Problem 156 - Applied Statistics and Probability for Engineers

Question

Assume that the life of a roller bearing follows a Weibull distribution with parameters \(\beta=2\) and \(\delta=10,000\) hours. (a) Determine the probability that a bearing lasts at least 8000 hours. (b) Determine the mean time until failure of a bearing. (c) If 10 bearings are in use and failures occur independently, what is the probability that all 10 bearings last at least 8000 hours?

Step-by-Step Solution

Answer

(a) 0.5273, (b) 8862 hours, (c) 0.002647

Step 1: Understand the Weibull Distribution

The Weibull distribution is a continuous probability distribution used in reliability analysis. It is defined by two parameters: shape parameter \( \beta \) and scale parameter \( \delta \). The probability density function is given by \( f(t) = \frac{\beta}{\delta} \left( \frac{t}{\delta} \right)^{\beta - 1} e^{-\left( \frac{t}{\delta} \right)^\beta} \). The cumulative distribution function (CDF) is \( F(t) = 1 - e^{-\left( \frac{t}{\delta} \right)^\beta} \), and the reliability function (survival function) is \( R(t) = e^{-\left( \frac{t}{\delta} \right)^\beta} \).

Step 2: Calculate the Probability a Bearing Lasts At Least 8000 Hours

To find the probability that a bearing lasts at least 8000 hours, use the reliability function \( R(t) = e^{-\left( \frac{t}{\delta} \right)^\beta} \). Substitute \( t = 8000 \), \( \beta = 2 \), and \( \delta = 10000 \) into the equation:\[ R(8000) = e^{-\left( \frac{8000}{10000} \right)^2} = e^{-\left( 0.8 \right)^2} = e^{-0.64} \]Now calculate \( e^{-0.64} \).

Step 3: Compute the Exponential Function

Calculate \( e^{-0.64} \) using a calculator:\[ e^{-0.64} \approx 0.5273 \]So, the probability that a bearing lasts at least 8000 hours is approximately 0.5273.

Step 4: Calculate the Mean Time Until Failure

For a Weibull distribution, the mean time until failure (MTTF) is given by \( \delta \Gamma\left(1 + \frac{1}{\beta}\right) \). In this case, \( \beta = 2 \) and \( \delta = 10000 \).Calculate \( \Gamma\left(1 + \frac{1}{2}\right) = \Gamma\left(1.5\right) \), which is equivalent to \( \frac{\sqrt{\pi}}{2} \). Use this to find the MTTF:\[ MTTF = 10000 \cdot \frac{\sqrt{\pi}}{2} \approx 10000 \cdot 0.8862 \approx 8862 \text{ hours} \]

Step 5: Determine Probability All 10 Bearings Last at Least 8000 Hours

Assuming each bearing lasts independently at least 8000 hours with a probability of 0.5273 (calculated previously), the probability that all 10 bearings last at least 8000 hours is given by:\[ (0.5273)^{10} \]Calculate this probability.

Step 6: Compute Probability for 10 Bearings

Calculate \( (0.5273)^{10} \) using a calculator:\[ (0.5273)^{10} \approx 0.002647 \]Thus, the probability that all 10 bearings last at least 8000 hours is approximately 0.002647.

Key Concepts

Weibull Distribution

The Weibull distribution is a versatile and widely-used probability distribution in the field of reliability analysis. It helps model the life of products like bearings, encapsulating how long they last before failure. This distribution is particularly suitable because it can represent various types of life data behavior.
It is characterized by two key parameters: the shape parameter \( \beta \) and the scale parameter \( \delta \). These parameters influence the behavior of the distribution. The shape parameter \( \beta \) determines the failure rate tendency, where \( \beta = 2 \) implies that the failure rate increases linearly over time. The scale parameter \( \delta \) represents the characteristic life of the product—in this case, roller bearings—which means the point in time by which 63.2% of the units would have failed in a population.
The reliability function, \( R(t) = e^{-\left( \frac{t}{\delta} \right)^\beta} \), expresses the probability that a device will last beyond time \( t \). In the exercise, substituting the values of \( t = 8000 \), \( \delta = 10000 \), and \( \beta = 2 \) allows us to calculate \( R(t) \), resulting in approximately 0.5273, indicating a 52.73% chance the bearings last at least 8000 hours.

Mean Time Until Failure

Mean Time Until Failure (MTTF) is a critical metric in evaluating the expected time for a product to fail. In the context of the Weibull distribution, the MTTF provides an average estimate of the operating life of a product. For Weibull distributions, the formula to compute MTTF is \( \delta \cdot \Gamma \left(1 + \frac{1}{\beta} \right) \).
Here, \( \Gamma \) is the gamma function, which extends the factorial concept to non-integer values. Specifically, for \( \beta = 2 \), the gamma value is \( \Gamma(1.5) \), equal to \( \frac{\sqrt{\pi}}{2} \). Given \( \delta = 10000 \), the calculation \( 10000 \cdot \frac{\sqrt{\pi}}{2} \approx 8862 \) reveals that on average, each bearing is expected to last around 8862 hours before failure occurs.
MTTF is an essential parameter because it provides consumers and manufacturers insights into product longevity, facilitating better design, usage, and replacement strategies.

Probability Calculations

Probability calculations are central to understanding reliability and risk, especially when multiple units or components are considered.
In the scenario where 10 roller bearings are in use, and we wish to calculate the probability that all 10 survive at least 8000 hours, we employ the multiplication rule. Each bearing lasting independently for 8000 hours has a probability of approximately 0.5273. Therefore, the probability that all 10 last at least 8000 hours is obtained by raising this probability to the power of 10: \( (0.5273)^{10} \). Calculating this expression gives approximately 0.002647.
Thus, there is only a 0.26% chance that all 10 bearings will survive beyond 8000 hours without failures, highlighting the challenges in systems with numerous components due to compounded risk of failure. Understanding these probabilities helps in decision-making regarding resource allocation, maintenance scheduling, and risk management.