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The lifetime \(X\) (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=3\). Compute the following: a. \(E(X)\) and \(V(X)\) b. \(P(X \leq 6)\) c. \(P(1.5 \leq X \leq 6)\) (This Weibull distribution is suggested as a model for time in service in "On the Assessment of Equipment Reliability: Trading Data Collection Costs for Precision," J. of Engr. Manuf., 1991: 105-109.)

Short Answer

Expert verified
a. \(E(X) \approx 2.6586\), \(V(X) \approx 1.9356\) b. \(P(X \leq 6) \approx 0.9817\) c. \(P(1.5 \leq X \leq 6) \approx 0.7605\)

Step by step solution

01

Understanding Weibull Distribution

The Weibull distribution for a random variable X is defined with a scale parameter \(\beta\) and shape parameter \(\alpha\). For this problem, \(\alpha = 2\) and \(\beta = 3\). The probability density function (pdf) is given by: \[ f(x) = \frac{\alpha}{\beta} \left(\frac{x}{\beta}\right)^{\alpha - 1} e^{-(x/\beta)^\alpha} \text{ for } x > 0\]. The expectation and variance for a Weibull distribution are calculated using specific formulas.
02

Calculating Expectation \(E(X)\)

The expectation for a Weibull distribution is given by \( E(X) = \beta \Gamma(1 + \frac{1}{\alpha}) \). Here, \(\Gamma\) is the gamma function. Substitute the values given: \(\beta = 3\) and \(\alpha = 2\), then \[E(X) = 3 \Gamma(1 + \frac{1}{2}) = 3 \Gamma(1.5)\]. Using the property \(\Gamma(1.5) = \frac{\sqrt{\pi}}{2}\), we have \[E(X) = 3 \times \frac{\sqrt{\pi}}{2} \approx 3 \times 0.8862 \approx 2.6586\].
03

Calculating Variance \(V(X)\)

Variance for a Weibull distribution is calculated using \( V(X) = \beta^2 \left[ \Gamma(1 + \frac{2}{\alpha}) - (\Gamma(1 + \frac{1}{\alpha}))^2 \right]\). Substitute \(\alpha = 2\) and \(\beta = 3\): \(\Gamma(1 + \frac{2}{2}) = \Gamma(2) = 1! = 1\), and \((\Gamma(1.5))^2 = \left(\frac{\sqrt{\pi}}{2}\right)^2 = \frac{\pi}{4}\). Therefore, \[V(X) = 3^2 \left[1 - \frac{\pi}{4}\right] = 9 \left[1 - 0.7854\right] \approx 1.9356\].
04

Calculating \(P(X \leq 6)\)

The CDF of the Weibull distribution, \( F(x) \), is \( F(x) = 1 - e^{-(x/\beta)^\alpha} \). For \(P(X \leq 6)\), \(x = 6\): \[P(X \leq 6) = 1 - e^{-(6/3)^2} = 1 - e^{-4}\]. Calculating the exponential function, \( e^{-4} \approx 0.0183 \), thus \[ P(X \leq 6) \approx 1 - 0.0183 = 0.9817\].
05

Calculating \(P(1.5 \leq X \leq 6)\)

To compute the probability over an interval, use the CDF: \[ P(1.5 \leq X \leq 6) = F(6) - F(1.5) \]. From earlier, \(F(6) \approx 0.9817\). Now calculate \(F(1.5)\): \[F(1.5) = 1 - e^{-(1.5/3)^2} = 1 - e^{-0.25}\]. \(e^{-0.25} \approx 0.7788\), so \(F(1.5) \approx 1 - 0.7788 = 0.2212\). Thus, \[ P(1.5 \leq X \leq 6) \approx 0.9817 - 0.2212 = 0.7605\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expectation and Variance
We focus on the concepts of expectation and variance, which are essential in statistics, especially for probability distributions like the Weibull distribution. The **expectation** or expected value of a random variable is a measure of the center of its distribution. It can be thought of as the "average" value we might expect from observing the random variable in the long run.
The formula for expectation in Weibull distribution is given by:
  • \( E(X) = \beta \Gamma(1 + \frac{1}{\alpha}) \)
This formula relies on the Gamma function \( \Gamma \), an extension of the factorial function used for non-integers.
For variance, which gives a measure of the spread of the distribution, the formula is:
  • \( V(X) = \beta^2 \left[ \Gamma(1 + \frac{2}{\alpha}) - (\Gamma(1 + \frac{1}{\alpha}))^2 \right] \)
Using these formulas, we can plug in the given parameters \( \alpha = 2 \) and \( \beta = 3 \), yielding an expectation of approximately 2.659 and a variance of about 1.936.Understanding these concepts helps us summarize the distribution's central tendency and variability, which are crucial for analyzing and interpreting data.
CDF of Weibull Distribution
The **Cumulative Distribution Function** (CDF) is a crucial concept in probability, displaying the probability that a random variable \( X \) takes a value less than or equal to \( x \). For the Weibull distribution, the CDF is given by:
  • \( F(x) = 1 - e^{-(x/\beta)^\alpha} \)
In this form:
  • \( \beta \) is the scale parameter, adjusting how stretched or compressed the distribution is.
  • \( \alpha \) is the shape parameter, influencing the distribution's form, such as its skew.
For tasks like calculating \( P(X \leq 6) \), we substitute \( x = 6 \) into the CDF formula. This results in finding \( F(6) \), showing the cumulative probability up to 6. Similarly, finding the probability that a variable lies in an interval, say between 1.5 and 6, involves subtracting \( F(1.5) \) from \( F(6) \).
This approach highlights how the CDF provides a powerful tool for evaluating probabilities across intervals, aiding in deeper insights into how random variables behave under specific Weibull distribution parameters.
Gamma Function
The **Gamma function** \( \Gamma(n) \) is a critical mathematical concept used extensively in calculating expectations and variances for certain probability distributions, such as the Weibull. It serves as a generalization of the factorial function for real and complex numbers.
In simpler terms, for a positive integer \( n \), the Gamma function fulfills the equation:
  • \( \Gamma(n) = (n-1)! \)
More generally, it's defined as an integral:
  • \( \Gamma(z) = \int_0^{\infty} t^{z-1} e^{-t} \, dt \) for \( z > 0 \).
One of the convenient properties of the Gamma function is \( \Gamma(1/2) = \sqrt{\pi} \). This property is particularly useful when calculating expectations and variances with non-integer parameters, such as \( \Gamma(1.5) \), which translates to \( \frac{\sqrt{\pi}}{2} \).
The incorporation of the Gamma function in Weibull-related calculations makes this mathematical tool indispensable for a comprehensive understanding of continuous probability models.

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