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A cumulative distribution function, or CDF, describes the probability that a random variable takes on a value less than or equal to a specific point. For the Weibull distribution, this is particularly useful because it captures the probability for various outcomes given the specific parameters of the distribution, which are \(\alpha\) and \(\beta\). Here, the Weibull distribution CDF is given by:\[F_Y(y) = 1 - e^{-(y/\beta)^\alpha}\]When we consider the excess \(X-3.5\), we adjust the formula for the shift in the variable to reflect the original random variable,\(X\). This means the cumulative distribution function for \(X\) becomes:\[F_X(x) = 1 - e^{-((x-3.5)/1.5)^2}\]This equation gives us the probability that the value of \(X\) is less than or equal to a specified number, considering the parameters of the Weibull distribution.

The expected value is essentially the mean or average of a random variable, providing a central value around which data points are spread. For a Weibull-distributed variable like our \(Y = X - 3.5\), the formula to calculate the expected value is:\[E(Y) = \beta \Gamma\left(1 + \frac{1}{\alpha}\right)\]Where\(\Gamma\) represents the gamma function. By substituting\(\beta = 1.5\) and\(\alpha = 2\), and using \(\Gamma(1.5) = \sqrt{\pi}/2\), we calculate:\[E(X-3.5) = 1.5 \times \frac{\sqrt{\pi}}{2} \approx 1.329\]To find the expected return time of the original random variable\(X\), we simply add back the constant\(3.5\):\[E(X) = E(X-3.5) + 3.5 = 4.829\]Thus, the average return time is calculated to be approximately 4.829. This value is critical in understanding the centrality of the distribution.

Variance gives insight into how spread out the values of a random variable are around the expected value. For a Weibull distribution, variance is calculated as follows:\[V(Y) = \beta^2 \left[\Gamma\left(1 + \frac{2}{\alpha}\right) - \left(\Gamma\left(1 + \frac{1}{\alpha}\right)\right)^2\right]\]In our example, use \(\beta = 1.5\), and the values of\(\Gamma(1.5)\) and\(\Gamma(2) = 1\). This leads to:\[V(X-3.5) = 1.5^2 \left[1 - \left(\frac{\sqrt{\pi}}{2}\right)^2\right] \approx 0.291\]Since adding or subtracting a constant does not affect the dispersion of data, the variance of\(X\) (original variable) remains:\[V(X) = V(X-3.5) \approx 0.291\]This tells us how much the return times vary from the expected value, helping to grasp the variability inherent in the data. Understanding variance is essential for gauging risk and reliability in predictions.