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The life (in hours) of a magnetic resonance imaging machine (MRI) is modeled by a Weibull distribution with parameters \(\beta=2\) and \(\delta=500\) hours. (a) Determine the mean life of the MRI. (b) Determine the variance of the life of the MRI. (c) What is the probability that the MRI fails before 250 hours?

Short Answer

Expert verified
(a) Mean: 443.1 hours. (b) Variance: 53675 hours squared. (c) Probability of failure before 250 hours: 22.12%.

Step by step solution

01

Identify the Mean Formula

The mean of a Weibull distribution is given by \( \delta \cdot \Gamma(1 + 1/\beta) \). To find the mean life, we need to substitute the provided parameters \( \delta = 500 \) and \( \beta = 2 \) into the formula.
02

Compute the Mean

Substitute \( \delta = 500 \) and \( \beta = 2 \) into the mean formula: \( 500 \cdot \Gamma(1 + 1/2) = 500 \cdot \Gamma(1.5) \). The gamma function \( \Gamma(1.5) = \sqrt{\pi}/2 \approx 0.8862 \). Thus, the mean is \( 500 \times 0.8862 \approx 443.1 \) hours.
03

Identify the Variance Formula

The variance of a Weibull distribution is \( \delta^2 \cdot \left( \Gamma(1 + 2/\beta) - [\Gamma(1 + 1/\beta)]^2 \right) \). We will use \( \delta = 500 \) and \( \beta = 2 \) to find the variance.
04

Compute the Variance

For \( \beta = 2 \), we need \( \Gamma(1 + 2/2) = \Gamma(2) = 1 \). Previously, \( \Gamma(1.5) = 0.8862 \). Substitute these into the variance formula: \( 500^2 \times (1 - 0.8862^2) \). Calculate \( 0.8862^2 \approx 0.7853 \). So variance \( = 250000 \times (1 - 0.7853) \approx 250000 \times 0.2147 = 53675 \) hours squared.
05

Establish the Probability Formula

The cumulative distribution function (CDF) of the Weibull distribution gives the probability that a random variable \( X \) fails before \( t \) hours. It's given by \( 1 - e^{-(t/\delta)^\beta} \). Substitute \( t = 250 \), \( \delta = 500 \), and \( \beta = 2 \).
06

Calculate the Probability

Substitute into the CDF formula: \( 1 - e^{-(250/500)^2} \). Simplify: \( 1 - e^{-0.5^2} = 1 - e^{-0.25} \). Calculate \( e^{-0.25} \approx 0.7788 \). Thus, the probability is \( 1 - 0.7788 = 0.2212 \) or 22.12%.

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Key Concepts

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

Mean Computation
The mean of a Weibull distribution provides insight into the expected lifespan of an MRI machine. To compute the mean, we use the formula: \( \delta \cdot \Gamma(1 + 1/\beta) \).
For this exercise, \( \delta = 500 \) and \( \beta = 2 \). By substituting these values, we have: \( 500 \cdot \Gamma(1 + 1/2) = 500 \cdot \Gamma(1.5) \).
The gamma function \( \Gamma(1.5) \) is approximately equal to \( \sqrt{\pi}/2 \), which is about \( 0.8862 \).
Therefore, the computation becomes \( 500 \times 0.8862 \approx 443.1 \) hours. This denotes the average time before the MRI is likely to require maintenance or begin to fail.
Variance Calculation
Variance provides a measure of the variability or spread in the lifespan estimates of the MRI machine. It is calculated using the formula: \( \delta^2 \cdot \left( \Gamma(1 + 2/\beta) - [\Gamma(1 + 1/\beta)]^2 \right) \).
Given that \( \delta = 500 \) and \( \beta = 2 \), we need \( \Gamma(2) \) and the square of \( \Gamma(1.5) \).
- \( \Gamma(2) = 1 \)- \( \Gamma(1.5) \approx 0.8862 \)Squaring \( \Gamma(1.5) \) gives \( 0.8862^2 \approx 0.7853 \).
Substitute these into the variance formula: \( 500^2 \times (1 - 0.7853) \approx 250000 \times 0.2147 \).
The variance calculates to \( 53675 \) hours squared, reflecting how spread out the potential failure times are around the mean.
Probability Assessment
Evaluating the probability of early failure helps in risk assessment and management of the MRI machine. The Weibull distribution's cumulative distribution function (CDF) enables us to calculate the likelihood that the MRI will fail before a specified time.
The CDF is expressed as \( 1 - e^{-(t/\delta)^\beta} \). To find the probability of failure before 250 hours with \( \delta = 500 \) and \( \beta = 2 \), we substitute: \( t = 250 \).
This simplifies to \( 1 - e^{-(250/500)^2} = 1 - e^{-0.25} \).
Calculating \( e^{-0.25} \approx 0.7788 \), the probability becomes \( 1 - 0.7788 = 0.2212 \), or 22.12%.
This indicates that there is a 22.12% chance the MRI will experience an issue before reaching 250 hours of operation.
Failure Analysis
Failure analysis helps to understand the reliability of an MRI machine and its components over time. The Weibull distribution is a popular choice for modeling such lifespans owing to its flexibility in characterizing various failure rates.
With parameters \( \beta = 2 \) and \( \delta = 500 \), this distribution is particularly suited for products that exhibit a "wear-out" stage, meaning the failure rate increases over time.
Key aspects of failure analysis include:
  • Determining the Mean Time to Failure (MTTF), which is effectively the mean of the distribution calculated here as 443.1 hours.
  • Understanding Variance, which indicates how widely lifespans are spread around the mean. A variance of 53675 hours squared shows potential variation within the lifespan expectations.
  • Assessing Probabilities of failure at different intervals to plan maintenance schedules effectively. For instance, knowing a higher probability of failure before 250 hours can help in planning early check-ups.
Using these insights, stakeholders can design better maintenance schedules, manage inventories, and ensure higher reliability of equipment.

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