/*! 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 129 Let \(X\) denote the time to fai... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) denote the time to failure (in years) of a hydraulic component. Suppose the pdf of \(X\) is \(f(x)=32 /(x+4)^{3}\) for \(x>0\). a. Verify that \(f(x)\) is a legitimate pdf. b. Determine the cdf. c. Use the result of part (b) to calculate the probability that time to failure is between 2 and 5 years. d. What is the expected time to failure? e. If the component has a salvage value equal to \(100 /(4+x)\) when its time to failure is \(x\), what is the expected salvage value?

Short Answer

Expert verified
a. Legitimate pdf b. CDF: \( 1 - \frac{16}{(x+4)^2} \) c. Probability: \( \frac{20}{81} \) d. Expected time: 12 years e. Expected salvage: 16.67

Step by step solution

01

Verify Legitimacy of PDF

To verify that a function is a legitimate probability density function (pdf), we need to check two things: (1) The function must be non-negative for all permissible values of the random variable. In this case, since the function is defined for \(x > 0\), we see that \( (x+4)^3 > 0 \), making \( f(x) = \frac{32}{(x+4)^3} > 0 \). (2) The integral of the function over the domain must equal 1. For this function:\[\int_{0}^{\infty} \frac{32}{(x+4)^3} \, dx = 1\]To solve this, we perform the substitution \( u = x + 4 \), which transforms the limits of integration to \( u = 4 \) and \( u = \infty \). The integral becomes:\[\int_{4}^{\infty} \frac{32}{u^3} \, du = 32 \left[ -\frac{1}{2u^2} \right]_{4}^{\infty} = 32 \left( 0 - \left(-\frac{1}{2(4)^2}\right) \right) = 32 \times \frac{1}{32} = 1\]Thus, \(f(x)\) is a legitimate pdf.
02

Determine the CDF

The cumulative distribution function (CDF), \( F(x) \), is the integral of the pdf from 0 to \( x \):\[ F(x) = \int_{0}^{x} \frac{32}{(t+4)^3} \, dt \]Substituting \( t+4 = u \), so \( dt = du \), and modifying the limits gives:\[ F(x) = 32 \int_{4}^{x+4} \frac{1}{u^3} \, du = 32 \left[ -\frac{1}{2u^2} \right]_{4}^{x+4} \]Solving the integral gives:\[ F(x) = 32 \left( -\frac{1}{2(x+4)^2} + \frac{1}{2 \times 16} \right) = 1 - \frac{32}{2(x+4)^2} \]After further simplification, we have:\[ F(x) = 1 - \frac{16}{(x+4)^2} \]
03

Calculate Probability of Failure Between 2 and 5 Years

Using the CDF derived in the previous step, calculate the probability that the time to failure is between 2 and 5 years:\[ P(2 \leq X \leq 5) = F(5) - F(2) \]Calculate \( F(5) \):\[ F(5) = 1 - \frac{16}{(5+4)^2} = 1 - \frac{16}{81} = \frac{65}{81} \]Calculate \( F(2) \):\[ F(2) = 1 - \frac{16}{(2+4)^2} = 1 - \frac{16}{36} = \frac{20}{36} = \frac{5}{9} \]Finally, compute the probability:\[ P(2 \leq X \leq 5) = \frac{65}{81} - \frac{5}{9} = \frac{65}{81} - \frac{45}{81} = \frac{20}{81} \]
04

Calculate Expected Time to Failure

The expected value, \( E(X) \), is the integral of \( x \cdot f(x) \) over the domain. Calculate:\[ E(X) = \int_{0}^{\infty} x \cdot \frac{32}{(x+4)^3} \, dx \]By using substitution, \( u = x + 4 \), \( x = u - 4 \), \( dx = du \), the limits change from \( 4 \) to \( \infty \):\[ E(X) = 32 \int_{4}^{\infty} \frac{u-4}{u^3} \, du = 32 \left( \int_{4}^{\infty} \frac{1}{u^2} \, du - \int_{4}^{\infty} \frac{4}{u^3} \, du \right) \]Calculate separately and add:\[ = 32 \left( \left[ -\frac{1}{u} \right]_4^{\infty} + 4\left[ -\frac{1}{2u^2} \right]_4^{\infty} \right) \]Simplifying yields:\[ = 32 \left( 0 + \frac{1}{4} + 4(0 + \frac{1}{32}) \right) = 8 + 4 = 12\]Thus, the expected time to failure is 12 years.
05

