Answer: 126.10

## Explanation To find the difference between the 70th and 30th percentiles, we need to calculate each percentile separately using the given probability density function (PDF). ### Step 1: Find the cumulative distribution function (CDF) The PDF is given as: $$f_X(x) = \frac{2(200)^2}{x^3} \text{ for } x > 200$$ To find the CDF, we integrate the PDF from 200 to x: $$F(x) = \int_{200}^{x} \frac{2(200)^2}{t^3} dt$$ ### Step 2: Calculate the 30th percentile (q₃₀) We need to find q₃₀ such that F(q₃₀) = 0.3: $$\int_{200}^{q_{30}} \frac{2(200)^2}{t^3} dt = 0.3$$ Solving the integral: $$\int \frac{2(200)^2}{t^3} dt = -\frac{(200)^2}{t^2} + C$$ So: $$\left[-\frac{(200)^2}{t^2}\right]_{200}^{q_{30}} = 0.3$$ $$-\frac{(200)^2}{(q_{30})^2} + \frac{(200)^2}{(200)^2} = 0.3$$ $$-\frac{(200)^2}{(q_{30})^2} + 1 = 0.3$$ $$1 - \frac{(200)^2}{(q_{30})^2} = 0.3$$ $$\frac{(200)^2}{(q_{30})^2} = 0.7$$ $$(q_{30})^2 = \frac{(200)^2}{0.7}$$ $$q_{30} = \frac{200}{\sqrt{0.7}} \approx 239.05$$ ### Step 3: Calculate the 70th percentile (q₇₀) Similarly, for q₇₀ where F(q₇₀) = 0.7: $$\int_{200}^{q_{70}} \frac{2(200)^2}{t^3} dt = 0.7$$ $$1 - \frac{(200)^2}{(q_{70})^2} = 0.7$$ $$\frac{(200)^2}{(q_{70})^2} = 0.3$$ $$(q_{70})^2 = \frac{(200)^2}{0.3}$$ $$q_{70} = \frac{200}{\sqrt{0.3}} \approx 365.15$$ ### Step 4: Calculate the difference $$q_{70} - q_{30} = 365.15 - 239.05 = 126.10$$ Therefore, the difference between the 70th and 30th percentiles is **126.10**, which corresponds to option B. **Key Concepts:** - Percentile calculation from a continuous probability distribution - Integration of PDF to obtain CDF - Solving for quantiles using the inverse CDF method - This is a Pareto distribution with shape parameter α = 2 and scale parameter θ = 200

Author: Nikitesh Somanthe