《2025年小题狂做高中数学选择性必修第三册人教版》

1. [2025 湖北恩施期末]某田径协会组织开展竞走的步长和步频之间关系的课题研究，得到相应的试验数据，如下表：

(1)若步频和步长近似为线性相关关系，当$m = 101$时，$\sum_{i = 1}^{5}x_{i}^{2}=0.451$，$\sum_{i = 1}^{5}x_{i}y_{i}=$
150.56，根据表中数据，求出$Y$关于$x$的经验回归方程；
(2)记$\hat{e}_{i}=y_{i}-\hat{y}_{i}=y_{i}-\hat{b}x_{i}-\hat{a}$，其中$y_{i}$为观测值，$\hat{y}_{i}$为预测值，$\hat{e}_{i}$为对应$(x_{i},y_{i})$的残差，根据表中数据，若得出$Y$关于$x$的经验回归方程为$\hat{y}=600x+\hat{a}$，且计算出在样本点$(0.30,99)$处的残差为$-1.8$，求实数$m$的值.
附：在经验回归方程$\hat{y}=\hat{b}x+\hat{a}$中，$\hat{b}=\frac{\sum_{i = 1}^{n}x_{i}y_{i}-n\bar{x}\bar{y}}{\sum_{i = 1}^{n}x_{i}^{2}-n\bar{x}^{2}}$，$\hat{a}=\bar{y}-\hat{b}\bar{x}$.

答案： 解：
(1)$\bar{x} = \frac{1}{5}\sum_{i = 1}^{5}x_{i} = 0.3$,$\bar{y} = \frac{1}{5}\sum_{i = 1}^{5}y_{i} = 100$，又由$\sum_{i = 1}^{5}x_{i}^{2} = 0.451$,$\sum_{i = 1}^{5}x_{i}y_{i} = 150.56$，可得$\hat{b} = \frac{\sum_{i = 1}^{5}x_{i}y_{i} - 5\bar{x}\bar{y}}{\sum_{i = 1}^{5}x_{i}^{2} - 5\bar{x}^{2}} =$
$\frac{150.56 - 5 × 0.3 × 100}{0.451 - 5 × {0.3}^{2}} = 560$，则$\hat{a} = 100 - 560 × 0.3 =$
$- 68$，所以经验回归方程为$\hat{y} = 560x - 68$.
(2)由题设，将残差带入经验回归方程可得$99 - \hat{y} = 99 -$
$(600 × 0.3 + \hat{a}) = - 1.8$，解得$\hat{a} = - 79.2$.因为$\bar{x} =$
$0.3$,$\bar{y} = \frac{90 + 95 + 99 + m + 115}{5} = \frac{399 + m}{5}$，
所以$\frac{399 + m}{5} = 600 × 0.3 - 79.2$，解得$m = 105$.