91Ó°ÊÓ

A system consists of five identical components connected in series as shown: As soon as one component fails, the entire system will fail. Suppose each component has a lifetime that is exponentially distributed with \(\lambda=.01\) and that components fail independently of one another. Define events \(A_{i}=\\{i\) th component lasts at least \(t\) hours \(\\}, i=1, \ldots, 5\), so that the \(A_{i}\) 's are independent events. Let \(X=\) the time at which the system fails-that is, the shortest (minimum) lifetime among the five components. a. The event \(\\{X \geq t\\}\) is equivalent to what event involving \(A_{1}, \ldots, A_{5}\) ? b. Using the independence of the five \(A_{i}\) 's, compute \(P(X \geq t)\). Then obtain \(F(t)=P(X \leq t)\) and the pdf of \(X\). What type of distribution does \(X\) have? c. Suppose there are \(n\) components, each having exponential lifetime with parameter \(\lambda\). What type of distribution does \(X\) have?

Step by step solution

01

Express the Event in Terms of Component Events

The event \( \{ X \geq t \} \) means that all components last at least \( t \) hours. Therefore, it is equivalent to the event \( A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5 \).

02

Calculate the Probability of the Event

By independence of the events \( A_i \), we have:\[P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5) = P(A_1) \times P(A_2) \times P(A_3) \times P(A_4) \times P(A_5)\]Since each component has an exponentially distributed lifetime, \( P(A_i) = P(\text{component lasts at least } t) = e^{-\lambda t} \) for each \( A_i \). Thus:\[ P(X \geq t) = e^{-5\lambda t} \]

03

Compute the CDF and PDF of X

The cumulative distribution function (CDF) of \( X \) is:\[F(t) = P(X \leq t) = 1 - P(X \geq t) = 1 - e^{-5\lambda t}\]The probability density function (PDF) is obtained by differentiating the CDF:\[f(t) = \frac{d}{dt}F(t) = 5\lambda e^{-5\lambda t}\]

04

Identify the Type of Distribution

The distribution of \( X \) is exponential with parameter \( 5\lambda \), which is an exponential distribution with rate \( 5 \times 0.01 = 0.05 \).

05

Generalize to n Components

For \( n \) components, each with exponential lifetime \( \lambda \), the time \( X \) until the system fails has an exponential distribution with rate \( n\lambda \). Therefore, \( X \) follows an exponential distribution with parameter \( n\lambda \).

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Key Concepts

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

System Reliability

System reliability refers to the probability that a system will perform its intended function without failure for a specified period of time. In the case of a series system, as described in the original exercise, all components must function correctly for the system to be operational. If any component fails, the system's operation is compromised, leading to failure even if only one of these components malfunctions.
For our system with five components, the reliability can be calculated based on the reliability of each individual component. When components are connected in series, the overall reliability is the product of the reliabilities of each component. In simple terms:

• If each component functions with probability 0.99, then for a series of 5 components, the system reliability is \(0.99^5 \approx 0.95\).
• This decreasing reliability demonstrates how each additional component in a series contributes to the potential failure of the system as a whole.

This is why improving the reliability of each individual component or using parallel systems can significantly enhance overall system reliability.

Independent Events

Independent events in probability are those whose occurrence does not affect the probability of occurrence of another event. In the context of our exercise, the lifetime of each component is considered independent of the others. This implies that the failure of one component does not alter the likelihood of another component failing.
Understanding independence is crucial when calculating the probability of combined events. If the events are independent, the probability of all events occurring is the product of their individual probabilities:

• For component events \( A_1, A_2, A_3, A_4, \text{and} A_5 \), \( P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5) = P(A_1) \times P(A_2) \times P(A_3) \times P(A_4) \times P(A_5) \).
• This product rule simplifies the otherwise complex calculation required for dependent events, where one event's occurrence influences another.

Recognizing and correctly identifying independence helps avoid errors in probability analysis involving multiple events.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a fundamental concept in statistics and probability. It represents the probability that a random variable is less than or equal to a certain value. For an exponential distribution like the one in the exercise, the CDF can be derived from the complementary probability of the system remaining operational for at least time \( t \).
The CDF for the system's failure time \( X \), given that each component has an exponential lifetime distribution, is:

• \( F(t) = P(X \leq t) = 1 - e^{-5\lambda t} \).
• This formula captures the entire range of the random variable, showing how the system's failure probability accumulates over time.

The CDF is important because it provides a comprehensive picture of the distribution of system failure times, helping in planning and reliability assessments.

Probability Density Function

The Probability Density Function (PDF) describes the likelihood of a random variable taking on a specific value. For continuous variables, it is the derivative of the Cumulative Distribution Function (CDF).
In our system's scenario, after finding the CDF for time \( X \) until failure, differentiating it provides the PDF:

• \( f(t) = \frac{d}{dt}[1 - e^{-5\lambda t}] = 5\lambda e^{-5\lambda t} \).
• This expression shows at what rate the probability mass of time \( X \) is allocated, illustrating the system's expected failure behavior.

Understanding the PDF is essential for interpreting how likely the system is to fail at different time intervals, critical for maintenance scheduling and risk assessment.