Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

A random variable \(X\) has a Weibull distribution if it has probability density function $$ f(x)= \begin{cases}\frac{\beta}{\theta}\left(\frac{x}{\theta}\right)^{\beta-1} e^{-(x / \theta)^{\beta}} & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{cases} $$ (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

(a) Verified. (b) Mean: 2.6586, Variance: 1.9426. (c) Probability: 35.88%.

Step 1: Verify that Total Probability is 1

We need to show that the total probability for the variable is 1. This is expressed by the integral \( \int_0^{\infty} f(x) \, dx = 1 \). For the Weibull distribution, this becomes \[ \int_0^{\infty} \frac{\beta}{\theta}\left(\frac{x}{\theta}\right)^{\beta-1} e^{-(x / \theta)^{\beta}} \, dx = 1 \].To solve, perform the substitution \( u = \left(\frac{x}{\theta}\right)^{\beta} \) which simplifies this to a gamma function: \( \int_0^{\infty} u^{\frac{1}{\beta}-1} e^{-u} \, du = \Gamma\left(\frac{1}{\beta}\right) \). Known properties of the gamma function tell us \( \Gamma(x) = 1 \) for integer values and \( \beta > 1 \). Hence, the initial integral equals 1.Therefore, the Weibull distribution is valid as its PDF integrates to 1 over the real numbers.

Step 2: Calculate the Mean for the Weibull Distribution

To find the mean \( \mu \), we use the formula for the expectation of the Weibull distribution: \( \mu = \theta \Gamma\left(1 + \frac{1}{\beta}\right) \).Given \( \theta = 3 \) and \( \beta = 2 \), the mean is \( \mu = 3 \Gamma\left(1 + \frac{1}{2}\right) \). Using \( \Gamma\left(1.5\right) = \frac{\sqrt{\pi}}{2} \approx 0.8862 \), we get \( \mu \approx 3 \times 0.8862 = 2.6586 \).

Step 3: Calculate the Variance for the Weibull Distribution

The variance \( \sigma^2 \) is found by \( \sigma^2 = \theta^2 \left(\Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2\right) \).Given \( \theta = 3 \), \( \beta = 2 \), and using Gamma function values: \( \Gamma\left(2\right) = 1 \) and \( \Gamma\left(1.5\right) = 0.8862 \), we calculate the variance: \[ \sigma^2 = 3^2 \left(1 - 0.8862^2\right) \approx 9 \times 0.21584 = 1.9426 \].

Step 4: Find the Probability of Failure Before 2 Years

The probability that a monitor fails before 2 years is given by the cumulative distribution function (CDF) of the Weibull distribution: \( P(X < 2) = 1 - e^{-(\frac{x}{\theta})^{\beta}} \) evaluated at \( x = 2 \).Using \( \theta = 3 \) and \( \beta = 2 \), this becomes: \[ P(X < 2) = 1 - e^{-(\frac{2}{3})^2} = 1 - e^{-\frac{4}{9}} \].Calculating, we find \( P(X < 2) = 1 - e^{-0.4444} \approx 1 - 0.6412 \approx 0.3588 \). Therefore, the probability is approximately 35.88%.

Key Concepts

Probability Density Function

The probability density function (PDF) plays a critical role in understanding the probability distribution of a random variable. In the context of the Weibull Distribution, the PDF is defined as follows for a random variable \( X \):

• For \( x > 0 \), \( f(x) = \frac{\beta}{\theta}\left(\frac{x}{\theta}\right)^{\beta-1} e^{-(x / \theta)^{\beta}} \).
• For \( x \leq 0 \), \( f(x) = 0 \).

This function essentially provides the "height" of the probability density for different values of \( x \).
This means the likelihood of the variable taking on a particular value or falling within a specific range.

To ensure it is a valid probability distribution, the total area under the PDF curve over the entire space must equal 1. Mathematically, this is expressed as \( \int_{0}^{\infty} f(x) \, dx = 1 \).
For the Weibull distribution, using an appropriate substitution reduces this integral to a gamma function, which confirms that our Weibull PDF is properly normalized when \( \beta > 1 \). That establishes it as a legitimate probability function.

Expected Value

The expected value, or mean, provides a measure of the central tendency of a distribution. For the Weibull distribution, it is calculated using the formula:
\( \mu = \theta \Gamma\left(1 + \frac{1}{\beta}\right) \).
This accounts for the shape \( \beta \) and scale \( \theta \) parameters of the distribution.

Given \( \theta = 3 \) and \( \beta = 2 \) from our specific exercise:

• The expected value, \( \mu \), becomes \( 3 \Gamma\left(1 + \frac{1}{2}\right) \).
• Using the gamma function \( \Gamma(1.5) = \frac{\sqrt{\pi}}{2} \approx 0.8862 \),
• we find \( \mu \approx 3 \times 0.8862 = 2.6586 \).

This value represents the "average" time until an event (like failure) occurs in the context of a Weibull-distributed process.

Variance

Variance measures how much the values of the random variable differ from the mean. For the Weibull distribution, it is given by the formula:
\( \sigma^2 = \theta^2 \left(\Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2\right) \).

Let's delve into an example with \( \theta = 3 \) and \( \beta = 2 \):

• The variance formula becomes \( 3^2 \left(\Gamma(2) - \Gamma(1.5)^2\right) \).
• Using \( \Gamma(2) = 1 \) and \( \Gamma(1.5) \approx 0.8862 \),
• we calculate \( \sigma^2 = 9 \left(1 - 0.8862^2\right) \).
• This results in \( \sigma^2 \approx 9 \times 0.21584 = 1.9426 \).

Variance helps to understand the spread of the distribution around its mean, or how consistent the variable behaves under repeated experiments.

Cumulative Distribution Function

The cumulative distribution function (CDF) tells us the probability that a random variable will take a value less than or equal to a specific number.
For the Weibull distribution, the CDF is given by:
\( F(x) = 1 - e^{-(x / \theta)^{\beta}} \).

It essentially accumulates the probability from negative infinity to a point \( x \).
Let's explore the scenario where the lifetime of a computer monitor is represented by a Weibull distribution with \( \theta = 3 \) and \( \beta = 2 \):

• We want to find the probability that the monitor fails before 2 years.
• We use \( F(2) = 1 - e^{-(2 / 3)^2} \).
• Calculation \, yields \( F(2) = 1 - e^{-0.4444} \approx 1 - 0.6412 \approx 0.3588 \).

Thus, the probability that the monitor fails in under 2 years is about 35.88%.
The CDF provides valuable insight into the likelihood of different outcomes over a range of values.