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In the realm of probability and statistics, the expected value is a crucial concept. It tells us the average outcome we can anticipate from a random variable over many trials. For a lognormal distribution, which is a probability distribution of a random variable whose logarithm is normally distributed, the expected value is calculated in a unique way. To determine the expected value of a lognormal distribution, you can use the formula: \[ E(X) = e^{\mu + \frac{\sigma^2}{2}} \] Here, \( \mu \) and \( \sigma \) are the mean and standard deviation of the underlying normal distribution. In our given problem, we have \( \mu = 5 \) and \( \sigma = 0.1 \). Substituting these values, the equation becomes \( e^{5.005} \), which approximates to 148.41.This result signifies the average ductile strength expected for the material modeled by this distribution. Understanding expected value helps in anticipating the behavior of various processes and making informed predictions.

Probability calculations involve determining the likelihood of certain events within the framework of a given statistical distribution. When dealing with lognormal distributions, these calculations often require conversion to a standard normal distribution to make use of standard probability tables.For instance, if we want to find the probability that the ductile strength \( X \) is greater than 125, denoted as \( P(X > 125) \), we first convert \( X \) to a normal random variable. The conversion uses the formula:\[ Z = \frac{\log(125) - \mu}{\sigma} \]Substituting \( \mu = 5 \) and \( \sigma = 0.1 \), we compute \( Z \approx -0.34 \). Using the standard normal distribution table, \( P(Z > -0.34) \approx 0.6331 \). Thus, there's approximately a 63.31% chance that a sample has a strength over 125.Similarly, for a range such as \( P(110 \leq X \leq 125) \), calculate two \( Z \) values: one for 110 and one for 125. The difference between these probabilities gives the result for this interval, further analyzing the behavior of this distribution.

Variance is another key concept in statistics, representing the degree of spread in a set of data. It measures how much individual numbers in a dataset vary from the expected value. For the lognormal distribution, variance is found via:\[ V(X) = (e^{\sigma^2} - 1)e^{2\mu + \sigma^2} \]Using the problem's parameters, \( \mu = 5 \) and \( \sigma = 0.1 \), we calculate this variance:\[ V(X) = (e^{0.01} - 1)e^{10.01} \approx 220.32 \]This result shows us the variability of the ductile strength around its mean value. A higher variance indicates a wider spread of values around the mean, suggesting more unpredictability. Understanding variance allows us to gauge the reliability and consistency of the material's strength in practical applications.