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In probability and statistics, the expected value is a fundamental concept. It represents the mean or average outcome of a random variable over the long run. Specifically, for the Weibull distribution, the expected value can be calculated using a specific formula. The expected value, denoted as \( E(X) \), is calculated by \( \frac{1}{\alpha} \cdot \Gamma\left(1 + \frac{1}{\beta}\right) \) where \( \alpha \) and \( \beta \) are parameters of the Weibull distribution. In this exercise, for the given Weibull distribution with \( \alpha = 2 \) and \( \beta = 3 \), substitution into the formula gives us \[ E(X) \approx \frac{0.893}{2} = 0.4465 \] suggesting that on average, the lifetime of a vacuum tube is 44.65 hours.

The variance of a random variable provides insight into its spread or variability around the expected value. For the Weibull distribution, the variance \( V(X) \) is calculated as \[ V(X) = \left(\frac{1}{\alpha^2}\right) \cdot \left[\Gamma\left(1 + \frac{2}{\beta}\right) - (\Gamma\left(1 + \frac{1}{\beta}\right))^2\right] \]This formula lets us assess how much the actual values differ from the expected value. In our example, with parameters \( \alpha = 2 \) and \( \beta = 3 \), the variance turns out to be approximately \( 0.13925 \) indicating a moderate deviation from the expected lifetime. This tells us that while the average lifetime is 44.65 hours, there will be some tubes with shorter and longer lifespans, and variance quantifies how much that spread can be expected.

The cumulative distribution function (CDF) is a crucial concept in probability as it gives the probability that a random variable is less than or equal to a certain value. For the Weibull distribution, the CDF is given by: \[ F(x) = 1 - e^{-(\alpha x)^\beta} \] This function helps us see how the probability accumulates over time. In our exercise, when evaluating at \( X = 6 \), we find that \( F(6) \approx 1 \). This suggests that almost certainly a vacuum tube will fail by 600 hours, which helps in predicting reliability of components in engineering tasks.

Probability is the measure of how likely an event is to occur. In the context of our example, it involves the likelihood of the vacuum tube's lifetime falling within specified ranges.From step 5, we computed the probability of the tube lasting between 1.5 to 6 units, which corresponds to 150 to 600 hours. Using the cumulative distribution function, it turns out to be 0, indicating that such a range is practically negligible. Since \( e^{-27} \) and larger exponents approach zero, the probability for such uncommon events becomes null, serving as a precise measure of circumstances' likelihood.

Understanding statistical distributions, such as the Weibull distribution, is essential in engineering because it aids in the modeling and prediction of component lifetimes and reliability. Engineering statistics blend statistical methods with engineering problems to enhance decision making. For example, using the Weibull distribution, engineers can forecast the lifespan of various products, manage maintenance, and anticipate failures. By understanding parameters such as the expected value and variance, engineers can design more reliable systems, reduce costs, and improve safety. This fusion of statistical understanding with practical application defines the value of statistics in engineering disciplines, guiding effective strategy and design.