To solve the problem, we need to calculate the expectation $E(X)$ and the variance $\text{Var}(X)$ of the random variable $X$ using the provided probability distribution.

(a) Finding the Expectation $E(X)$

The expectation (or mean) $E(X)$ of a random variable $X$ is calculated using the formula: $E(X) = \sum_{i} x_i \cdot P(X = x_i)$ Given the table:

\hline \text{Value } x_i \text{ of } X & P(X = x_i) \\ \hline -1 & 0.10 \\ 0 & 0.15 \\ 1 & 0.35 \\ 2 & 0.15 \\ 3 & 0.25 \\ \hline \end{array}$$ Now, compute $$E(X)$$: $$E(X) = (-1) \cdot 0.10 + 0 \cdot 0.15 + 1 \cdot 0.35 + 2 \cdot 0.15 + 3 \cdot 0.25$$ $$E(X) = -0.10 + 0 + 0.35 + 0.30 + 0.75 = 1.30$$ So, $$E(X) = 1.30$$. ### (b) Finding the Variance $$\text{Var}(X)$$ The variance $$\text{Var}(X)$$ is calculated using the formula: $$\text{Var}(X) = E(X^2) - [E(X)]^2$$ First, we need to calculate $$E(X^2)$$, where: $$E(X^2) = \sum_{i} x_i^2 \cdot P(X = x_i)$$ Compute $$E(X^2)$$: $$E(X^2) = (-1)^2 \cdot 0.10 + 0^2 \cdot 0.15 + 1^2 \cdot 0.35 + 2^2 \cdot 0.15 + 3^2 \cdot 0.25$$ $$E(X^2) = 1 \cdot 0.10 + 0 + 1 \cdot 0.35 + 4 \cdot 0.15 + 9 \cdot 0.25$$ $$E(X^2) = 0.10 + 0 + 0.35 + 0.60 + 2.25 = 3.30$$ Now, compute the variance $$\text{Var}(X)$$: $$\text{Var}(X) = E(X^2) - [E(X)]^2 = 3.30 - (1.30)^2 = 3.30 - 1.69 = 1.61$$ So, $$\text{Var}(X) = 1.61$$. ### Summary - $$E(X) = 1.30$$ - $$\text{Var}(X) = 1.61$$ Would you like to go into further details, or do you have any questions? ### Related Questions: 1. What is the interpretation of the expectation $$E(X)$$ in this context? 2. How would the variance change if one of the probabilities were adjusted? 3. What is the significance of variance in probability theory? 4. Can you compute the standard deviation of $$X$$ from the variance? 5. How do you calculate the expectation for a continuous random variable? 6. Why is it important to square the deviation when calculating variance? 7. What are the properties of the expectation operator $$E(\cdot)$$? 8. How does variance relate to the spread of the distribution? **Tip:** Understanding the expectation and variance of a random variable helps in predicting outcomes and assessing the variability of the data.