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Let \(X\) denote the time to failure (in years) of a certain 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. Yes, f(x) is legitimate. b. Cdf: \( F(x) = 1 - \frac{16}{(x+4)^2} \). c. Probability: \( \frac{20}{81} \). d. Expected time: 4 years. e. Expected salvage: 8.

Step by step solution

01

Check if f(x) is a legitimate pdf

To verify that \( f(x) \) is a legitimate probability density function (pdf), we must ensure that the integral of \( f(x) \) over all possible values of \( x \) equals 1. This involves calculating: \[ \int_{0}^{\infty} \frac{32}{(x+4)^3} \, dx \] Utilizing substitution, let \( u = x + 4 \), then \( du = dx \), and the limits change from \( x = 0 \) to \( u = 4 \) and \( x = \infty \) to \( u = \infty \). Thus, the integral becomes: \[ \int_{4}^{\infty} \frac{32}{u^3} \, du \] Compute the integral: \[ = 32 \left[ -\frac{1}{2u^2} \right]_{4}^{\infty} = 32 \left( 0 + \frac{1}{32} \right) = 1 \] Thus, \( f(x) \) is a legitimate pdf.
02

Determine the cdf

The cumulative distribution function (cdf) \( F(x) \) is found by integrating the pdf from 0 to \( x \): \[ F(x) = \int_{0}^{x} \frac{32}{(t+4)^3} \, dt \] Use substitution as before where \( u = t + 4 \), so \( du = dt \), giving: \[ = 32 \int_{4}^{x+4} \frac{1}{u^3} \, du = 32 \left[ -\frac{1}{2u^2} \right]_{4}^{x+4} \] Calculate: \[ = 32 \left( -\frac{1}{2(x+4)^2} + \frac{1}{32} \right) \] Simplify to get: \[ F(x) = 1 - \frac{16}{(x+4)^2} \] This is the cdf for the pdf given.
03

Calculate P(2 < X < 5)

Use the cdf to find the probability that \( X \) is between 2 and 5 years: \[ P(2 < X < 5) = F(5) - F(2) \] Using the cdf: \[ F(5) = 1 - \frac{16}{(5+4)^2} = 1 - \frac{16}{81} = \frac{65}{81} \] \[ F(2) = 1 - \frac{16}{(2+4)^2} = 1 - \frac{16}{36} = \frac{20}{36} = \frac{5}{9} \] Thus, \[ P(2 < X < 5) = \frac{65}{81} - \frac{5}{9} = \frac{20}{81} \]
04

Find the expected time to failure

The expected value (mean) \( E(X) \) is given by \( E(X) = \int_{0}^{\infty} x f(x) \, dx \): \[ E(X) = \int_{0}^{\infty} x \frac{32}{(x+4)^3} \, dx \] Using integration by parts where \( u = x \) and \( dv = \frac{32}{(x+4)^3} \, dx \), compute the integral. After computing: \[ E(X) = 4 \] Thus, the expected time to failure is 4 years.
05

Find the expected salvage value

The expected salvage value \( E(S) \) is calculated as: \[ E(S) = \int_{0}^{\infty} \frac{100}{4 + x} f(x) \, dx \] \[ = \int_{0}^{\infty} \frac{100}{4 + x} \cdot \frac{32}{(x+4)^3} \, dx \] Evaluate the integral through substitution or using tables. After calculation: \[ E(S) = 8 \] Therefore, the expected salvage value is 8.

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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 a key concept in probability, indicating the probability that a random variable is less than or equal to a certain value. For a given probability density function (pdf), the cdf can be found by integrating the pdf from the lower bound, often zero, to a variable upper limit. This integration effectively sums the probabilities for all outcomes less than or equal to the specified value, giving a comprehensive view of the distribution up to that point.

In the given exercise, the pdf of the time to failure, given by \( f(x) = \frac{32}{(x+4)^3} \), serves as the basis for deriving the cdf. The integration process transforms this pdf into the cdf, \( F(x) = 1 - \frac{16}{(x+4)^2} \), showing the accumulation of probabilities for \( X \), the time to failure, up to a point \( x \). This transformation allows us to easily compute the probability of failure within specific time intervals.
Expected Value
Expected Value, often represented as \( E(X) \), is an essential statistical measure that provides the average or mean value of a random variable over numerous experiments or trials. It's calculated by integrating the variable's value, weighted by its probability density, over all possible outcomes.

In the context of the exercise, the expected time to failure is calculated as \( E(X) = \int_{0}^{\infty} x \frac{32}{(x+4)^3} \, dx \). Through integration by parts methods, this integral evaluates to 4, indicating that, on average, the hydraulic component will fail after 4 years. Understanding \( E(X) \) is vital, as it provides predictive insights about the system's behavior, informing operational expectations and maintenance scheduling.
Integration by Parts
Integration by parts is a mathematical technique often used to solve integrals involving products of functions. It transforms the integral of a product of functions into simpler parts, which are easier to evaluate. The method relies on the formula: \[ \int u \, dv = uv - \int v \, du \], where \( u \) and \( dv \) are chosen parts of the integrand.

For the expected value in the exercise, integration by parts is used where \( u = x \) and \( dv = \frac{32}{(x+4)^3} \, dx \). This selection simplifies the integration process, allowing us to derive the mean time to failure, which is crucial for understanding the performance and reliability of the component. Mastery of techniques like integration by parts is invaluable for solving complex probability integrals efficiently.
Probability Calculation
Probability calculation within certain intervals is an important aspect of understanding distribution functions. The probability that a random event falls within a specific range can be efficiently calculated using the cdf. In this exercise, the probability that the time to failure falls between 2 and 5 years is calculated using the derived cdf: \( P(2 < X < 5) = F(5) - F(2) \).

By substituting into the formula, we obtain \( F(5) = \frac{65}{81} \) and \( F(2) = \frac{5}{9} \), leading to the calculated probability of \( \frac{20}{81} \). Understanding this process helps in evaluating risk and planning for potential failures, providing valuable insights for decision-making in maintenance and reliability engineering.

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Most popular questions from this chapter

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