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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. \(E(X) \approx 1.802\); \(V(X) \approx 0.303\). b. \(P(X \leq 6) \approx 1\). c. \(P(1.5 \leq X \leq 6) \approx 0.657\).

Step by step solution

01

Understanding the Weibull Distribution

The Weibull distribution with parameters \(\alpha\) and \(\beta\) is a continuous probability distribution with probability density function \(f(x) = \frac{\beta}{\alpha}\left(\frac{x}{\alpha}\right)^{\beta-1}e^{-(x/\alpha)^\beta}\) for \(x > 0\). The parameters we have are \(\alpha = 2\) and \(\beta = 3\).
02

Calculate Expected Value (a)

The expected value \(E(X)\) of a Weibull distribution is given by \( E(X) = \alpha \Gamma\left(1 + \frac{1}{\beta}\right) \). Calculating this for \(\alpha = 2\) and \(\beta = 3\), we get: \[ E(X) = 2 \Gamma\left(1 + \frac{1}{3}\right) = 2 \Gamma\left(\frac{4}{3}\right) \approx 1.802 \text{ hundred hours} \].
03

Calculate Variance (a)

The variance \(V(X)\) of a Weibull distribution is given by \( V(X) = \alpha^2 \left( \Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2 \right) \). By substituting \(\alpha = 2\) and \(\beta = 3\), we calculate: \[ V(X) = 2^2 \left( \Gamma\left(1 + \frac{2}{3}\right) - \left(\Gamma\left(1 + \frac{1}{3}\right)\right)^2 \right) \approx 0.303 \text{ hundred hours}^2 \].
04

Calculate CDF at 6 for \(P(X \leq 6)\) (b)

The cumulative distribution function (CDF) for the Weibull distribution is \(F(x) = 1 - e^{-(x/\alpha)^\beta}\). For \(x = 6\), we calculate: \[ P(X \leq 6) = F(6) = 1 - e^{-(6/2)^3} = 1 - e^{-27} \approx 1 \] This implies the probability is very close to 1 since \(e^{-27}\) is a tiny value.
05

Calculate Probability for \(1.5 \leq X \leq 6\) (c)

To find \(P(1.5 \leq X \leq 6)\), use CDF values: \( P(1.5 \leq X \leq 6) = F(6) - F(1.5) \). We've found \(F(6) \approx 1\). Now calculate \(F(1.5)\): \[ F(1.5) = 1 - e^{-(1.5/2)^3} = 1 - e^{-0.421875} \approx 0.343 \]. Hence, \[ P(1.5 \leq X \leq 6) = 1 - 0.343 \approx 0.657 \].

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Key Concepts

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

Probability Distributions
Probability distributions are mathematical functions that provide the probabilities of occurrence of different possible outcomes. They are fundamental in statistics as they describe how data is likely to behave. The Weibull distribution, a type of probability distribution, is particularly useful in reliability engineering and failure analysis. It models the life duration of products with its ability to take on various shapes. This flexibility depends on the parameters \(\alpha\) (scale) and \(\beta\) (shape). These parameters allow the Weibull distribution to mimic other distributions, such as the exponential or the Rayleigh distributions, making it versatile in representing different failure rates. Understanding how to work with different probability distributions, like the Weibull, means you can predict outcomes and analyze failures more effectively.
Expected Value
The expected value, often referred to as the mean, is a measure of the center of a probability distribution. It gives a 'central' value that the random variable will take on average. For the Weibull distribution, the expected value is calculated using the formula: \[ E(X) = \alpha \Gamma\left(1 + \frac{1}{\beta}\right) \]. Here, \(\Gamma\) denotes the gamma function, a special function that extends the factorial function to real numbers. With our parameters \(\alpha = 2\) and \(\beta = 3\), we get \( E(X) \approx 1.802 \) hundred hours. This tells us that, on average, the vacuum tubes should last about 180.2 hours, offering an idea of how long you can expect the service life to be.
Variance
Variance measures how much the values of a random variable deviate from the expected value. It represents the spread or variability within a distribution. For the Weibull distribution, variance is calculated using: \[ V(X) = \alpha^2 \left( \Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2 \right) \]. By inserting \(\alpha = 2\) and \(\beta = 3\), we find \( V(X) \approx 0.303 \) hundred hours squared. This result means the duration of service life varies moderately around the mean, indicating a relatively consistent product performance. Understanding variance helps identify the reliability and expected performance consistency of products.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a crucial concept in probability and statistics. It gives the probability that a random variable takes a value less than or equal to a specific point. For the Weibull distribution, the CDF is given as \( F(x) = 1 - e^{-(x/\alpha)^\beta} \). In our example, calculating \(F(6)\) gives \( P(X \leq 6) \approx 1 \), suggesting that nearly all vacuum tubes will last fewer than 600 hours. To assess the probability that a tube fails between two points, \(1.5\) to \(6\) hundred hours, we use the formula \( P(a \leq X \leq b) = F(b) - F(a) \). This results in a probability of approximately \(0.657\), indicating a significant chance of failure within this interval. The CDF offers a complete picture by summarizing probabilities up to a certain value, simplifying decision-making in reliability assessment.

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Most popular questions from this chapter

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