/*! 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 69 The length of life \(Y\) for fus... [FREE SOLUTION] | 91Ó°ÊÓ

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The length of life \(Y\) for fuses of a certain type is modeled by the exponential distribution, with $$f(y)=\left\\{\begin{array}{ll}(1 / 3) e^{-y / 3}, & y>0 \\\0, & \text { elsewhere }\end{array}\right.$$ (The measurements are in hundreds of hours.) a. If two such fuses have independent lengths of life \(Y_{1}\) and \(Y_{2}\), find the joint probability density function for \(Y_{1}\) and \(Y_{2}\). b. One fuse in part (a) is in a primary system, and the other is in a backup system that comes into use only if the primary system fails. The total effective length of life of the two fuses is then \(Y_{1}+Y_{2} .\) Find \(P\left(Y_{1}+Y_{2} \leq 1\right)\).

Short Answer

Expert verified
The joint PDF is \(\frac{1}{9} e^{-(y_1+y_2)/3}\). The probability is \(\frac{1}{3} - \frac{4}{9} e^{-1/3}\).

Step by step solution

01

Define Joint PDF

Since the fuses have independent lengths of life and each follows an exponential distribution, the joint probability density function (PDF) for \(Y_1\) and \(Y_2\) is the product of their individual PDFs. Given function: \(f(y) = \frac{1}{3} e^{-y/3}\) for \(y > 0\).Thus, the joint PDF is:\[ f_{Y_1,Y_2}(y_1, y_2) = \left(\frac{1}{3} e^{-y_1/3}\right) \left(\frac{1}{3} e^{-y_2/3}\right) = \frac{1}{9} e^{-(y_1 + y_2)/3} \quad \text{for } y_1, y_2 > 0.\]
02

Define the Integral for P(Y_1 + Y_2 ≤ 1)

For the second part, we want to find \(P(Y_1 + Y_2 \leq 1)\). This means we integrate the joint PDF over the region \(y_1 + y_2 \leq 1\) with both \(y_1, y_2 > 0\). This forms a triangular region in the \(y_1\)-\(y_2\) plane.
03

Set the Limits for Integration

Since \(0 < y_1 < 1\) and \(0 < y_2 < 1 - y_1\), we express this as:\[ \int_{0}^{1} \int_{0}^{1-y_1} \frac{1}{9} e^{-(y_1 + y_2)/3} \, dy_2 \, dy_1.\]
04

Integrate with Respect to y_2 First

Integrate the inner integral first with respect to \(y_2\):\[ \int_{0}^{1-y_1} \frac{1}{9} e^{-(y_1 + y_2)/3} \, dy_2.\]The antiderivative is\[-\frac{1}{3} e^{-(y_1 + y_2)/3}\], evaluated from \(y_2 = 0\) to \(y_2 = 1 - y_1\):\[\left[-\frac{1}{3} e^{-(y_1 + y_2)/3}\right]_{0}^{1-y_1} = -\frac{1}{3} e^{-1/3} + \frac{1}{3} e^{-y_1/3}.\]
05

Integrate with Respect to y_1

Now integrate the result with respect to \(y_1\):\[ \int_{0}^{1} \left(-\frac{1}{3} e^{-1/3} + \frac{1}{3} e^{-y_1/3}\right) \, dy_1.\]\[-\frac{1}{3} e^{-1/3} y_1 + 1 - e^{-1/3}\], evaluated from \(y_1 = 0\) to \(y_1 = 1\), yields:\[ -\frac{1}{9}e^{-1/3} + \frac{1}{3}(1 - e^{-1/3}).\]
06

Calculate and Simplify the Result

Simplify:\[ -\frac{1}{9}e^{-1/3} + \frac{1}{3} - \frac{1}{3}e^{-1/3} = \frac{1}{3} - \frac{4}{9}e^{-1/3}.\]Therefore, the probability is equal to:\[ \frac{1}{3} - \frac{4}{9} e^{-1/3}.\]

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Key Concepts

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

Joint Probability Density Function
In probability theory, the Joint Probability Density Function (PDF) is a function used to describe the likelihood of two or more random variables occurring simultaneously. For independent random variables, calculating the joint PDF involves multiplying the individual PDFs. In the case of exponential distributions, like the ones described in the problem, each random variable is described by its own exponential PDF.

To find the Joint PDF of two exponential random variables, say \(Y_1\) and \(Y_2\), one simply takes the product of their individual densities. For our problem, each random variable follows \(f(y) = \frac{1}{3}e^{-y/3}\) for \(y > 0\). Hence, the joint PDF becomes:
  • \(f_{Y_1,Y_2}(y_1, y_2) = \left(\frac{1}{3}e^{-y_1/3}\right) \left(\frac{1}{3}e^{-y_2/3}\right)\).
  • Resulting in \(\frac{1}{9}e^{-(y_1 + y_2)/3}\) for \(y_1, y_2 > 0\).
Understanding joint PDFs is critical, especially when dealing with correlated events in statistics, as they lay the groundwork for calculating probabilities involving multiple random variables.
Independent Random Variables
Random variables are termed "independent" if the occurrence of one provides no information about the occurrence of another. This means that any event related to one variable doesn't affect the likelihood of events related to the other.

