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In statistics, a random variable is a numerical outcome of a random phenomenon. It assigns a numeric value to each possible outcome in a sample space. For instance, when considering Ana's Skee Ball scores, we can define a random variable \( X \) that represents the possible scores Ana obtains in a game. The possible values for \( X \) are tied to specific probabilities of occurrence. For instance, there's a 32% chance Ana scores 10 points.Random variables can be discrete or continuous. Ana's score, \( X \), is a discrete random variable, as it takes on specific, countable values. When translating these scores into tickets, another random variable, \( T \), emerges. Here, \( T = \frac{X}{10} \) indicates tickets earned, also being discrete.The relationship between \( X \) and \( T \) reflects a transformation of the original variable, demonstrating how one random variable can influence another by a specific mathematical rule.

The expected value of a random variable gives an idea of its "average" numerical outcome if an experiment were repeated many times. For Ana's game, the expected number of points, \( \mu_X \), is 23.8, indicating what she'd average per roll over many tries.Moving to tickets, the expected value \( \mu_T \) is derived from \( \mu_X=23.8 \) by scaling it. Since one ticket is awarded per 10 points, the expected value \( \mu_T \) becomes \( \frac{23.8}{10} = 2.38 \). This suggests Ana earns about 2.38 tickets per roll on average.Interpreting \( \mu_T \), we understand the concept as a long-term average: should Ana play indefinitely, the mean number of tickets per game will stabilize near 2.38, aligning her occasional successes with more common lower scores.

Standard deviation measures the variability or spread of a set of values around the expected value. For Ana’s scores, the standard deviation is \( \sigma_X = 12.63 \). This number tells us how much individual scores differ from Ana's average score of 23.8 points.Converting scores into tickets, the standard deviation becomes \( \sigma_T = \frac{12.63}{10} = 1.263 \), showing the spread of tickets she collects when compared to \( \mu_T \), her average of 2.38 tickets.Understanding \( \sigma_T \) helps in realizing how ticket counts vary: most throws will result in a number close to 2.38 tickets but can fluctuate by about 1.263 tickets. In simpler terms, while Ana may often gain around 2 tickets, occasional surprises may see her earn more or fewer.