/*! 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 78 The reliability function \(R(t)\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 78

The reliability function \(R(t)\) gives the probability that \(x > t\) For the pdf of a light bulb, this is the probability that the bulb lasts at least \(t\) hours. Compute \(R(t)\) for a general exponential pdf \(f(x)=c e^{-c x}\)

Short Answer

Expert verified
The reliability function \(R(t)\) for a general exponential pdf \(f(x)=c e^{-c x}\) is \(R(t)= e^{-c t}\).

Step by step solution

01

Compute the Cumulative Distribution Function (CDF)

The CDF \(F(t)\) of a random variable is given by integrating its pdf from the lower limit to t: \(F(t) = \int_{0}^{t} f(x) dx\). For the given problem, the pdf \(f(x)=c e^{-c x}\) is provided. Integrating this from 0 to t, we get: \[F(t) = \int_{0}^{t} c e^{-c x} dx\]. Solving the integral gives us \(-e^{-c x}\) evaluated at 0 and t, yielding: \[F(t)= 1 - e^{-c t}\].
02

Derive the reliability function from the CDF

The reliability function \(R(t)\) is related to the CDF by the equation: \(R(t)= 1-F(t)\). Thus, substituting the CDF \(F(t)\) obtained from Step 1, we have \(R(t)= 1-(1 - e^{-c t})\).
03

Simplify the expression

Simplify the expression obtained in Step 2 to get the final answer. The terms 1 cancel out, resulting in: \(R(t)= e^{-c t}\). This is the reliability function for a general exponential pdf.

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.

Reliability Function
In probability and statistics, the reliability function represents the likelihood that a particular system or component will function beyond a specified time, denoted as \( t \).
The reliability function, often represented as \( R(t) \), is crucial in fields such as engineering to predict the lifespan and effectiveness of components like light bulbs.
  • Reliability function is defined as \( R(t) = P(X > t) \), indicating the probability that a random variable \( X \) survives longer than time \( t \).
  • It is calculated using the formula \( R(t) = 1 - F(t) \), where \( F(t) \) is the cumulative distribution function (CDF).
For an exponential distribution with a probability density function \( f(x) = c e^{-c x} \), the reliability function becomes \( R(t) = e^{-c t} \).
This derivation shows that the time \( t \) a system remains operational can be modeled using exponential decay, a common assumption for certain types of systems and processes.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) of a random variable provides the probability that this variable takes a value less than or equal to \( t \).
This function is essential as it gives a complete overview of the probability distribution of a random variable.
  • The CDF \( F(t) \) is derived from the probability density function (PDF) by integrating the PDF from a lower bound up to \( t \).
  • Mathematically, for a PDF \( f(x) = c e^{-c x} \), the CDF is \( F(t) = \int_{0}^{t} c e^{-c x} \, dx \).
By evaluating this integral, we find \( F(t) = 1 - e^{-c t} \), which clearly shows the accumulative nature of probabilities up to time \( t \).
Understanding \( F(t) \) helps in analyzing the distribution and behavior of the system or process modeled by the exponential distribution.
Probability Density Function (PDF)
The probability density function (PDF) is a core concept that describes how the values of a continuous random variable are distributed. In the context of the exponential distribution, knowing the PDF helps to understand where and how the values of a random variable are concentrated.
  • The PDF for an exponential distribution is generally represented as \( f(x) = c e^{-c x} \).
  • This function indicates the likelihood of the random variable taking on a specific value \( x \).
The exponential PDF is characterized by a constant rate \( c \), responsible for the sharp decline in the probability density as \( x \) increases.
The PDF captures how rapid this decrease is, making it particularly useful for modeling time-to-failure in reliability analysis and other processes that exhibit a constant failure rate over time.

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

The excellent movie Stand and Deliver tells the story of mathematics teacher Jaime Escalante, who developed a remarkable AP calculus program in inner-city Los Angeles. In one scene, Escalante shows a student how to evaluate the integral \(\int x^{2} \sin x d x .\) He forms a chart like the following: $$\begin{array}{|c|c|c|} \hline & \sin x & \\ \hline x^{2} & -\cos x & \+ \\ \hline 2 x & -\sin x & \- \\ \hline 2 & \cos x & \+ \\ \hline \end{array}$$ Multiplying across each full row, the antiderivative is \(-x^{2} \cos x+2 x \sin x+2 \cos x+c .\) Explain where each column comes from and why the method works on this problem.

Being careful to use limits for the improper integrals, use the substitution \(u=\frac{\pi}{2}-x\) to show that (a) \(\int_{0}^{\pi / 2} \ln (\sin x) d x=\int_{0}^{\pi / 2} \ln (\cos x) d x .\) Add \(\int_{0}^{\pi / 2} \ln (\sin x) d x\) to both sides of this equation and simplify the right-hand side with the identity \(\sin 2 x=2 \sin x \cos x .\) (b) Use this result to show that \(2 \int_{0}^{\pi / 2} \ln (\sin x) d x=-\frac{\pi}{2} \ln 2+\frac{1}{2} \int_{0}^{\pi} \ln (\sin x) d x\) (c) Show that \(\int_{0}^{\pi} \ln (\sin x) d x=2 \int_{0}^{\pi / 2} \ln (\sin x) d x .\) (d) Use parts (b) and (c) to evaluate \(\int_{0}^{\pi / 2} \ln (\sin x) d x\)

In exercises find the partial fractions decomposition and an antiderivative. If you have a CAS available, use it to check $$\frac{x-5}{x^{2}-1}$$

Find out if your CAS has a special command (e.g., APART in Mathematica) to do partial fractions decompositions. Also, try \(\int \frac{x^{2}+2 x-1}{(x-1)^{2}\left(x^{2}+4\right)} d x\) and \, \(\int \frac{3 x}{\left(x^{2}+x+2\right)^{2}} d x.\)

In exercises find the partial fractions decomposition and an antiderivative. If you have a CAS available, use it to check your answer. $$\frac{x^{3}+x+2}{x^{2}+2 x-8}$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Pure 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.