2. Let $Q_{1}$ be the mean of $x, Q_{2}$ the sample standard deviation of $x, Q_{3}$ the mean of $y$, $Q_{4}$ the sample standard deviation of $y$, and $Q_{5}$ the linear correlation coefficient of $x$ and $y$, where the sequences $x$ and $y$ are as follows: \[ \begin{array}{l} x:-48,-52,-89,35,-18,-62,-96,-47,84,-93,98,38, \\ y:-82,-7,-63,14,-57,91,-15,36,6,-77,79,-52 . \end{array} \] Let $Q=\ln \left(3+\left|Q_{1}\right|+2\left|Q_{2}\right|+3\left|Q_{3}\right|+4\left|Q_{4}\right|+5\left|Q_{5}\right|\right)$. Then $T=5 \sin ^{2}(100 Q)$ satisfies: (A) $0 \leq T<1$. (B) $1 \leq T<2$. (C) $2 \leq T<3$. (D) $3 \leq T<4$. (E) $4 \leq T \leq 5$.

Step 1 ：Given the sequences x and y as follows: \[x: -48, -52, -89, 35, -18, -62, -96, -47, 84, -93, 98, 38\] and \[y: -82, -7, -63, 14, -57, 91, -15, 36, 6, -77, 79, -52\]

Step 2 ：Calculate the mean of x, denoted as \(Q_{1}\), which is approximately -20.833333333333332

Step 3 ：Calculate the sample standard deviation of x, denoted as \(Q_{2}\), which is approximately 68.2852750443226

Step 4 ：Calculate the mean of y, denoted as \(Q_{3}\), which is approximately -10.583333333333334

Step 5 ：Calculate the sample standard deviation of y, denoted as \(Q_{4}\), which is approximately 58.43015773839168

Step 6 ：Calculate the linear correlation coefficient of x and y, denoted as \(Q_{5}\), which is approximately 0.3717132891359829

Step 7 ：Substitute \(Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}\) into the formula \[Q = \ln(3 + |Q_{1}| + 2|Q_{2}| + 3|Q_{3}| + 4|Q_{4}| + 5|Q_{5}|)\] to get \(Q\) which is approximately 6.058499358094586

Step 8 ：Substitute \(Q\) into the formula \[T = 5 \sin^{2}(100Q)\] to get \(T\) which is approximately 1.0557590696303

Step 9 ：Finally, we can see that the value of \(T\) falls into the range \(1 \leq T < 2\)

Step 10 ：\(\boxed{\text{Final Answer: (B) } 1 \leq T < 2}\)