Problem 157 - Applied Statistics and Probability for Engineers

Question

The life (in hours) of a computer processing unit (CPU) is modeled by a Weibull distribution with parameters \(\beta=3\) and \(\delta=900\) hours. Determine (a) and (b): (a) Mean life of the CPU. (b) Variance of the life of the CPU. (c) What is the probability that the CPU fails before 500 hours?

Step-by-Step Solution

Answer

The mean life is about 803.68 hours, variance is 186,676 hours², and failure probability before 500 hours is 15.8%.

Step 1: Understanding the Weibull Distribution

The Weibull distribution is often used to model the life of products. It is characterized by two parameters: the shape parameter \(\beta\) and the scale parameter \(\delta\). In this exercise, \(\beta = 3\) and \(\delta = 900\).

Step 2: Calculating the Mean Life

The mean (expected value) of a Weibull distribution is given by \( E(X) = \delta \cdot \Gamma(1 + 1/\beta) \). Using \( \beta = 3 \) and \( \delta = 900 \), we calculate \( E(X) = 900 \cdot \Gamma(1 + 1/3) \). The gamma function \( \Gamma(1 + 1/3) \approx 0.89297951 \). Therefore, \( E(X) \approx 900 \times 0.89297951 \approx 803.68 \) hours.

Step 3: Calculating the Variance of Life

The variance of a Weibull distribution is given by \( \text{Var}(X) = \delta^2 \left( \Gamma(1 + 2/\beta) - (\Gamma(1 + 1/\beta))^2 \right) \). For \( \beta = 3 \), \( \Gamma(1 + 2/3) \approx 1.160529 \). Substituting values, \( \text{Var}(X) = 900^2 \left( 1.160529 - 0.797415^2 \right) \approx 900^2 \times (1.160529 - 0.636516) \approx 186,676 \) hours squared.

Step 4: Calculating Probability of Failure Before 500 Hours

The probability that a CPU fails before a certain time \( t \) is given by the cumulative distribution function: \( F(t) = 1 - e^{-(t/\delta)^\beta} \). For \( t = 500 \), substitute into the equation: \( F(500) = 1 - e^{-(500/900)^3} \). Calculating \( (500/900)^3 \approx 0.17146776 \), we find \( F(500) = 1 - e^{-0.17146776} \approx 1 - 0.842 \approx 0.158 \).

Step 5: Summary

We calculated the mean life of the CPU as approximately 803.68 hours and the variance as approximately 186,676 hours squared. The probability that the CPU fails before 500 hours is approximately 0.158 or 15.8%.

Key Concepts

Parameter Estimation

In Weibull distribution, parameter estimation is crucial for understanding the behavior of a product's lifespan. The two main parameters are the shape parameter, \( \beta \), and the scale parameter, \( \delta \).

• **Shape parameter (\( \beta \))**: Determines how the failure rate changes over time. A \( \beta \) less than 1 indicates a decreasing failure rate, \( \beta \) equal to 1 indicates a constant failure rate, and \( \beta \) greater than 1 indicates an increasing failure rate.
• **Scale parameter (\( \delta \))**: Influences the scale of the data, often related to the average life of the items.

For this exercise, we have \( \beta = 3 \) and \( \delta = 900 \) hours. This configuration typically indicates an increasing failure rate, suggesting that the risk of failure grows as the CPU approaches its lifespan. Accurate estimation of these parameters allows for better predictions and planning regarding replacements and maintenance schedules. Understanding these parameters provides the groundwork for further calculations such as mean life and failure probability.

Mean Life Calculation

Calculating the mean life of a CPU using the Weibull distribution helps us determine the expected duration the CPU will function before failing. The formula to find the mean life \( E(X) \) is:\[E(X) = \delta \cdot \Gamma(1 + 1/\beta) \]Where \( \Gamma \) is the gamma function. In our case:

• \( \delta = 900 \)
• \( \beta = 3 \)
• \( \Gamma(1 + 1/3) \approx 0.89297951 \)

Substituting these values, we compute the mean:\[E(X) = 900 \times 0.89297951 \approx 803.68 \text{ hours}\]This tells us that, on average, a CPU will last around 803.68 hours before it is likely to fail. This information is pivotal for product life cycle management and ensuring that replacements and spare parts are appropriately managed.

Variance Calculation

Variance provides insight into the variability of CPU lifespans. It shows how much the lifetimes differ from the mean. For a Weibull distribution, the variance \( \text{Var}(X) \) is given by:\[\text{Var}(X) = \delta^2 \left( \Gamma(1 + 2/\beta) - (\Gamma(1 + 1/\beta))^2 \right)\]In our problem:

• \( \delta = 900 \)
• \( \Gamma(1 + 2/3) \approx 1.160529 \)
• \( \Gamma(1 + 1/3) \approx 0.89297951 \)

We substitute these values into the variance formula:\[\text{Var}(X) = 900^2 \times (1.160529 - 0.797415) \approx 186,676 \text{ hours squared}\]A higher variance indicates greater variability in lifespans. This value allows manufacturers and managers to assess the consistency of the CPU's life and make decisions regarding quality and reliability enhancements.

Failure Probability

Understanding the failure probability of a CPU before a certain time period is vital for reliability calculations. The cumulative distribution function (CDF) for the Weibull distribution helps in calculating this probability. The equation is:\[F(t) = 1 - e^{-(t/\delta)^\beta}\]For our CPU, to find the probability of failure before 500 hours:

• \( t = 500 \)
• \( \delta = 900 \)
• \( \beta = 3 \)

First, compute \((500/900)^3 \approx 0.17146776\). Then, calculate:\[F(500) = 1 - e^{-0.17146776} \approx 1 - 0.842 \approx 0.158\]This result means there's a 15.8% chance of the CPU failing before it reaches 500 hours of operation. This probability helps in assessing risks related to product durability and planning for maintenance or warranty terms.