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 \(A_{l}=\\{i\) th component lasts at least \(t\) hours \(\\}, i=1, \ldots, 5\), so that the \(A s\) 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 \(A_{f}^{\text {'s, }}\), 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

Understand the Equivalent Event

The event \(\{X \geq t\}\) happens if all components last at least \(t\) hours. This means every event \(A_i = \{i\text{th component lasts at least } t\}\) must occur simultaneously for all components, i.e., \(X \geq t\) is equivalent to \(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5\).

02

Calculate the Probability \(P(X \geq t)\)

Since the events \(A_i\) are independent, \(P(X \geq t) = P(A_1) \cdot P(A_2) \cdot P(A_3) \cdot P(A_4) \cdot P(A_5)\). For each component, the probability that it lasts at least \(t\) hours is given by the exponential distribution formula: \(P(A_i) = e^{-\lambda t}\). Therefore, \(P(X \geq t) = (e^{-\lambda t})^5 = e^{-5\lambda t}\).

03

Compute \(F(t)\) and Identify the Distribution of \(X\)

The cumulative distribution function (CDF) of \(X\) is given by \(F(t) = P(X \leq t) = 1 - P(X \geq t) = 1 - e^{-5\lambda t}\). The probability density function (pdf) is obtained by differentiating the CDF: \(f(t) = \frac{d}{dt}[1 - e^{-5\lambda t}] = 5\lambda e^{-5\lambda t}\). Therefore, \(X\) has an exponential distribution with parameter \(5\lambda\).

04

Generalize to \(n\) Components

When there are \(n\) components, each having an exponential lifetime with parameter \(\lambda\), \(X\) represents the minimum lifetime, so \(P(X \geq t) = (e^{-\lambda t})^n = e^{-n\lambda t}\). 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.

Independent Events

In probability and statistics, independent events are events where the occurrence of one event does not affect the occurrence of another. This means that the probability of their intersection is simply the product of their individual probabilities.

For the problem at hand, each of the five components in the system fails independently of the others. This is due to each component having an exponentially distributed lifetime, which inherently defines them as independent events. If we denote by \(A_i\) the event that the \(i\)-th component lasts at least \(t\) hours, then the independence implies:

• \( P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5) = P(A_1) \cdot P(A_2) \cdot P(A_3) \cdot P(A_4) \cdot P(A_5) \)

Understanding independence is crucial when solving for the probability of the system lasting at least \(t\) hours, as it allows us to multiply individual probabilities.

Minimum Lifetime

The minimum lifetime in the context of this system refers to the shortest time one of the components lasts before failing, which directly affects when the entire system fails.

Since the system fails when the first component fails, the variable \(X\) is defined as the minimum lifetime among all five components. Mathematically, this is expressed as the smallest value among the lifetimes of the five components.

• Thus, \( X = \min(T_1, T_2, T_3, T_4, T_5) \) where \(T_i\) is the lifetime of the \(i\)-th component.

The concept of the minimum is central to determining the likelihood of the system's time to failure (i.e., \( P(X \geq t) \)). This is because if \(X \geq t\), all components must last at least \(t\), highlighting the intersection of events.

Probability Density Function

A Probability Density Function (PDF) describes the likelihood of a random variable taking on a particular value. In continuous probability distributions, the PDF is a function that shows the relative likelihood for a random variable to equal a specific value.

For an exponential distribution, the PDF captures how probabilities are distributed over time for the component lifetimes. With the exponential lifetime distribution parameter \( \lambda = 0.01 \), the individual component lifetime PDF is:

• \( f(t) = \lambda e^{-\lambda t} = 0.01 e^{-0.01 t} \)

The probability density function for the system, accounting for all five components, becomes \( 5\lambda e^{-5\lambda t} \) after differentiation of the cumulative distribution function, reflecting the system's minimum component lifetime.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a crucial concept in probability, representing the probability that a random variable takes on a value less than or equal to a specific point. It's a function denoted by \(F(t)\), indicating the cumulative probability up to time \(t\).

In this problem, we look at the CDF for the system's failure time. It's calculated as \(F(t) = P(X \leq t)\), which is equal to \( 1 - P(X \geq t) \). For the system comprised of the five components, we find:

• \( F(t) = 1 - e^{-5\lambda t} = 1 - e^{-0.05t} \)

This CDF indicates that the probability increases as time \(t\) progresses to capture scenarios where system failure is likely, by capturing all those instances when at least one component fails before time \(t\).