How to Solve Continuous Random Variables — Univariate RV — density normalization and tail probability Questions on Exam P
Sample Practice Problem
ID: #OYLXD
Let \(X\) be a delivery-size index with Pareto density \(3x^{-4}\) for \(x>1\). Calculate \(\Pr(X<\frac{9}{5}\mid X\ge \tfrac{3}{2})\).
📖 Worked Solution & Strategy
Integrating Pareto density \(3x^{-4}\) on \((x, \infty)\) gives survival function \(S_X(x) = x^{-3}\) for \(x > 1\). The probability of the interval \([\frac{3}{2}, \frac{9}{5})\) is \(\Pr(\frac{3}{2} \le X < \frac{9}{5}) = S_X(\frac{3}{2}) - S_X(\frac{9}{5})\). Conditioning on \(X \ge \frac{3}{2}\), \(\Pr(X < \frac{9}{5} \mid X \ge \frac{3}{2}) = \frac{\Pr(\frac{3}{2} \le X < \frac{9}{5})}{\Pr(X \ge \frac{3}{2})} = \frac{S_X(\frac{3}{2}) - S_X(\frac{9}{5})}{S_X(\frac{3}{2})} = 1 - \frac{S_X(\frac{9}{5})}{S_X(\frac{3}{2})} = \frac{91}{216}\). The evaluated result is \(\frac{91}{216}\), which matches the correct option.