In statistical modeling, this assumption simplifies computations significantly. When dealing with independent exponential random variables like in our exercise, their joint distribution directly reflects their individual characteristics without any interaction term. Therefore:
  • The joint PDF as a product \(f_{Y_1,Y_2}(y_1, y_2)\) confirms their independence.
  • Events involving such variables can be calculated separately and combined simply by multiplication.
This independence is key in applications such as reliability analysis. For example, modeling the life span of two separate fuses highlights their behaviors unaffected by each other. Recognizing independent random variables is essential for building accurate probabilistic models.
Integration Limits
When dealing with integration in probability, especially with joint PDFs, properly defining integration limits is crucial. These limits describe the region over which the PDF is integrated to find probabilities.

In our problem, the goal is to find \(P(Y_1 + Y_2 \leq 1)\), which leads to defining a region in the \(y_1\)-\(y_2\) plane to integrate over. Given:
  • The constraint \(Y_1 + Y_2 \leq 1\) forms a right-angled triangle with vertices at \((0,0)\), \((1,0)\), and \((0,1)\).
  • The integration limits become \(0 < y_1 < 1\) and \(0 < y_2 < 1 - y_1\).
Correctly setting these boundaries ensures the calculation accounts for all relevant probability in the specified region. Visualizing these limits often helps, making it easier to handle complex multi-variable integrals that arise in joint distributions. Understanding integration limits is critical for successful application of probability theory in real-world scenarios.

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Most popular questions from this chapter

In Exercise \(5.16, Y_{1}\) and \(Y_{2}\) denoted the proportions of time that employees I and II actually spent working on their assigned tasks during a workday. The joint density of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} y_{1}+y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ Employee I has a higher productivity rating than employee II and a measure of the total productivity of the pair of employees is \(30 Y_{1}+25 Y_{2}\). Find the expected value of this measure of productivity.

Let \(Y_{1}\) and \(Y_{2}\) be jointly distributed random variables with finite variances. a. Show that \(\left[E\left(Y_{1} Y_{2}\right)\right]^{2} \leq E\left(Y_{1}^{2}\right) E\left(Y_{2}^{2}\right) .\) [Hint: Observe that \(E\left[\left(t Y_{1}-Y_{2}\right)^{2}\right] \geq 0\) for any real number t or, equivalently, $$t^{2} E\left(Y_{1}^{2}\right)-2 t E\left(Y_{1} Y_{2}\right)+E\left(Y_{2}^{2}\right) \geq 0$$ This is a quadratic expression of the form \(A t^{2}+B t+C\); and because it is nonnegative, we must have \(B^{2}-4 A C \leq 0 .\) The preceding inequality follows directly.] b. Let \(\rho\) denote the correlation coefficient of \(Y_{1}\) and \(Y_{2} .\) Using the inequality in part (a), show that \(\rho^{2} \leq 1\)

Let \(Y_{1}\) denote the weight (in tons) of a bulk item stocked by a supplier at the beginning of a week and suppose that \(Y_{1}\) has a uniform distribution over the interval \(0 \leq y_{1} \leq 1\). Let \(Y_{2}\) denote the amount (by weight) of this item sold by the supplier during the week and suppose that \(Y_{2}\) has a uniform distribution over the interval \(0 \leq y_{2} \leq y_{1},\) where \(y_{1}\) is a specific value of \(Y_{1}\) a. Find the joint density function for \(Y_{1}\) and \(Y_{2}\) b. If the supplier stocks a half-ton of the item, what is the probability that she sells more than a quarter-ton? c. If it is known that the supplier sold a quarter-ton of the item, what is the probability that she had stocked more than a half-ton?

Suppose that we are to observe two independent random samples: \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denoting a random sample from a normal distribution with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2} ;\) and \(X_{1}, X_{2}, \ldots, X_{m}\) denoting a random sample from another normal distribution with mean \(\mu_{2}\) and variance \(\sigma_{2}^{2} .\) An approximation for \(\mu_{1}-\mu_{2}\) is given by \(\bar{Y}-\bar{X}\), the difference between the sample means. Find \(E(\bar{Y}-\bar{X})\) and \(V(\bar{Y}-\bar{X})\)

A quality control plan calls for randomly selecting three items from the daily production (assumed large) of a certain machine and observing the number of defectives. However, the proportion \(p\) of defectives produced by the machine varies from day to day and is assumed to have a uniform distribution on the interval (0,1) . For a randomly chosen day, find the unconditional probability that exactly two defectives are observed in the sample.

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