Calculate Expected Salvage Value

The expected salvage value is given by \( E\left(\frac{100}{4+X}\right) \). Compute:\[ E\left(\frac{100}{4+X}\right) = \int_{0}^{\infty} \frac{100}{4+x} \cdot \frac{32}{(x+4)^3} \, dx \]Using substitution \( u = x+4 \), the integral simplifies equivalently to:\[ = 3200 \int_{4}^{\infty} \frac{1}{u^4} \, du = 3200 \left[-\frac{1}{3u^3}\right]_4^{\infty} \]Evaluating gives:\[ = 3200 \left(0 + \frac{1}{192} \right) = \frac{3200}{192} \approx 16.67\]Therefore, the expected salvage value is approximately 16.67.

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Key Concepts

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

Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is an essential concept in probability theory. It provides the probability that a random variable will take a value less than or equal to a certain number. For a continuous random variable like the one presented in the exercise, the CDF is obtained by integrating the Probability Density Function (PDF) from the lower bound up to the point of interest.
In this problem, the PDF is given as \(f(x) = \frac{32}{(x+4)^3}\). To find the CDF \(F(x)\), we calculate:
  • The definite integral from 0 to \(x\) of the PDF.
  • Use substitution to make the integration more manageable.
After solving, the CDF is \(F(x) = 1 - \frac{16}{(x+4)^2}\).
This function allows us to calculate the probability of failure of the hydraulic component within any interval by subtracting the CDF values at the interval's boundaries. For instance, to find the probability that failure occurs between 2 and 5 years, subtract \(F(2)\) from \(F(5)\). This computation is straightforward once the CDF is known.
Expected Value
The Expected Value (or expectation) is a fundamental concept that provides the long-term average of random variable outcomes. It is especially significant when evaluating the average time to failure for a component in reliability studies, such as in this exercise.

For a continuous random variable, the expected value \(E(X)\) is calculated using the integral of \(x \cdot f(x)\) over the entire range of \(x\):
  • Given the PDF, \(f(x) = \frac{32}{(x+4)^3}\), we integrate \(x \cdot f(x)\) from 0 to \( \infty \).
  • Use substitution to simplify the integration process.
After performing the integration, the expected time to failure for the hydraulic component is found to be 12 years.
This expected value can guide maintenance schedules and inventory management for replacement components in an engineering context.
Integration
Integration is a mathematical technique used to calculate the area under a curve, which in probability theory, gives us meaningful values like probabilities and expected values. It is heavily used in this exercise to verify the PDF, compute the CDF, and determine expected values.
When verifying a PDF, we check:
  • The integral of the PDF over its domain equals 1, ensuring the function is a valid probability distribution.
  • This exercise involved computing \(\int_{0}^{\infty} \frac{32}{(x+4)^3} \, dx = 1\) to confirm the legitimacy of PDF.
For CDF calculations, integrate from 0 to \(x\). And for expected value, integrate the product of \(x\) and the PDF.
Integration requires methods such as substitution to simplify solving integrals, as seen in this problem where \(u = x+4\) is used. Mastering integration techniques is crucial in solving many problems in probability theory.
Probability Theory
Probability Theory provides the foundation for understanding and modeling randomness and uncertainty. This exercise involves several key components of probability theory, including probability distributions, expected values, and the interpretation of PDFs and CDFs.
Here:
  • The Probability Density Function (PDF) \(f(x)\) is used to describe the distribution of a continuous random variable, indicating how likely different outcomes are.
  • The Cumulative Distribution Function (CDF) then helps derive probabilities over intervals.
  • Expected value calculations provide insights into the average behavior of the random variable, essential for planning and decision-making processes.
Understanding these concepts helps in modeling scenarios like the time to failure for components, crucial in fields like engineering and risk management. Probability theory equips learners with tools to approach and solve real-world problems involving uncertainty systematically.

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