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In probability and statistics, a random variable is a numerical description of the outcomes of a random process or experiment. In our exercise, when rolling two dice, we define the random variable \(X\) as the absolute difference in the number of dots that appear on the top faces of the dice. A random variable can take on various discrete values depending on the outcome of the experiment. Here, the possible values for \(X\) are {0, 1, 2, 3, 4, 5}. Each of these values represents a different scenario in which the difference in dice rolls leads to a particular result. Understanding random variables is crucial because they form the foundation for probability distributions, allowing us to study different possible outcomes and their probabilities.

The mean, also known as the expected value, provides us with an average outcome of the random variable over a large number of trials. For our random variable \(X\), the mean \(\mu\) is calculated using the formula:\[\mu = \sum (x \cdot P(x))\]This basically involves multiplying each value of the random variable by its probability, then adding all those products. It gives us a central value around which the other outcomes are distributed. The standard deviation \(\sigma\) measures the amount of variation or dispersion in the set of values. It is computed from the variance \(\sigma^2\), using the formula:\[\sigma = \sqrt{\sigma^2}\]Variance is calculated as the average of the squared differences from the mean. A higher standard deviation indicates that the data points are spread out over a wider range of values, whereas a lower standard deviation indicates that they are closer to the mean.

A discrete probability distribution represents the probabilities of all possible values of a discrete random variable. In the context of our exercise, it involves creating a probability distribution for the random variable \(X\), derived from rolling two dice. Each value that \(X\) can take corresponds to a probability, calculated by dividing the number of successful occurrences by the total number of outcomes. For \(X\), those probabilities are:- \(P(X = 0) = \frac{1}{6}\)- \(P(X = 1) = \frac{5}{18}\)- \(P(X = 2) = \frac{2}{9}\)- \(P(X = 3) = \frac{1}{6}\)- \(P(X = 4) = \frac{1}{9}\)- \(P(X = 5) = \frac{1}{18}\)This distribution helps us understand the likelihood of each potential outcome, so we can predict future probabilities based on past events. Knowing how to build and interpret these distributions helps in various fields such as risk analysis, finance, and operations management.