/*! 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 15 Consider a system consisting of ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component 2 or component 3 functions. Let \(X_{1}, X_{2}\), and \(X_{3}\) denote the lifetimes of components 1,2 , and 3 , respectively. Suppose the \(X_{i}\) 's are independent of each other and each \(X_{i}\) has an exponential distribution with parameter \(\lambda\). a. Let \(Y\) denote the system lifetime. Obtain the cumulative distribution function of \(Y\) and differentiate to obtain the pdf. [Hint: \(F(y)=P(Y \leq y) ;\) express the event \(\\{Y \leq y\\}\) in terms of unions and/or intersections of the three events \(\left\\{X_{1} \leq y\right\\},\left\\{X_{2} \leq y\right\\}\), and \(\left.\left\\{X_{3} \leq y\right\\} .\right]\) b. Compute the expected system lifetime.

Short Answer

Expert verified
(a) CDF: \(1 - e^{-\lambda y}\); PDF: \(\lambda e^{-\lambda y}\). (b) Expected lifetime: \(\frac{1}{\lambda}\).

Step by step solution

01

Define the System Lifetime Event

For the system to fail, either Component 1 fails or both Components 2 and 3 must fail. Therefore, the event that the system fails within time \(y\) is given by the union of the events \(\{X_1 \leq y\}\) and \(\{X_2 > y\} \cap \{X_3 > y\}\). We have \(F(y) = P\left(X_1 \leq y \text{ or } (X_2 > y \text{ and } X_3 > y)\right)\).
02

Separate the Events

Using probability properties for independent events, we have to express \(F(y)\) in terms of individual probabilities: \(F(y) = P(X_1 \leq y) + P(X_2 > y)P(X_3 > y) - P(X_1 \leq y)P(X_2 > y)P(X_3 > y)\).
03

Calculate the Complementary Probabilities

Since each \(X_i\) is exponentially distributed with parameter \(\lambda\), the probability that \(X_i\) is greater than \(y\) is given by \(1 - P(X_i \leq y) = e^{-\lambda y}\). Therefore, \(P(X_2 > y) = e^{-\lambda y}\) and \(P(X_3 > y) = e^{-\lambda y}\).
04

Substitute the Probabilities in CDF Expression

The probability \(P(X_1 \leq y) = 1 - e^{-\lambda y}\). Hence, \(F(y) = (1 - e^{-\lambda y}) + e^{-\lambda y} e^{-\lambda y} - (1 - e^{-\lambda y}) e^{-\lambda y} e^{-\lambda y}\).
05

Simplify the CDF Expression

The simplified CDF becomes \(F(y) = 1 - e^{-\lambda y} e^{-\lambda y} = 1 - e^{-\lambda y}\). This is the cumulative distribution function of \(Y\).
06

Differentiate CDF to Find PDF

The probability density function (pdf) is the derivative of the CDF: \(f(y) = \frac{d}{dy}(1 - e^{-\lambda y}) = \lambda e^{-\lambda y}\).
07

Calculate the Expected Lifetime

The expected value of an exponential distribution with parameter \(\lambda\) is \(\frac{1}{\lambda}\). Since the system lifetime \(Y\) is also exponentially distributed with parameter \(\lambda\), the expected system lifetime is \(E(Y) = \frac{1}{\lambda}\).

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Key Concepts

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

Exponential Distribution
Let's talk about the exponential distribution, a fundamental concept in probability theory. This type of distribution is often used to model the time until an event occurs, such as the lifetime of a component in a system. In our case, each component's lifetime is modeled by an exponential distribution with a parameter, commonly denoted as \( \lambda \).
This parameter \( \lambda \) represents the rate at which the events occur.
  • If \( \lambda \) is large, events happen quickly.
  • If \( \lambda \) is small, events occur more slowly.
The probability density function (PDF) for an exponential distribution is given by \[ f(x) = \lambda e^{-\lambda x}\quad \text{for } x \geq 0.\] The exponential distribution is memoryless, meaning the probability of an event occurring in the next interval is independent of the past. This property significantly simplifies calculations involving independent events, as we'll see with the system discussed here.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a crucial tool to express the likelihood of a random variable being less than or equal to a certain value. When it comes to our system's lifetime, it helps us find the probability that the system fails within a given time \( y \).
For the system to fail, Component 1 must fail, or both Components 2 and 3 must fail. Given the independent nature of each component's lifetime, our CDF can be expressed as:
\[ F(y) = P(X_1 \leq y) + P(X_2 > y)P(X_3 > y) - P(X_1 \leq y)P(X_2 > y)P(X_3 > y) \]This formula combines different possible failure events. More specifically, with exponentially distributed lifetimes, the CDF simply becomes:
\[ F(y) = 1 - e^{-\lambda y} \]This expression is used to ascertain the probability of the system failing within the time \( y \), giving us a very handy result for further analysis, such as the PDF.
Probability Density Function
The probability density function (PDF) shows us the likelihood of the random variable taking on a specific value. It is the derivative of the CDF, which offers a deeper insight into the distribution of probabilities over an interval.
In terms of our system's lifetime, the PDF describes how the probabilities are distributed over time that a particular component will fail.
To find the PDF of the system's lifetime, take the derivative of the CDF we obtained:\[f(y) = \frac{d}{dy}(1 - e^{-\lambda y}) = \lambda e^{-\lambda y}\] This PDF indicates the likelihood that our system will last exactly a certain amount of time, with the exponential form reflecting the rapidity of failure events following a random, memoryless process.
Independent Events
In probability, independent events are those whose occurrences do not affect each other. This concept is essential when dealing with systems with multiple components, like ours.
Each component's lifetime, denoted \(X_1, X_2,\) and \(X_3\), are independent events. Thus, the probability of one component failing is unaffected by the status of the other components.
This independence simplifies calculations significantly, particularly with exponential distributions. For example:
  • The probability of two independent events both occurring is the product of their individual probabilities.
  • This property allows us to straightforwardly compute complex joint probability scenarios, simplifying our overall system lifetime calculations.
Understanding such independence is crucial in establishing reliable models for system operation and failure, especially for real-world applications.

