Problem 159 - Applied Statistics and Probability for Engineers

Question

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. Determine the following: (a) Mean life of the MRI (b) Variance of the life of the MRI (c) Probability that the MRI fails before 250 hours.

Step-by-Step Solution

Answer

(a) 443.1 hours, (b) 53675 hours², (c) 22.1%

Step 1: Understand the Weibull Distribution

The Weibull distribution is a continuous probability distribution used to model reliability data and life data. It is defined by two parameters, shape parameter \( \beta \) and scale parameter \( \delta \). The life distribution characteristics can be calculated using these parameters.

Step 2: Mean of Weibull Distribution

The mean or expected value \( E(X) \) of the Weibull distribution is given by the formula: \[ E(X) = \delta \cdot \Gamma\left(1 + \frac{1}{\beta}\right) \] where \( \Gamma \) is the gamma function. For \( \beta = 2 \), the gamma function can be calculated using \( \Gamma(1 + 0.5) = \Gamma(1.5) \approx 0.8862 \). Thus, \[ E(X) = 500 \times 0.8862 \approx 443.1 \text{ hours} \].

Step 3: Variance of Weibull Distribution

The variance \( Var(X) \) is given by the formula: \[ Var(X) = \delta^2 \cdot \left[\Gamma\left(1 + \frac{2}{\beta}\right) - \Gamma\left(1 + \frac{1}{\beta}\right)^2 \right] \] For \( \beta = 2 \), \( \Gamma(1 + 1) = 1 \) and therefore: \[ Var(X) = 500^2 \times (1 - 0.8862^2) = 250000 \times (1 - 0.7853) \approx 53675 \text{ hours}^2 \].

Step 4: Probability of Failure Before 250 Hours

The probability that the MRI fails before a certain time \( t \) is given by \( F(t) \), the cumulative distribution function: \[ F(t) = 1 - e^{-(t/\delta)^\beta} \] For \( t = 250, \beta = 2, \delta = 500 \), it becomes: \[ F(250) = 1 - e^{-(250/500)^2} = 1 - e^{-0.25} \approx 0.221 \].

Step 5: Summarize Results

To summarize the solutions: (a) The mean life of the MRI is approximately 443.1 hours, (b) The variance of the life is approximately 53675 hours², and (c) The probability that the MRI machine fails before 250 hours is approximately 22.1%.

Key Concepts

Mean Life Calculation

The mean life of a system described by the Weibull distribution offers insights into the expected operational time. In the Weibull model, the mean or expected life, \( E(X) \), is calculated using the formula: \[E(X) = \delta \cdot \Gamma\left(1 + \frac{1}{\beta}\right)\]Here, \( \delta \) is the scale parameter, and \( \beta \) is the shape parameter. The gamma function \( \Gamma \) is a key aspect of the calculation, addressing complex mathematical functions. In our example, with \( \beta = 2 \) and \( \delta = 500 \) hours:

• Calculate \( \Gamma(1.5) \) which is approximately 0.8862.
• Plug it into the formula: \( E(X) = 500 \times 0.8862 \)

Thus, the mean life is approximately 443.1 hours. This tells us that on average, the MRI machine operates for about 443 hours before failing.

Variance Calculation

Understanding the spread or variability in the life of the system often requires looking at the variance. For the Weibull distribution, the variance \( Var(X) \) is defined as:\[Var(X) = \delta^2 \cdot \left[\Gamma\left( 1 + \frac{2}{\beta} \right) - \Gamma\left( 1 + \frac{1}{\beta} \right)^2 \right]\]This equation uses the gamma function twice, once for \( 1 + \frac{2}{\beta} \) and then squared for \( 1 + \frac{1}{\beta} \). In our exercise:

• For \( \beta = 2 \), calculate \( \Gamma(2) = 1 \).
• Use \( \Gamma(1.5) \), calculated earlier as 0.8862.

Putting it all together: \[Var(X) = 500^2 \times (1 - 0.8862^2) = 250000 \times (1 - 0.7853) \]This yields a variance of about 53675 hours², guiding us on the life span variability.

Cumulative Distribution Function

The cumulative distribution function (CDF) is crucial in determining the likelihood that an event happens before a given time. For a Weibull distribution, the CDF \( F(t) \) is expressed as:\[F(t) = 1 - e^{-(t/\delta)^\beta}\]This function calculates the probability that the life of the equipment is less than or equal to a specific time \( t \). With \( t = 250 \), \( \beta = 2 \), and \( \delta = 500 \):

• Calculate \( \left( \frac{250}{500} \right)^2 = 0.25 \).
• Then evaluate: \( F(250) = 1 - e^{-0.25} \)

This results in a probability of approximately 0.221 or 22.1%. This indicates the chance that the MRI fails before reaching 250 hours of use.

Reliability Data Modeling

Reliability data modeling is the cornerstone of understanding equipment longevity using statistical tools. The Weibull distribution allows us to model and predict the reliability of devices like the MRI machine, due to its flexibility in accommodating diverse life span behaviors.

• It gives insights into both early failure rates (if \( \beta < 1 \)), constant failure rates (if \( \beta = 1 \)), and wear-out failure rates (if \( \beta > 1 \)).
• The parameters \( \beta \) and \( \delta \) allow for detailed modeling tailored to specific equipment characteristics.

The Weibull distribution is often employed in industries to make informed decisions about maintenance scheduling, warranty assessments, and overall quality improvement. By accurately predicting reliability, businesses can optimize performance and reduce unexpected downtimes, reinforcing strategic planning.