1. The problem is to understand the mean and variance of a discrete probability distribution. 2. The mean (or expected value) $\mu$ of a discrete random variable $X$ with possible values $x_i$ and probabilities $p_i$ is given by the formula: $$\mu = E(X) = \sum_i x_i p_i$$ This means you multiply each value by its probability and add all those products. 3. The variance $\sigma^2$ measures how spread out the values are around the mean. It is given by: $$\sigma^2 = Var(X) = E\left[(X - \mu)^2\right] = \sum_i (x_i - \mu)^2 p_i$$ This means you find the squared difference between each value and the mean, multiply by the probability, and sum. 4. Alternatively, variance can be computed using: $$\sigma^2 = E(X^2) - (E(X))^2 = \sum_i x_i^2 p_i - \mu^2$$ This is often easier to calculate. 5. Important rules: - Probabilities $p_i$ must satisfy $0 \leq p_i \leq 1$ and $\sum_i p_i = 1$. - Mean gives the center or average value. - Variance gives the spread or variability. 6. To compute mean and variance, list all $x_i$ and $p_i$, calculate $\mu$, then use either formula for variance. This explanation covers the formulas and concepts for mean and variance of a discrete probability distribution.