breakdowns per week).
  • \( e \) is the base of the natural logarithm, approximately \( 2.71828 \).
  • \( x \) denotes the number of events for which you want to calculate the probability.
  • \( x! \) is the factorial of \( x \), representing the product of all positive integers up to \( x \).
Understanding this function allows you to calculate the probability of a specific number of events, given the average rate, effectively capturing how likely or common various outcomes are.
Machine Breakdowns
In the domain of industrial operations, machine breakdowns can significantly impact productivity. Understanding the distribution of these breakdowns over time helps in planning and resource allocation, reducing downtime and ensuring efficient operations.

The scenario presented involves a packaging machine that, on average, breaks down twice a week. To determine the probability of different breakdown occurrences, the Poisson distribution is utilized as it suits scenarios where events happen independently at a constant average rate. This kind of statistical insight can guide maintenance schedules and parts stocking.

By analyzing the breakdown pattern, for instance, it becomes clearer if unexpected situations frequently disrupt plant operations or if breakdowns align with the historical average. Organizations can leverage such data-driven insights to implement predictive maintenance, where maintenance efforts are more precise and targeted, eventually maintaining productivity at optimal levels.
Probability Calculation
Probability calculation is an intrinsic part of statistics that quantifies the likelihood of particular outcomes. In exercises like determining machine breakdown probabilities, understanding and calculating probabilities allows decision-makers to anticipate various outcomes credibly.

For example, to find the probability of no machine breakdowns in a week, we set \( x = 0 \) in the Poisson PMF, which results in:
  • \[ P(0) = \frac{e^{-2} \cdot 2^0}{0!} \approx 0.1353 \]
This result implies around a 13.53% chance of experiencing no breakdowns in the week.

Similarly, to find the probability of no more than two breakdowns, calculations sum the probabilities for breakdowns of 0, 1, and 2:
  • \[ P(x \leq 2) = P(0) + P(1) + P(2) \]
  • \[ P(x \leq 2) = 0.1353 + 0.2707 + 0.2707 \approx 0.6767 \]
This provides a 67.67% likelihood of having two or fewer breakdowns in a week, aiding in realistic expectation setting for maintenance needs. Understanding this process strengthens clarity on how likely various operational scenarios are, equipping teams to prepare strategically for potential disruptions.

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

A Snapshot in USA Today shows that \(60 \%\) of consumers say they have become more conservative spenders. \({ }^{12}\) When asked "What would you do first if you won \(\$ 1\) million tomorrow?" the answers had to do with somewhat conservative measures like "hire a financial advisor," or "pay off my credit card," or "pay off my mortgage.' Suppose a random sample of \(n=15\) consumers is selected and the number \(x\) of those who say they have become conservative spenders recorded. a. What is the probability that more than six consumers say they have become conservative spenders? b. What is the probability that fewer than five of those sampled have become conservative spenders? c. What is the probability that exactly nine of those sampled are now conservative spenders.

Americans are really getting away while on vacation. In fact, among small business owners, more than half \((51 \%)\) say they check in with the office at least once a day while on vacation; only \(27 \%\) say they cut the cord completely. \({ }^{7}\) If 20 small business owners are randomly selected, and we assume that exactly half check in with the office at least once a day, then \(n=20\) and \(p=.5 .\) Find the following probabilities. a. Exactly 16 say that they check in with the office at least once a day while on vacation. b. Between 15 and 18 (inclusive) say they check in with the office at least once a day while on vacation. c. Five or fewer say that they check in with the office at least once a day while on vacation. Would this be an unlikely occurrence?

The recession has caused many people to use their credit cards far less. In fact, in the United States, \(60 \%\) of consumers say they are committed to living with fewer credit cards. \({ }^{15} \mathrm{~A}\) sample of \(n=400\) consumers with credit cards are randomly selected. a. What is the average number of consumers in the sample who said they are committed to living with fewer credit cards? b. What is the standard deviation of the number in the sample who said they are committed to living with fewer credit cards? c. Within what range would you expect to find the number in the sample who said they are committed to living with fewer credit cards? d. If only 200 of the sample of consumers said they were committed to living with fewer credit cards, would you consider this unusual? Explain. What conclusion might you draw from this sample information?

Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected and checked for whitefly. a. What is the average number of fields sampled that are infested with whitefly? b. Within what limits would you expect to find the number of infested fields, with probability approximately \(95 \% ?\) c. What might you conclude if you found that \(x=25\) fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.

Suppose the four engines of a commercial aircraft are arranged to operate independently and that the probability of in-flight failure of a single engine is .01. What is the probability of the following events on a given flight? a. No failures are observed. b. No more than one failure is observed.

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.