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