Let X have PDF f(x) = x/24 for 1 6).

P(X > 6) is the integral of the PDF over (6,7). For a continuous random variable, probabilities are found by integrating the PDF over the desired interval; the PDF must be nonnegative and integrate to 1 over its support. Compute the probability: f(x) = x/24 on 1 6) = ∫_{6}^{7} (x/24) dx. Find the antiderivative: ∫ x/24 dx = x^2/(48). Evaluate the definite integral: P(X > 6) = [x^2/48]_{6}^{7} = (7^2/48) - (6^2/48) = (49/48) - (36/48) = (49 - 36)/48 = 13/48. Convert to decimal: 13/48 ≈ 0.2708333… ≈ 0.2708. As a consistency check, verify the PDF is normalized: ∫_{1}^{7} (x/24) dx = [x^2/48]_{1}^{7} = (49 - 1)/48 = 48/48 = 1, so the PDF is valid. Why other options are incorrect: - 0.0417 ≈ 1/24: this would be f(1) or a mistaken single-point value; probabilities for intervals require integration, not evaluating the density at a point. - 0.25 = 1/4: no step in the correct integration produces 1/4; this is not equal to 13/48. - 0.2917 ≈ 14/48 = 7/24: this corresponds to a mistaken numerator (14 instead of 13) and does not match the correct integral. Conclusion: P(X > 6) = 13/48 ≈ 0.2708, which matches the provided correct answer 0.2708.