1. **Problem statement:** Given the probability density function (pdf) of a continuous random variable $X$: $$f(x) = \begin{cases} \frac{x^3}{20} & 1 \leq x \leq 3 \\ 0 & \text{otherwise} \end{cases}$$ We are asked to find: (a) The expected value of $X^2$, $E(X^2)$, using algebraic integration. (b) The variance $\mathrm{Var}(X)$ given $E(X) = 2.42$. 2. **Formula and rules:** - The expected value of a function $g(X)$ is: $$E(g(X)) = \int_{-\infty}^{\infty} g(x) f(x) \, dx$$ - Variance is: $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2$$ 3. **Step (a): Find $E(X^2)$** We use $g(x) = x^2$: $$E(X^2) = \int_1^3 x^2 \cdot \frac{x^3}{20} \, dx = \int_1^3 \frac{x^5}{20} \, dx$$ Calculate the integral: $$E(X^2) = \frac{1}{20} \int_1^3 x^5 \, dx = \frac{1}{20} \left[ \frac{x^6}{6} \right]_1^3 = \frac{1}{20} \left( \frac{3^6}{6} - \frac{1^6}{6} \right)$$ Evaluate powers: $$3^6 = 729, \quad 1^6 = 1$$ So: $$E(X^2) = \frac{1}{20} \cdot \frac{729 - 1}{6} = \frac{1}{20} \cdot \frac{728}{6} = \frac{728}{120}$$ Simplify fraction by dividing numerator and denominator by 4: $$\frac{728}{120} = \frac{\cancel{728}/4}{\cancel{120}/4} = \frac{182}{30}$$ Further simplify by dividing numerator and denominator by 2: $$\frac{182}{30} = \frac{\cancel{182}/2}{\cancel{30}/2} = \frac{91}{15}$$ Convert to decimal: $$E(X^2) = \frac{91}{15} \approx 6.0667$$ 4. **Step (b): Find $\mathrm{Var}(X)$** Given $E(X) = 2.42$, use variance formula: $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = 6.0667 - (2.42)^2$$ Calculate square: $$(2.42)^2 = 5.8564$$ Subtract: $$\mathrm{Var}(X) = 6.0667 - 5.8564 = 0.2103$$ Rounded to 3 significant figures: $$\mathrm{Var}(X) = 0.210$$