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The lifetime \(X\) (in hundreds of hours) of a 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. Engrg. Manuf., 1991:

Short Answer

Expert verified
a. \(E(X) \approx 3.77\), \(V(X) \approx 3.52\); b. \(P(X \leq 6) \approx 0.9817\); c. \(P(1.5 \leq X \leq 6) \approx 0.7605\).

Step by step solution

01

Understand Weibull Distribution

The Weibull distribution is characterized by parameters \(\alpha\) (shape) and \(\beta\) (scale). For this problem, \(\alpha = 2\) and \(\beta = 3\). The probability density function is \(f(x) = \frac{\alpha}{\beta}\left(\frac{x}{\beta}\right)^{\alpha-1}e^{-(x/\beta)^\alpha}\) for \(x \geq 0\).
02

Compute Expected Value \(E(X)\)

The expected value of a Weibull distributed random variable \(X\) is given by \(E(X) = \beta \Gamma(1+\frac{1}{\alpha})\), where \(\Gamma\) is the gamma function. For \(\alpha=2\), it simplifies to \(E(X) = 3 \Gamma(1+\frac{1}{2}) = 3 \cdot \sqrt{\pi}/2\approx 3.77\).
03

Compute Variance \(V(X)\)

The variance of \(X\) is given by \(V(X) = \beta^2 \left[\Gamma(1+\frac{2}{\alpha}) - (\Gamma(1+\frac{1}{\alpha}))^2\right]\). Calculating for \(\alpha = 2\), \(V(X) = 3^2 \left[\Gamma(1) - (\Gamma(1.5))^2 \right]\approx 3.52\).
04

Compute Probability \(P(X \leq 6)\)

The cumulative distribution function (CDF) of \(X\) is \(F(x) = 1 - e^{-(x/\beta)^\alpha}\). Substitute \(x=6\): \(F(6) = 1 - e^{-(6/3)^2} = 1 - e^{-4} \approx 0.9817\).
05

Compute Probability \(P(1.5 \leq X \leq 6)\)

The probability \(P(a \leq X \leq b)\) is \(F(b) - F(a)\). Calculate \(F(1.5)\) and \(F(6)\): \(F(1.5) = 1 - e^{-(1.5/3)^2} = 1 - e^{-0.25}\approx 0.2212\) and \(F(6)\) was calculated previously as \(0.9817\). 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.

Expected Value
The expected value, or mean, of a random variable is a central concept in probability. It gives us a measure of the 'center' of a distribution. For the Weibull distribution, the expected value depends on the parameters
  • \(\alpha\) (shape)
  • \(\beta\) (scale)
where we use the formula:\[E(X) = \beta \Gamma\left(1+\frac{1}{\alpha}\right)\]. The gamma function, denoted \(\Gamma\), is an extension of the factorial function with its argument shifted down by 1 for continuous variables.
The expected value helps to determine the average lifetime of the vacuum tube in this problem, providing an estimate that can be of practical importance. With \(\alpha = 2\) and \(\beta = 3\), the expected value is \(3 \times \frac{\sqrt{\pi}}{2} \approx 3.77\). This number suggests that, on average, the vacuum tube lasts for about 377 hours, given that time was measured in hundreds of hours.
Variance
Variance is another critical aspect of probability and statistics, capturing the degree of spread of a distribution. It essentially measures how much individual observations of a random variable differ from the expected value. In the context of the Weibull distribution, the variance can be calculated as:\[V(X) = \beta^2 \left[\Gamma\left(1+\frac{2}{\alpha}\right) - \left(\Gamma\left(1+\frac{1}{\alpha}\right)\right)^2\right]\].
Using the parameters \(\alpha = 2\) and \(\beta = 3\), the variance computes to approximately 3.52. This value tells us how varied the lifespans of these vacuum tubes can be when compared to the estimated mean. The smaller a variance, the more tightly clustered around the mean the lifetimes are. A higher variance indicates a larger spread.
Cumulative Distribution Function
The cumulative distribution function (CDF) provides the probability that a random variable is less than or equal to a specific value. For the Weibull distribution, the CDF is given by:\[F(x) = 1 - e^{-(\frac{x}{\beta})^\alpha}\].
This function allows us to calculate the probability of the vacuum tube lasting up to a certain number of hours. For example, the probability that the lifetime \(X\) is less than or equal to 600 hours (or \(X \leq 6\) in hundreds of hours) is found by plugging 6 into the CDF, resulting in approximately 0.9817. This indicates a high probability that a tube will fail before reaching this mark, which is important for reliability assessments.
Gamma Function
The gamma function is a broad generalization of the factorial function for complex numbers. While the factorial \(n!\) is defined only for natural numbers, the gamma function \(\Gamma(n)\) extends this concept to real (and complex) numbers:\[\Gamma(n) = \int_0^\infty t^{n-1}e^{-t} \, dt\].
For positive integers, \(\Gamma(n) = (n-1)!\). It plays a vital role in the calculations of both the expected value and variance for the Weibull distribution. Specifically, it helps in adjusting the scale and spread due to the shape parameter. Understanding the gamma function is essential for manipulating probabilities in advanced statistical distributions like the Weibull, accommodating various shapes of data distributions.

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