1988 AMC 12 第 29 题

Question

把三位朋友的体重

(y)

对身高

(x)

作图，得到点

(x_1,y_1)

、

(x_2,y_2)

、

(x_3,y_3)

。若

\begin{aligned} x_1&\lt x_2\lt x_3,\\ x_3-x_2&=x_2-x_1 \end{aligned}\text{，}

则下列哪一个式子一定是数据最佳拟合直线的斜率？“最佳拟合”是指数据点到直线的竖直距离的平方和小于对任何其他直线所得的平方和。

You plot weight

(y)

against height

(x)

for three of your friends and obtain the points

(x_1,y_1),

(x_2,y_2),

(x_3,y_3).

If

\begin{aligned} x_1&\lt x_2\lt x_3,\\ x_3-x_2&=x_2-x_1, \end{aligned}

which of the following is necessarily the slope of the line which best fits the data? “Best fits” means that the sum of the squares of the vertical distances from the data points to the line is smaller than for any other line.

\frac{y_3-y_1}{x_3-x_1}

\frac{(y_2-y_1)-(y_3-y_2)}{x_3-x_1}

\frac{2y_3-y_1-y_2}{2x_3-x_1-x_2}

\frac{y_2-y_1}{x_2-x_1}+\frac{y_3-y_2}{x_3-x_2}

以上均不是

none of these

答案：A

知识点：least squares 斜率 symmetric coordinates

难度评级：2480

小提示：

平移并缩放

x

坐标，使其变为

-1,0,1

Translate and scale the

x

-coordinates to

-1,0,1

大提示：

对于

x

值对称的最小二乘直线，利用协方差的分子计算斜率

For a least-squares line through symmetric

x

-values, compute the slope from the covariance numerator

解答：

进行平移和缩放，使

x

坐标为

-d,0,d

。它们的平均值为

0

，所以最小二乘直线的斜率为

\begin{aligned} \frac{(-d)y_1+0y_2+dy_3}{d^2+0+d^2} &=\frac{y_3-y_1}{2d}\\ &=\frac{y_3-y_1}{x_3-x_1} \end{aligned}\text{。}

因此，正确答案是 A。

Translate and scale so the

x

-coordinates are

-d,0,d.

Their mean is

0,

so the least-squares slope is

\begin{aligned} \frac{(-d)y_1+0y_2+dy_3}{d^2+0+d^2} &=\frac{y_3-y_1}{2d}\\ &=\frac{y_3-y_1}{x_3-x_1}. \end{aligned}

Thus the correct answer is A.

1988 AMC 12 第 29 题

答案：A

难度评级：2480

小提示：

大提示：

解答：

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