Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

A random variable \(X\) has a Weibull distribution if it has probability density function $$ f(x)=\left\\{\begin{array}{ll} \frac{\beta}{\theta}\left(\frac{x}{\theta}\right)^{\beta-1} e^{-(x / \theta)^{\beta}} & \text { if } x>0 \\ 0 & \text { if } x \leq 0 \end{array}\right. $$ (a) Show that \(\int_{-\infty}^{\infty} f(x) d x=1\). (Assume \(\beta>1\).) (b) If \(\theta=3\) and \(\beta=2\), find the mean \(\mu\) and the variance \(\sigma^{2}\). (c) If the lifetime of a computer monitor is a random variable \(X\) that has a Weibull distribution with \(\theta=3\) and \(\beta=2\) (where age is measured in years) find the probability that a monitor fails before two years.

Step-by-Step Solution

Answer

1. Integral of the PDF is \(1\). 2. Mean \(\mu = 3 \cdot \frac{\sqrt{\pi}}{2}\). 3. Probability of failure before 2 years: \(1 - e^{-(2/3)^2}\).

Step 1: Integrate the PDF over the Entire Range

To show that \(\int_{-\infty}^{\infty} f(x) \, dx = 1\), we only need to integrate from 0 to \(\infty\) because \(f(x) = 0\) for \(x \leq 0\). Thus, \(\int_{0}^{\infty} \frac{\beta}{\theta}\left(\frac{x}{\theta}\right)^{\beta-1} e^{-(x /\theta)^{\beta}} \, dx\) must equal 1. Let \(u = \left(\frac{x}{\theta}\right)^\beta\), then \(du = \frac{\beta}{\theta}\left(\frac{x}{\theta}\right)^{\beta-1} \, dx\). The integral becomes \(\int_{0}^{\infty} e^{-u} \, du = 1\), confirming the total probability is 1.

Step 2: Calculate the Mean of the Weibull Distribution

For a Weibull distribution, the mean \(\mu\) is \(\mu = \theta \Gamma(1 + \frac{1}{\beta})\). Given \(\theta = 3\) and \(\beta = 2\), the mean is \(\mu = 3 \Gamma(1.5)\). The gamma function \(\Gamma(1.5) = \frac{\sqrt{\pi}}{2}\), so \(\mu = 3 \cdot \frac{\sqrt{\pi}}{2}\).

Step 3: Calculate the Variance of the Weibull Distribution

The variance \(\sigma^2\) is given by \(\sigma^2 = \theta^2 \left(\Gamma(1 + \frac{2}{\beta}) - (\Gamma(1 + \frac{1}{\beta}))^2\right)\). For \(\beta = 2\), \(\Gamma(2) = 1\) and the calculation becomes \(9 (1 - (\frac{\sqrt{\pi}}{2})^2) \).

Step 4: Calculate the Probability of Failure Before Two Years

The probability that a monitor fails before \(t\) years is computed using \(P(X < t) = 1 - e^{-(t/\theta)^\beta}\). Substituting \(t = 2\), \(\theta = 3\), and \(\beta = 2\) gives \(P(X < 2) = 1 - e^{-(2/3)^2}\). Compute this to find the probability.

Key Concepts

Probability Density Function

The probability density function (PDF) is a fundamental concept for continuous random variables. It describes the likelihood of different outcomes in a continuous dataset. For a Weibull distribution, the PDF is given by the function:

• \[ f(x) = \frac{\beta}{\theta}\left(\frac{x}{\theta}\right)^{\beta-1} e^{-(x /\theta)^{\beta}} \quad \text{, if } x > 0 \]
• 0 \quad \text{ , if } x \leq 0

This function depends on two parameters, \(\theta\) (scale parameter) and \(\beta\) (shape parameter). It ensures the probability of all possible outcomes sums up to one. If you integrate the PDF over its entire range, it should always equal one, signifying the total probability. For the Weibull distribution, this requires integrating from zero to infinity, since the PDF is zero for any non-positive \(x\). By using a substitution method, such as \(u = \left(\frac{x}{\theta}\right)^{\beta}\), the integration simplifies to \(\int_{0}^{\infty} e^{-u} du = 1\), verifying our PDF is valid.

Gamma Function

The gamma function \(\Gamma(z)\) is an extension of the factorial function to complex numbers. For positive integers, it satisfies \(\Gamma(n) = (n-1)!\). In the context of the Weibull distribution, it helps in calculating the mean and variance.
The gamma function is used in the calculation of the mean of a Weibull distribution: \(\mu = \theta \Gamma(1 + \frac{1}{\beta})\). For \(\beta = 2\), this simplifies to \(\Gamma(1.5)\), which equals \(\frac{\sqrt{\pi}}{2}\).
Additionally, in variance calculations, the gamma function appears in terms like \(\Gamma(1 + \frac{2}{\beta})\). Understanding how \(\Gamma(z)\) works is crucial for working with the Weibull distribution, as it appears in these foundational equations.

Mean and Variance Calculation

Understanding and calculating the mean and variance are key aspects of using the Weibull distribution effectively. The mean \(\mu\) can be calculated with the formula:

• \[ \mu = \theta \Gamma(1 + \frac{1}{\beta}) \]

Given \(\theta = 3\) and \(\beta = 2\), we replace to find \(\mu = 3 \Gamma(1.5) = 3 \cdot \frac{\sqrt{\pi}}{2}\).
The variance \(\sigma^2\) measures how much the data is spread out and is calculated by:

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

Substituting the values gives \(\sigma^2 = 9 (1 - (\frac{\sqrt{\pi}}{2})^2)\).
These calculations are important to describe the distribution of lifetime data in various applications, including reliability engineering.

Probability of Failure

Calculating the probability of failure within a certain time frame is an essential application of the Weibull distribution. For a device with reliability characterized by Weibull parameters, the probability \(P\) that it fails before a time \(t\) is determined by:

• \[ P(X < t) = 1 - e^{-(t/\theta)^\beta} \]

When \(\theta = 3\) and \(\beta = 2\), and if you want to find the probability of failure before two years, substitute the values into the formula:

• \[ P(X < 2) = 1 - e^{-(2/3)^2} \]

Calculating this expression provides you with the likelihood of failure within that time period. This probability gives valuable insights into planning maintenance schedules or estimating product lifespans.