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In the context of the Weibull Distribution, the Cumulative Distribution Function (CDF) is a fundamental concept that helps us measure the probability that a random variable takes a value less than or equal to a particular number. For Weibull distributions, especially when dealing with excess variables, the CDF takes the form: \( F_Y(y) = 1 - e^{-(y/1.5)^2} \). In this exercise, we extend this to our actual variable \(X\), where the CDF becomes \( F_X(x) = 1 - e^{-((x-3.5)/1.5)^2} \) for \( x > 3.5 \). This means the probability of \(X\) being less than or equal to \(x\) is given by this expression.
For any specific value \(x\), the CDF calculates how likely the event is to occur within that range. Understanding the CDF allows us to calculate the probability of a variable falling within a certain interval, which is essential for problems like finding \(P(X > 5)\) or \(P(5 \leq X \leq 8)\) in this exercise.

The expected value, often known as the mean, and variance are statistical measures that provide insights into the characteristics of a probability distribution. For a Weibull-distributed variable, the expected value \(E(Y)\) and variance \(V(Y)\) can be calculated using specific formulas. The expected value for our exercise involves \( E(Y) = \beta \Gamma(1 + 1/\alpha) \), where \(\beta = 1.5\) and \(\alpha = 2\). This results in approximately \(1.3293\).
The variance is determined by \( V(Y) = \beta^2 [\Gamma(1 + 2/\alpha) - (\Gamma(1 + 1/\alpha))^2] \), yielding roughly \(0.4241\). Since the actual variable \(X\) is \(Y + 3.5\), its expected value and variance are \(E(X) = 4.8293\) and \(V(X) = 0.4241\).

• Expected Value: Indicates the average return time of the product based on the distribution.
• Variance: Measures the spread of the return times around the expected value.
  This understanding helps analyze variations and predict future probabilities effectively.

Probability calculations are used to determine the likelihood of an event occurring within certain parameters, based on the CDF. For instance, to calculate \(P(X > 5)\), we use the complement rule: \(P(X > 5) = 1 - F_X(5)\). Here, we substitute into the CDF formula, obtaining \( e^{-1} \), approximately \(0.3679\).
For a range, such as \(P(5 \leq X \leq 8)\), we utilize the difference between the CDF values at the boundaries: \(F_X(8) - F_X(5)\). Calculating, we find approximately \(0.6321\).
These calculations are critical for statistical analysis as they help in:

• Predicting the return times.
• Making informed decisions based on statistical outcomes.
  In real-world applications, such probability assessments guide quality control and risk management.

Statistical distributions like the Weibull distribution are used extensively in various fields for modeling data. They are particularly useful when analyzing life data, failure times, and return times as in this exercise.
The Weibull distribution is versatile, offering a simplified way to handle data with different shapes depending on its parameters. Here, we have a practical example focused on understanding the return time of a defective product. This real-life application demonstrates how statistical methods provide clarity and predictability to seemingly random events.

• Industries involved include: Manufacturing, reliability engineering.
• Common applications: Predicting equipment failures, analyzing customer return patterns, planning maintenance schedules.

By using Weibull distribution, organizations can improve operations, set quality benchmarks, and optimize processes effectively.