/*! 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 27 If the probability that a fluore... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the probability that a fluorescent light has a useful life of at least 800 hours is \(0.9,\) find the probabilities that among 20 such lights (a) exactly 18 will have a useful life of at least 800 hours; (b) at. least 15 will have a useful life of at least 800 hours; (c) at least. 2 will not have a useful life of at least 800 hours.

Short Answer

Expert verified
The probability that, among 20 lights, (a) exactly 18 will have a useful life of at least 800 hours, (b) at least 15 will have a useful life of at least 800 hours, and (c) at least 2 will not have a useful life of at least 800 hours can be calculated using the binomial probability formula.

Step by step solution

01

Identify the parameters

In this case, the number of trials n is 20, because we have 20 fluorescents and the probability of a fluorescent bulb having a life at least 800 hours, p, is 0.9.
02

Calculate the probability for (a) exactly 18 will have a useful life of at least 800 hours.

Use the binomial probability formula to solve this. P(X=18) = C(20, 18) * (0.9^18) * ((1 - 0.9)^(20 - 18))
03

Calculate the probability for (b) at least 15 will have a useful life of at least 800 hours.

This means 15, 16, 17, 18, 19, or 20 fluorescents can be successful, which needs to be calculated each using the binomial probability formula and added together.
04

Calculate the probability for (c) at least 2 will not have a useful life of at least 800 hours.

This is equivalent to no more than 18 will have a useful life of more than 800 hours. Calculate this by subtracting the probability of 19 or 20 bulbs having a useful life of at least 800 hours from 1.

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 for Engineering
In engineering contexts, probability plays a critical role in assessing the reliability and performance of systems and components. Consider the example of fluorescent lights with an 80% likelihood of functioning beyond 800 hours. Engineers determine the reliability of these lights in large quantities by using probability. The probability that exactly 18 out of 20 lights will work for the desired time frame is computed via a specific type of calculation: binomial probability.

Understanding these probabilities enables engineers to predict performance, plan for maintenance, and manage warranties. The calculations inform decisions such as how many spare bulbs to stock and predict the service intervals for replacement, ensuring efficient operation and customer satisfaction.
Statistics for Scientists
Statistics for scientists encompass tools and methods for analyzing data, including understanding the distribution of data and making inferences based on samples. In the context of the fluorescent light example, a scientist may investigate the longevity of lights from a data-driven perspective. The probability that at least 15 will last the required 800 hours is evaluated using the fundamental principles of descriptive and inferential statistics.

Scientists use such statistical measures to present their findings confidently, understand the variability in experiments, and build predictive models. This knowledge is paramount, especially when the cost of failure is high, and the need for precision is paramount.
Probability Distribution Calculations
Probability distribution calculations involve determining the likelihood of various outcomes in a process. These calculations provide a framework to quantify the random variables in a problem. When calculating the probability of at least 2 out of 20 fluorescent lights not lasting 800 hours, we are considering the cumulative probability of multiple binomial events. Here, the binomial distribution model efficiently condenses what would otherwise be a complex system with a large number of trials into a manageable calculation.

By mastering these calculations, students and professionals can forecast outcomes with greater accuracy. It is important to understand that even though each light bulb independently has a high probability of working beyond the 800 hours, there's a compound effect when considering them collectively.
Binomial Distribution
The binomial distribution is a discrete probability distribution that applies to scenarios with two possible outcomes (success or failure) and a fixed number of independent trials. It is a cornerstone in understanding processes that emerge in engineering, sciences, and daily decision-making.

Applying the binomial formula, as shown in the exercise, involves combinations and powers to calculate the probability of a specific number of successes. It enables the computation of various scenarios, such as the probability of exactly 18 successes, or at least 15, by summing up individual probabilities. Recognizing and applying the binomial distribution paves the way for streamlined and powerful reasoning when dealing with yes-or-no type data in a range of real-world applications.

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

An automobile manufacturer is concerned about a fault in the braking mechanism of a particular model. The fault can, on rare occasions, cause a catastrophe at high speed. The distribution of the number of cars per year that will experience the fault is a Poisson random variable with \(\mathrm{A}=5\) (a) What is the probability that at most 3 cars per year will experience a catastrophe? (b) What is the probability that more than 1 car per year will experience a catastrophe?

Find the probability of being dealt a bridge hand of 13 cards containing 5 spades, 2 hearts. 3 diamonds. and 3 clubs.

A random committee of size 3 is selected from 4 doctors and 2 nurses. Write a formula for the probability distribution of the random variable \(X\) representing the number of doctors on the committee. Find \(P(2 \leq X \leq 3)\).

The surface of a circular dart board has a small center circle called the bull's-eye and 20 pie-shaped regions numbered from 1 to \(20 .\) Each of the pie-shaped regions is further divided into three parts such that a person throwing a dart that lands on a specified number scores the value of the number, double the number, or triple the number, depending on which of the three parts the dart falls. If a person hits the bull's-eye with probability 0.01 . hits a double with probability 0.10 , hits a triple with probability \(0.05,\) and misses the dart board with probability \(0.02,\) what is the probability that 7 throws will result in no bull's-eyes, no triples, a double twice, and a complete miss once?

An electronic switching device occasionally malfunctions and may need to be replaced. It is known that the device is satisfactory if it makes, on the average, no more than 0.20 error per hour. A particular 5-hour period is chosen as a "test" on the device. If no more than 1 error occurs, the device is considered satisfactory. (a) What is the probability that a satisfactory device will be considered unsatisfactory on the basis of the rest? Assume that a Poisson process exists. (b) What is the probability that a device will be accepted as satisfactory when, in fact, the mean number of errors is \(0.25 ?\) Again, assume that a Poisson process exists.

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.