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The expected value, often referred to as the mean of a random variable, serves as a measure of the central tendency for a probability distribution. It provides an idea of the average outcome we might expect if we were to repeat an experiment a large number of times. In our exercise, the random variable is represented by \(Y\), which indicates the number of moving violations a car insurance holder might have. To calculate the expected value \(E(Y)\), we multiply each possible value of \(Y\) by its corresponding probability from the probability mass function (PMF) and sum them up.Here's how the formula to calculate the expected value looks: - \(E(Y) = \sum_{i=0}^{n} y_i \cdot p(y_i)\)For this specific problem, it involves: - Note all potential moving violation counts the driver might have: 0, 1, 2, 3. - Use their accompanying probabilities from the PMF, which are 0.60, 0.25, 0.10, and 0.05, respectively.By computing each term’s product and summing them, we get \(E(Y) = 0 \times 0.60 + 1 \times 0.25 + 2 \times 0.10 + 3 \times 0.05\). Upon evaluation, this yields an expected value of 0.60 violations.

A random variable is a mapping from possible outcomes of a statistical experiment to numerical values. It helps quantify outcomes in a way that facilitates statistical analysis. Random variables can be either discrete or continuous depending on the nature of the data they represent. In our context, \(Y\) is the discrete random variable.- **Discrete vs. Continuous:** - Discrete: takes finite or countably infinite values (like our example, 0 to 3 violations). - Continuous: takes uncountably infinite values, often any real number within a range (e.g., weight, height).The random variable \(Y\) denotes the number of moving violations within the past three years obtained by an individual with car insurance. Each possible outcome (0, 1, 2, or 3 violations) has a chance of occurring, described by the probability mass function (PMF).This variable helps in assessing risks associated with an insurance holder and allows actuaries to compute expected costs and surcharges accordingly through associated functions like \(100Y^2\) in this problem.

A probability distribution gives a comprehensive view of how the probabilities are spread over possible outcomes for a random variable. For discrete random variables, like \(Y\) in our exercise, the probability mass function (PMF) specifically outlines this distribution.In our scenario, the PMF is provided as:- \(\begin{array}{l|cccc}y & 0 & 1 & 2 & 3 \\hlinep(y) & 0.60 & 0.25 & 0.10 & 0.05\end{array}\)This means: - The probability that no violations have occurred (\(Y=0\)) is 0.60. - For 1 violation (\(Y=1\)), it is 0.25.- For 2 violations (\(Y=2\)), it is 0.10.- For 3 violations (\(Y=3\)), it is 0.05.Understanding this distribution is critical for calculating expected values and other statistical measures. It allows us to make informed predictions and decisions, integrating that into strategies like risk assessment and premium calculations in the insurance industry.