/*! 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 178 An automated egg carton loader h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 178

An automated egg carton loader has a \(1 \%\) probability of cracking an egg, and a customer will complain if more than one egg per dozen is cracked. Assume that each egg load is an independent event. (a) What is the distribution of cracked eggs per dozen? Include parameter values. (b) What is the probability that a carton of a dozen eggs results in a complaint? (c) What are the mean and standard deviation of the number of cracked eggs in a carton of a dozen eggs?

Short Answer

Expert verified
(a) Binomial distribution, n=12, p=0.01; (b) Probability of complaint: 0.007; (c) Mean = 0.12, Standard Deviation 鈮� 0.345.

Step by step solution

01

Identify the Distribution

The situation describes a scenario where each egg can either be cracked or not cracked, and the number of eggs cracked out of a fixed number of eggs follows a binomial distribution. Therefore, the number of cracked eggs per dozen follows a Binomial distribution with parameters \( n = 12 \) (since there are 12 eggs in a dozen) and \( p = 0.01 \) (the probability of an individual egg being cracked).
02

Calculate the Probability of a Complaint

The customer complains if more than one egg per dozen is cracked. So, we need to find the probability that 2 or more eggs are cracked in a dozen. For a binomial distribution \( X \) with parameters \( n = 12 \) and \( p = 0.01 \), we calculate:\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \]First, calculate \( P(X = 0) \) and \( P(X = 1) \):\[ P(X = 0) = \binom{12}{0} (0.01)^0 (0.99)^{12} \approx 0.886 \]\[ P(X = 1) = \binom{12}{1} (0.01)^1 (0.99)^{11} \approx 0.107 \]Thus,\[ P(X \geq 2) = 1 - 0.886 - 0.107 \approx 0.007 \]
03

Calculate the Mean and Standard Deviation

For a binomial distribution with parameters \( n \) and \( p \), the mean \( \mu \) is given by \( \mu = np \), and the standard deviation \( \sigma \) is given by \( \sigma = \sqrt{np(1-p)} \).For \( n = 12 \) and \( p = 0.01 \):\[ \mu = 12 \times 0.01 = 0.12 \]\[ \sigma = \sqrt{12 \times 0.01 \times 0.99} \approx 0.345 \]

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 91影视!

Key Concepts

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

Probability
Probability refers to the likelihood of an event occurring. In the context of our egg carton loader, we're interested in the probability of cracking eggs while being loaded into a carton. Since each egg has a 1% chance of being cracked, we represent this probability as \( p = 0.01 \). For understanding complaints, we want to calculate the probability of more than one egg being cracked in a dozen. Using the binomial probability formula, we can find the probability distribution for different scenarios, such as none, one, or two eggs being cracked.
Mean
The mean, often referred to as the expected value, is the average outcome we expect. For a binomial distribution, the mean can be calculated using the formula \( \mu = np \), where \( n \) is the number of trials, and \( p \) is the probability of success in each trial. So for our egg scenario, where \( n = 12 \) and \( p = 0.01 \), the mean number of cracked eggs per carton is \( 0.12 \). This means on average, we expect 0.12 eggs to be cracked per carton, which aligns with the low probability of cracking any given egg.
Standard deviation
The standard deviation tells us how much variation or "spread" exists from the mean. In the binomial distribution context, it is calculated as \( \sigma = \sqrt{np(1-p)} \). For our eggs, \( n = 12 \) and \( p = 0.01 \), yielding a standard deviation of approximately \( 0.345 \). This indicates that while the mean number of cracked eggs is low, there's a predictable range in which the number of cracked eggs can vary in any given dozen. A smaller standard deviation would suggest less variability, while a larger one would indicate more variability.
Independent events
In probability, independent events are those whose outcomes do not affect each other. For example, in our scenario, each egg being cracked is an independent event. The outcome of one egg cracking does not influence another's outcome. This is a key reason why a binomial distribution is appropriate for this situation. Independence ensures that the probability of each event remains at \( 0.01 \), unaffected by the success or failure of previous or subsequent events.
Parameter values
Parameter values are critical in defining the scope of a binomial distribution. In our case, these include \( n \), the number of trials (or eggs in a carton), and \( p \), the probability of an egg cracking. Here, \( n = 12 \) and \( p = 0.01 \). These parameters shape the distribution, allowing us to calculate other metrics such as probability, mean, and standard deviation. They essentially describe the framework within which we can predict and understand the cracking behavior of eggs during loading.

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 congested computer network has a \(1 \%\) chance of losing a data packet, and packet losses are independent events. An e-mail message requires 100 packets. (a) What distribution of data packets must be re-sent? Include the parameter values. (b) What is the probability that at least one packet must be re-sent? (c) What is the probability that two or more packets must be re-sent? (d) What are the mean and standard deviation of the number of packets that must be re-sent? (e) If there are 10 messages and each contains 100 packets, what is the probability that at least one message requires that two or more packets be re- sent?

Determine the constant \(c\) so that the following function is a probability mass function: \(f(x)=c x\) for \(x=1,2,3,4\).

Let the random variable \(X\) be equally likely to assume any of the values \(1 / 8,1 / 4,\) or \(3 / 8 .\) Determine the mean and variance of \(X\).

Assume that each of your calls to a popular radio station has a probability of 0.02 of connecting, that is, of not obtaining a busy signal. Assume that your calls are independent. (a) What is the probability that your first call that connects is your 10 th call? (b) What is the probability that it requires more than five calls for you to connect? (c) What is the mean number of calls needed to connect?

A Web site randomly selects among 10 products to discount each day. The color printer of interest to you is discounted today. (a) What is the expected number of days until this product is again discounted? (b) What is the probability that this product is first discounted again exactly 10 days from now? (c) If the product is not discounted for the next five days, what is the probability that it is first discounted again 15 days from now? (d) What is the probability that this product is first discounted again within three or fewer days?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
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.