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Most popular questions from this chapter

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let \(X=\) the number of points earned on the first part and \(Y=\) the number of points earned on the second part. Suppose that the joint pmf of \(X\) and \(Y\) is given in the accompanying table. $$ \begin{array}{ll|cccc} & &{y} \\ p(x, y) & 0 & 5 & 10 & 15 \\ \hline{x} & 0 & .02 & .06 & .02 & .10 \\ & 5 & .04 & .15 & .20 & .10 \\ & 10 & .01 & .15 & .14 & .01 \end{array} $$ a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score \(E(X+Y)\) ? b. If the maximum of the two scores is recorded, what is the expected recorded score?

Suppose the amount of rainfall in one region during a particular month has an exponential distribution with mean value 3 in., the amount of rainfall in a second region during that same month has an exponential distribution with mean value 2 in., and the two amounts are independent of each other. What is the probability that the second region gets more rainfall during this month than does the first region?

Let \(X_{1}\) and \(X_{2}\) be independent, each having a standard normal distribution. The pair \(\left(X_{1}, X_{2}\right)\) corresponds to a point in a two-dimensional coordinate system. Consider now changing to polar coordinates via the transformation, $$ \begin{gathered} Y_{1}=X_{1}^{2}+X_{2}^{2} \\ Y_{2}=\left\\{\begin{array}{cl} \arctan \left(\frac{X_{2}}{X_{1}}\right) & X_{1}>0, X_{2} \geq 0 \\ \arctan \left(\frac{X_{2}}{X_{1}}\right)+2 \pi & X_{1}>0, X_{2}<0 \\ \arctan \left(\frac{X_{2}}{X_{1}}\right)+\pi & X_{1}<0 \\ 0 & X_{1}=0 \end{array}\right. \end{gathered} $$ from which \(X_{1}=\sqrt{Y_{1}} \cos \left(Y_{2}\right), X_{2}=\sqrt{Y_{1}} \sin \left(Y_{2}\right)\). Obtain the joint pdf of the new variables and then the marginal distribution of each one. [Note: It would be nice if we could simply let \(Y_{2}=\arctan\) \(\left(X_{2} / X_{1}\right)\), but in order to insure invertibility of the arctan function, it is defined to take on values only between \(-\pi / 2\) and \(\pi / 2\). Our specification of \(Y_{2}\) allows it to assume any value between 0 and \(2 \pi\).]

Each front tire of a vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable \(-X\) for the right tire and \(Y\) for the left tire, with joint pdf $$ f(x, y)=\left\\{\begin{array}{cl} K\left(x^{2}+y^{2}\right) & 20 \leq x \leq 30, \quad 20 \leq y \leq 30 \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the value of \(K\) ? b. What is the probability that both tires are underfilled? c. What is the probability that the difference in air pressure between the two tires is at most 2 psi? d. Determine the (marginal) distribution of air pressure in the right tire alone. e. Are \(X\) and \(Y\) independent rv's?

A market has both an express checkout line and a superexpress checkout line. Let \(X_{1}\) denote the number of customers in line at the express checkout at a particular time of day, and let \(X_{2}\) denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of \(X_{1}\) and \(X_{2}\) is as given in the accompanying table. $$ \begin{array}{cc|cccc} & & & &{x_{2}} & \\ & & 0 & 1 & 2 & 3 \\ \hline & 0 & .08 & .07 & .04 & .00 \\ & 1 & .06 & .15 & .05 & .04 \\ x_{1} & 2 & .05 & .04 & .10 & .06 \\ & 3 & .00 & .03 & .04 & .07 \\ & 4 & .00 & .01 & .05 & .06 \end{array} $$ a. What is \(P\left(X_{1}=1, X_{2}=1\right)\), that is, the probability that there is exactly one customer in each line? b. What is \(P\left(X_{1}=X_{2}\right)\), that is, the probability that the numbers of customers in the two lines are identical? c. Let \(A\) denote the event that there are at least two more customers in one line than in the other line. Express \(A\) in terms of \(X_{1}\) and \(X_{2}\), and calculate the probability of this event. d. What is the probability that the total number of customers in the two lines is exactly four? At least four? e. Determine the marginal pmf of \(X_{1}\), and then calculate the expected number of customers in line at the express checkout. f. Determine the marginal pmf of \(X_{2}\). g. By inspection of the probabilities \(P\left(X_{1}=4\right)\), \(P\left(X_{2}=0\right)\), and \(P\left(X_{1}=4, X_{2}=0\right)\), are \(X_{1}\) and \(X_{2}\) independent random variables? Explain.

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