Step-by-step explanation:

1. Understanding the problem: We need to find the 80th percentile (q₈) of a continuous random variable X with probability density function (PDF) f_X(x) = (1/5)x² for 0 < x < 3.

2. Percentile definition: The 80th percentile q₈ satisfies: $\Pr(X < q_8) = 0.8$ This is equivalent to: $F(q_8) = 0.8$ where F(x) is the cumulative distribution function (CDF).

3. Finding the CDF: $F(x) = \int_0^x f_X(t)\,dt = \int_0^x \frac{1}{5}t^2\,dt$ = \frac{1}{5} \cdot \frac{t^3}{3} \Big|_0^x = \frac{x^3}{15}$`$4`. **Setting up the equation:** F(q_8) = \frac{(q_8)^3}{15} = 0.8$`$5. **Solving for q₈:** $$(q_8)^3 = 15 \times 0.8 = 12$$ $$q_8 = \sqrt[3]{12} = 12^{1/3}$$6`. **Calculating the value:**$$12^{1/3} \approx 2.289428... \approx 2.29$$

Verification:

• 2.25³ = 11.390625 (too small)
• 2.34³ = 12.812904 (too large)
• 2.67³ = 19.034163 (too large)
• 2.29³ = 12.009389 (closest to 12)

Therefore, the 80th percentile is approximately 2.29, which corresponds to option D.