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) $4 \leq T \leq 5$.
\(\boxed{\text{Final Answer: (B) } 1 \leq T < 2}\)
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 :\(\boxed{\text{Final Answer: (B) } 1 \leq T < 2}\)