Problem 34
Question
The reliability of the machine (the probability that it will work) in Exercise 33 is defined as $$ R(T)=1-\int_{0}^{T} 0.01 e^{-0.01 t} d t $$ where \(R(T)\) is the reliability at time \(T\). Write \(R(T)\) without using an integral.
Step-by-Step Solution
Verified Answer
The reliability function is \( R(T) = e^{-0.01T} \).
1Step 1: Understand the Integral
The given integral \( \int_{0}^{T} 0.01 e^{-0.01 t} \, dt \) represents the cumulative probability that the machine fails by time \( T \). The function inside the integral, \( 0.01 e^{-0.01 t} \), is the probability density function of an exponential distribution with a rate of 0.01.
2Step 2: Evaluate the Integral
To solve the integral \( \int_{0}^{T} 0.01 e^{-0.01 t} \, dt \), first recognize that it involves an exponential function, warranting substitution. The antiderivative of \( 0.01 e^{-0.01 t} \) is \(-e^{-0.01 t} \). Evaluating this from 0 to \( T \), we get:\[ -e^{-0.01T} + e^{0} = 1 - e^{-0.01T} \].
3Step 3: Substitute the Integral in R(T)
Substitute the evaluated integral back into the reliability function \( R(T) = 1 - \int_{0}^{T} 0.01 e^{-0.01 t} \, dt \). With the integral evaluated as \( 1 - e^{-0.01T} \), the reliability function is:\[ R(T) = 1 - (1 - e^{-0.01T}) \].
4Step 4: Simplify the Expression
Simplify the expression by distributing the subtraction:\[ R(T) = 0 + e^{-0.01T} \],therefore,\[ R(T) = e^{-0.01T} \].
Key Concepts
Reliability FunctionIntegral EvaluationProbability Density Function
Reliability Function
The reliability function is vital when dealing with systems like machines, as it tells us the probability that the system will perform successfully for a given time period, rather than failing. This is extremely important for understanding longevity and performance in both engineering and everyday applications.
Let’s break it down further:
In essence, the reliability function provides insights into how long a machine will last, which is crucial for design and maintenance planning.
Let’s break it down further:
- Purpose: The reliability function measures how long a machine is expected to function without failure.
- Expression: In this problem, it is expressed as \( R(T) = e^{-0.01T} \). This represents the probability that the machine will continue to work up to a time \( T \).
- Meaning of Exponential Term: The term \( e^{-0.01T} \) decreases as \( T \) increases, indicating that as more time passes, the probability of the machine surviving decreases.
In essence, the reliability function provides insights into how long a machine will last, which is crucial for design and maintenance planning.
Integral Evaluation
The process of integral evaluation is crucial in finding an explicit form of the reliability function without using integrals. Evaluating integrals can initially seem daunting, but with the exponential distribution, it simplifies nicely.
Here are the steps to evaluate the integral in this context:
Here are the steps to evaluate the integral in this context:
- Identify the Integral: Recognize the form \( \int_{0}^{T} 0.01 e^{-0.01 t} \, dt \) which indicates an exponential function integral.
- Use the Antiderivative: The antiderivative of \( 0.01 e^{-0.01 t} \) is \(-e^{-0.01 t} \), a standard result for integrals involving exponential decay.
- Apply Limits: By applying the limits of the integral from 0 to \( T \), you calculate it as \( 1 - e^{-0.01T} \).
Probability Density Function
The probability density function (PDF) of a random variable describes the likelihood of the variable taking particular values. For continuous data, this involves functions rather than discrete points.
In the context of the exponential distribution, the PDF is used to model the time between events like machine failures. Here’s how it works in this case:
In the context of the exponential distribution, the PDF is used to model the time between events like machine failures. Here’s how it works in this case:
- Exponential PDF: For this machine, the PDF is \( 0.01 e^{-0.01 t} \). This formula is typical for exponential distributions where 0.01 is the rate parameter.
- Role and Interpretation: The PDF describes how probabilities are distributed over time. In plain terms, it shows when the failures are most likely to occur within the given timeframe.
- Integral and PDF Relationship: The integral \( \int_{0}^{T} 0.01 e^{-0.01 t} \, dt \) gives the cumulative distribution function (CDF) up to time \( T \), representing the cumulative probability of failure by time \( T \).
Other exercises in this chapter
Problem 34
(a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the diffe
View solution Problem 34
A regulation football used in the National Football League (NFL) is 11 in. from tip to tip and 7 in. in diameter at its thickest (the regulations allow for slig
View solution Problem 34
Find the accumulated present value of an investment for which there is a perpetual continuous money flow of \(\$ 3500\) per year at an interest rate of \(4 \%,\
View solution Problem 35
A movie stuntman who receives an annual salary of \(\$ 180,000\) per year is injured and can no longer work. Through a settlement with an insurance company, he
View solution