Problem 201 - Applied Statistics and Probability for Engineers

Question

The life of a recirculating pump follows a Weibull distribution with parameters \(\beta=2\) and \(\delta=700\) hours. Determine for parts (a) and (b): (a) Mean life of a pump (b) Variance of the life of a pump (c) What is the probability that a pump will last longer than its mean?

Step-by-Step Solution

Answer

Mean life: 620.36 hours, Variance: 105,168.7, Probability: 0.44663.

Step 1: Understand the Weibull Distribution Parameters

The Weibull distribution is defined by two parameters, the shape parameter \(\beta\) and the scale parameter \(\delta\). In this exercise, we have \(\beta = 2\) and \(\delta = 700\) hours.

Step 2: Calculate the Mean of the Weibull Distribution

The mean \(\mu\) of a Weibull distribution is given by \(\mu = \delta \cdot \Gamma\left(1 + \frac{1}{\beta}\right)\), where \(\Gamma\) is the gamma function. Here: \(\mu = 700 \cdot \Gamma\left(1 + \frac{1}{2}\right)\). Using the Gamma function value \(\Gamma(1.5) = \frac{\sqrt{\pi}}{2} \approx 0.88623\), \(\mu \approx 700 \times 0.88623 \approx 620.36\) hours.

Step 3: Calculate the Variance of the Weibull Distribution

The variance \(\sigma^2\) is given by \(\sigma^2 = \delta^2 \left( \Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2 \right)\). Calculate each term: \(\Gamma(1 + \frac{2}{2}) = \Gamma(2) = 1! = 1\). So variance \(\sigma^2 = 700^2 \times (1 - 0.88623^2)\) \(\approx 490000 \times (1 - 0.78537)\) \(\approx 490000 \times 0.21463 \approx 105,168.7\).

Step 4: Calculate the Probability that a Pump Lasts Longer than Its Mean

The probability that a Weibull-distributed variable lasts longer than its mean life is not straightforward. However, we can compute it using the cumulative distribution function (CDF) \(F(t) = 1 - e^{-(t/\delta)^\beta}\). For \(t = \mu = 620.36\), compute: \(F(620.36) = 1 - e^{-(620.36/700)^2}\), \(\approx 1 - e^{-0.78537}\ \approx 0.55337\). The probability of exceeding the mean is \(P(T > 620.36) = 1 - 0.55337 \approx 0.44663\).

Key Concepts

Life Distribution Analysis

Understanding life distribution analysis is essential when dealing with reliability and survival studies, such as examining the lifespan of a recirculating pump. In this context, the Weibull distribution is a valuable tool. It is versatile and widely used because it can model various types of failure rates.

The pump's life follows a Weibull distribution parameterized by two numbers:

• The shape parameter (\(\beta\))
• The scale parameter (\(\delta\))

A shape parameter \(\beta = 2\) indicates a constant failure rate, neither increasing nor decreasing over time, which is typical for parts that experience random mechanical failures. Meanwhile, the scale parameter \(\delta = 700\) refers to the time scale of the process, suggesting the duration over which the examined life events occur.

Recognizing these parameters is crucial to understanding the distribution and applying it effectively to real-world situations, such as evaluating how long a piece of equipment might last or planning its maintenance schedule.

Mean and Variance Calculation

Calculating the mean and variance of a Weibull distribution gives us insights into the expected lifespan and the variability around that lifespan for a given system.

**Mean Calculation**

• The mean \(\mu\) of a Weibull distribution is calculated using the formula: \[\mu = \delta \cdot \Gamma\left(1 + \frac{1}{\beta}\right)\] where \(\Gamma\) is the gamma function.
• In our case: \(\mu = 700 \times \Gamma(1.5) \approx 700 \times 0.88623 \approx 620.36\) hours.

This computation involves the gamma function, a special function that generalizes factorials to non-integer values. It helps us convert the scale parameter into expected life.

**Variance Calculation**

• The variance, \(\sigma^2\), provides a measure of the spread of the pump lifespan and is determined by: \[\sigma^2 = \delta^2 \left( \Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2 \right)\]
• This leads to \(\sigma^2 \approx 490000 \times 0.21463 \approx 105,168.7\).

Variance shows us how much the pump life can differ from the average, emphasizing the variability inherent in the lifespan predictions.

Probability Calculations

Probability calculations in the context of life distributions help us determine the likelihood of specific life outcomes, like how long an item might last compared to its average life.

When considering a Weibull-distributed lifespan, the cumulative distribution function (CDF) is used to find probabilities of different endpoints. For example:

• The CDF is expressed as: \[ F(t) = 1 - e^{-(t/\delta)^\beta} \]
• To find the probability a pump lasts longer than its mean, we calculate: \(F(620.36) = 1 - e^{-(620.36/700)^2} \approx 0.55337\).

The value \(F(620.36)\) tells us the probability that the pump will fail at or before its mean life of 620.36 hours.

Therefore, the probability of the pump lasting longer than its mean life is \(1 - 0.55337 \approx 0.44663\), indicating it is less likely than not to exceed this lifespan.

Performing these calculations allows us to align maintenance schedules and decisions with the statistical likelihood of equipment failure, proving essential in risk management and operations planning.