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Chapter 4: Problem 73

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, 1985: 519-522) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens subjected to \(\mathrm{AC}\) voltage. The values of the parameters depend on the voltage and temperature; suppose \(\alpha=2.5\) and \(\beta=200\) (values suggested by data in the article). a. What is the probability that a specimen's lifetime is at most 250? Less than 250? More than 300 ? b. What is the probability that a specimen's lifetime is between 100 and 250 ? c. What value is such that exactly \(50 \%\) of all specimens have lifetimes exceeding that value?

Short Answer

Expert verified
a. 0.79, 0.79, 0.11; b. 0.69; c. 170.26 hours.

Step by step solution

01

Define the Weibull distribution

The Weibull distribution is given by the cumulative distribution function (CDF): \( F(t) = 1 - e^{-(t/\beta)^\alpha} \), where \( \alpha \) is the shape parameter, and \( \beta \) is the scale parameter. Here, \(\alpha = 2.5\) and \(\beta = 200\).
02

Calculate probability for lifetime at most 250

Using the CDF, compute \( F(250) = 1 - e^{-(250/200)^{2.5}} \). Plug in the values: \( F(250) = 1 - e^{-1.5625} \approx 0.79 \). Thus, the probability that a specimen's lifetime is at most 250 is approximately 0.79.
03

Calculate probability for lifetime less than 250

Since "at most 250" is the same as "less than or equal to 250," the result is the same: \( P(t < 250) \approx 0.79 \).
04

Calculate probability for lifetime more than 300

The probability that lifetime is more than 300 is \( 1 - F(300) \). Compute \( F(300) = 1 - e^{-(300/200)^{2.5}} \approx 0.89 \), hence \( P(t > 300) = 1 - 0.89 = 0.11 \).
05

Calculate probability for lifetime between 100 and 250

Find \( F(100) \) and \( F(250) \). Compute \( F(100) = 1 - e^{-(100/200)^{2.5}} \approx 0.10 \). Thus, \( P(100 \leq t \leq 250) = F(250) - F(100) = 0.79 - 0.10 = 0.69 \).
06

Find the median lifetime

To find the median lifetime, solve for \( t \) such that \( F(t) = 0.5 \). This implies \( 0.5 = 1 - e^{-(t/200)^{2.5}} \). Solving, we have \( e^{-(t/200)^{2.5}} = 0.5 \). Taking natural logs, \( -(t/200)^{2.5} = \ln(0.5) \), thus \( (t/200)^{2.5} = -\ln(0.5) \approx 0.6931 \). Finally, \( t = 200(-\ln(0.5))^{0.4} \approx 170.26 \). Hence, the median lifetime is approximately 170.26.

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Key Concepts

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

Statistical Modeling
Statistical modeling is a powerful tool that helps in understanding and predicting behaviors and outcomes by representing them mathematically using probabilistic structures. In the realm of engineering and reliability testing, statistical models like the Weibull distribution are especially useful.
They allow engineers to predict the life expectancy of materials under varying conditions. This understanding is crucial when planning for maintenance and anticipating failures.
The Weibull distribution, specifically, is chosen often because it can model a variety of life behaviors, from increasing failure rates (like that of a light bulb) to constant failure rates, and even decreasing rates, depending on its parameters.
Parameters like the shape parameter (\( \alpha \)) determine how the model represents aging and stress characteristics, making it adaptable to many different types of data.
Failure Analysis
Failure analysis using statistical models allows us to predict how and when materials might fail based on observed data. The Weibull distribution is a common choice in the study of failures because it provides a mathematical way to model the distribution of material lifetimes.
The authors in the exercise analyze the failure of solid insulating materials under electrical stress, showing how factors such as voltage and temperature affect material longevity.
By determining the shape and scale parameters of the Weibull distribution, engineers can establish critical insights into when failures are likely to occur, helping them to design safer, more reliable systems.
For example, using the distribution to find the probability that a specimen's lifetime is more than 300 hours, we calculate the cumulative distribution function (CDF) for 300 and subtract it from 1, revealing that only 11% will last this long.
Probability Calculation
Probability calculation within the Weibull distribution context allows us to determine the likelihood of different outcomes, such as a material failing within a particular time frame.
By using the Weibull CDF formula, \( F(t) = 1 - e^{-(t/\beta)^\alpha} \), we can find probabilities for specific time intervals. For instance, we calculate that the probability of a specimen lasting at most 250 hours is approximately 79%.
This calculation helps in design decisions and predicting future maintenance needs.
  • This foresight can prevent unexpected outages and optimize resource allocation effectively.
  • With precise calculations, maintenance schedules can be perfectly timed, reducing costs and improving reliability.
Median Lifetime
The median lifetime is a critical figure in failure analysis, indicating the time by which half of the products will have failed. In the Weibull distribution, the median is found by setting the CDF to 0.5 and solving for time \( t \).
From the example exercise, we set \( 0.5 = 1 - e^{-(t/200)^{2.5}} \) and solve for \( t \), yielding a median lifetime of approximately 170.26 hours.
This value is essential for reliability tests, as it provides a baseline for expected product performance.
Understanding the median helps manufacturers guarantee product longevity and also allows customers to have realistic expectations of product lifespan.
  • It aids in warranty calculations and helps in comparing the reliability of different products and materials.
  • Additionally, knowing the median helps companies optimize product replacements and improve designs.

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