Problem 164 - Applied Statistics and Probability for Engineers

Question

Suppose that the lifetime of a component (in hours), \(X\) is modeled with a Weibull distribution with \(\beta=0.5\) and \(\delta=4000\). Determine the following in parts (a) and (b): (a) \(P(X>3500)\) (b) \(P(X>6000 \mid X>3000)\) (c) Comment on the probabilities in the previous parts compared to the results for an exponential distribution. (d) Comment on the role of the parameter \(\beta\) in a lifetime model with the Weibull distribution.

Step-by-Step Solution

Answer

(a) 0.392 (b) 0.698. With Weibull, probabilities vary from exponential due to its shape parameter \(\beta\). \(\beta = 0.5\) indicates decreasing failure rate.

Step 1: Understanding the Weibull Distribution

The Weibull distribution is characterized by the probability density function (PDF):\[ f(x; \beta, \delta) = \frac{\beta}{\delta} \left(\frac{x}{\delta}\right)^{\beta - 1} e^{-(x/\delta)^\beta} \]where \( \beta \) is the shape parameter and \( \delta \) is the scale parameter. For this problem, \( \beta = 0.5 \) and \( \delta = 4000 \).

Step 2: Calculating P(X > 3500)

To find \( P(X > 3500) \), we use the cumulative distribution function (CDF), which for a Weibull distribution is given by:\[ F(x; \beta, \delta) = 1 - e^{-(x/\delta)^\beta} \]Thus, \( P(X > 3500) = 1 - F(3500; 0.5, 4000) \).\[ F(3500; 0.5, 4000) = 1 - e^{-(3500/4000)^{0.5}} \]Calculating inside the exponent:\[ \left(\frac{3500}{4000}\right)^{0.5} = \sqrt{0.875} \approx 0.935 \]\[ P(X > 3500) = 1 - (1 - e^{-0.935}) = e^{-0.935} \approx 0.392 \]

Step 3: Calculating P(X > 6000 | X > 3000)

The conditional probability \( P(X > 6000 | X > 3000) \) is calculated using the formula:\[ P(X > 6000 | X > 3000) = \frac{P(X > 6000)}{P(X > 3000)} \]Firstly, calculate \( P(X > 6000) \):\[ P(X > 6000) = e^{-(6000/4000)^{0.5}} = e^{-(1.5)^{0.5}} \approx e^{-1.225} \approx 0.294 \]Now, calculate \( P(X > 3000) \):\[ P(X > 3000) = e^{-(3000/4000)^{0.5}} \approx e^{-0.866} \approx 0.421 \]Finally, calculate the conditional probability:\[ P(X > 6000 | X > 3000) = \frac{0.294}{0.421} \approx 0.698 \]

Step 4: Comparing with Exponential Distribution

For an exponential distribution with the same mean as the Weibull distribution, the survival function is given by:\[ S(x) = e^{-x/\lambda} \]Because \(\beta = 0.5\), which implies a decreasing hazard rate, the Weibull distribution provides a different probability structure than an exponential distribution, which has a constant hazard rate. \( P(X > 3500) \) is typically larger for exponential distributions when \( \lambda \) matches the mean lifetime because the exponential model implies a memoryless property.

Step 5: Role of Parameter β in Weibull Distribution

The parameter \(\beta\) determines the shape of the distribution and affects the hazard function. When \(\beta = 1\), the Weibull distribution simplifies to the exponential distribution. For \(\beta < 1\), as in this problem, the hazard function decreases over time, reflecting that the component is more likely to fail at the beginning of its life cycle, which causes different behavior when compared to \(\beta > 1\).

Key Concepts

Lifetime Model

The Lifetime Model is a crucial concept in understanding how long a component, machine, or system will last under specified conditions, based on probability distributions like the Weibull distribution. The goal is to model the time until a failure occurs, evaluating the reliability and maintenance needs of a system.

The Weibull distribution is widely used in lifetime modeling due to its flexibility in adjusting to various data shapes. It is determined by two key parameters: the shape parameter ( β ) and the scale parameter ( δ ). In this example, β = 0.5 and δ = 4000 , indicating the model's adaptability in predicting component failures over time.

Understanding the Weibull Lifetime Model provides insights into when a system might fail and how to manage resources for repairs or replacements efficiently.

Probability Calculations

Probability Calculations in the context of the Weibull distribution involve determining the likelihood that an event, such as component failure, will occur within a specific timeframe.

For instance, calculating P(X > 3500) , the probability that the component lasts longer than 3500 hours, involves using the Weibull cumulative distribution function (CDF). The CDF helps estimate the probability of a component surviving past a certain time. In our example, the result was approximately 0.392 , meaning there is about a 39.2% chance the component lasts beyond 3500 hours.

Another crucial calculation is the conditional probability, like P(X > 6000 | X > 3000) . This form of probability tells us the chance that a component that has already lasted 3000 hours will continue to 6000 hours. The calculated probability was approximately 69.8% , indicating that once the component has reached 3000 hours, it is still in relatively good condition for making it to 6000 hours.

Exponential Distribution Comparison

When comparing the Weibull distribution with the Exponential distribution, it is important to note the fundamental differences in their probability structures. The Exponential distribution is a special case of the Weibull distribution, occurring when the shape parameter β = 1 . This distribution exhibits a constant hazard rate, implying that the rate of failure is the same regardless of time elapsed.

The Weibull model with β = 0.5 differs significantly because it suggests a decreasing hazard rate. This means the chance of failure decreases over time, unlike the memoryless property of the Exponential distribution where the failure rate stays constant. For example, P(X > 3500) is typically higher in an Exponential model if λ matches the Weibull's mean, highlighting the versatility and broader application of the Weibull model.

Shape Parameter Beta

The Shape Parameter Beta ( β ) in a Weibull distribution is critical in defining the curve shape of the probability distribution. It influences how the failure rate changes over time, which is essential for understanding component reliability and maintenance planning.

In this exercise, β = 0.5 indicates a decreasing hazard rate. This suggests components are more likely to fail early in their lifetime, but their reliability improves as they age, contrary to systems with β > 1 where the hazard rate increases over time.

With β < 1 , the Weibull distribution is suitable for 'infant mortality' scenarios, where initial failure is more probable. This understanding helps in creating strategies such as early life testing or burn-in procedures to uncover early failures before components are put into regular operation.