Problem 153 - Applied Statistics and Probability for Engineers

Question

Suppose that \(X\) has a Weibull distribution with \(\beta=0.2\) and \(\delta=100\) hours. Determine the mean and variance of \(X\).

Step-by-Step Solution

Answer

Mean is 12000 hours; Variance is 36144000000 hours squared.

Step 1: Understand Weibull Distribution Parameters

The Weibull distribution is parameterized by scale parameter \(\delta\) and shape parameter \(\beta\). In this exercise, \(\beta=0.2\) and \(\delta=100\). These parameters will be used to find the mean and variance.

Step 2: Use the Formula for Mean of Weibull Distribution

The mean of a Weibull distribution is given by the formula \( \mu = \delta \cdot \Gamma(1 + \frac{1}{\beta}) \), where \( \Gamma \) is the gamma function. Substitute \( \beta=0.2 \) and \( \delta=100 \) to find the mean.

Step 3: Calculate \( \Gamma(1 + \frac{1}{\beta}) \)

First, compute \(1 + \frac{1}{\beta} = 1 + \frac{1}{0.2} = 1 + 5 = 6 \). Then evaluate \( \Gamma(6) = 5! = 120 \), since the gamma function at a positive integer \(n\) is \((n-1)!\).

Step 4: Calculate the Mean

Substitute \(\Gamma(6) = 120\) and \(\delta = 100\) into the formula for the mean: \( \mu = 100 \cdot 120 = 12000 \). Therefore, the mean is 12000 hours.

Step 5: Use the Formula for Variance of Weibull Distribution

The variance of a Weibull distribution is given by \( \sigma^2 = \delta^2 \cdot \left( \Gamma(1 + \frac{2}{\beta}) - [\Gamma(1 + \frac{1}{\beta})]^2 \right) \). We need to calculate both gamma values to proceed.

Step 6: Calculate \( \Gamma(1 + \frac{2}{\beta}) \)

Compute \(1 + \frac{2}{\beta} = 1 + \frac{2}{0.2} = 1 + 10 = 11 \). Evaluate \( \Gamma(11) = 10! = 3628800 \). Use this along with the previous result \( \Gamma(6) = 120 \).

Step 7: Calculate the Variance

Substitute the gamma values into the variance formula: \( \sigma^2 = 100^2 \cdot (3628800 - 120^2) = 10000 \cdot (3628800 - 14400) = 10000 \cdot 3614400 = 36144000000 \). Thus, the variance is 36144000000 hours squared.

Key Concepts

Mean of Weibull Distribution

The mean of the Weibull distribution is an essential measure to understand the expected average outcome. The formula to calculate the mean is given by:

• \( \mu = \delta \cdot \Gamma(1 + \frac{1}{\beta}) \)

Here, \( \delta \) is the scale parameter and \( \beta \) is the shape parameter. The gamma function \( \Gamma(x) \) extends the concept of a factorial to real numbers. In practical terms, the mean actually provides the average time to failure for items under warranty.
In our problem, where \( \beta = 0.2 \) and \( \delta = 100 \), we first compute \( 1 + \frac{1}{\beta} = 6 \). Next, we evaluate \( \Gamma(6) \), which is equivalent to \( 5! \) and equals 120.
Finally, substituting into the mean formula, we find:

• \( \mu = 100 \cdot 120 = 12000 \)

This means the average time before the event occurs, such as failure, is 12000 hours.

Variance of Weibull Distribution

The variance of a Weibull distribution is a measure of how much outcomes deviate from the mean. A higher variance indicates more spread out values.

• Formula: \( \sigma^2 = \delta^2 \cdot \left( \Gamma(1 + \frac{2}{\beta}) - [\Gamma(1 + \frac{1}{\beta})]^2 \right) \)

To calculate this, we need to evaluate two gamma functions based on parameters \( \beta = 0.2 \) and \( \delta = 100 \).
First, compute \( 1 + \frac{2}{\beta} = 11 \) and evaluate \( \Gamma(11) \), which equals \( 10! \) or 3628800. We've already found \( \Gamma(6) = 120 \) from the mean calculation.
Substituting these into the variance formula:

• \( \sigma^2 = 100^2 \cdot (3628800 - 120^2) \)
• = 10000 \cdot (3628800 - 14400)
• = 10000 \cdot 3614400
• = 36144000000

Thus, the variance is 36144000000 hours squared, providing a sense of reliability or consistency.

Gamma Function

The gamma function \( \Gamma(x) \) is a fascinating mathematical concept that generalizes the factorial operation. Instead of only accepting non-negative integers like the factorial, the gamma function works for all complex numbers, except the negative integers.

• The formula is \( \Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \, dt \)

For positive integers \( n \), the gamma function follows \( \Gamma(n) = (n-1)! \).
The gamma function is particularly useful in continuous probability distributions, like the Weibull distribution. In our problem, we computed \( \Gamma(6) = 5! \) since 6 is an integer, resulting in the value 120. Similarly, when finding the variance, we used \( \Gamma(11) = 10! \) which equals 3628800.
By utilizing the gamma function, we can effectively calculate statistical properties involving non-integers or large factorials reliably, enhancing the study of statistical measures like the mean and variance within Weibull and other distributions.