课程： 5.7 线性内插和外推法

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线性内插和外推

Linear Interpolation and Extrapolation
::线性内插和外推

Katja’s sales figures were trending downward quickly at first, and she used a line of best fit to describe the numbers. But now they seem to be decreasing more slowly, and fitting the line less and less accurately. How can she make a more accurate prediction of what next week’s sales will be?
::Katja的销售数字一开始呈快速下降趋势,她用一条最合适的线来描述数字。但现在,这些数字似乎正在缓慢下降,并且越来越不准确。她怎样才能更准确地预测下周的销售情况呢?

The line of “best fit” is a good method to use when predicting values if the relationship between the dependent and the independent variables is linear. In this section you will learn other methods that are useful even when the relationship isn’t linear.
::如果依赖变量和独立变量之间的关系是线性,那么“最合适”的线条是预测数值的好方法。在本节中,你将学习其他有用的方法,即使关系不是线性。

Linear Interpolation
::线线内插

We use linear interpolation to fill in gaps in our data—that is, to estimate values that fall in between the values we already know. To do this, we use a straight line to connect the known data points on either side of the unknown point, and use the equation of that line to estimate the value we are looking for.
::我们用线性内插来填补数据中的空白,即估计我们已知值之间的数值。要做到这一点,我们用一条直线连接未知点两侧已知数据点,并用这条线的方程式来估计我们所要寻找的价值。

The following table shows the median ages of first marriage for men and women, as gathered by the U.S. Census Bureau.
::下表显示美国人口普查局收集的男女初婚平均年龄。

Year	Median age of males	Median age of females
1890	26.1	22.0
1900	25.9	21.9
1910	25.1	21.6
1920	24.6	21.2
1930	24.3	21.3
1940	24.3	21.5
1950	22.8	20.3
1960	22.8	20.3
1970	23.2	20.8
1980	24.7	22.0
1990	26.1	23.9
2000	26.8	25.1

Estimate the median age for the first marriage of a male in the year 1946.
::估计1946年男子第一次结婚的中位年龄。

We connect the two points on either side of 1946 with a straight line and find its equation. Here’s how that looks on a scatter plot:
::我们把1946年两边的两点与一条直线连接起来,找到它的等式。

We find the equation by plugging in the two data points:
::我们通过插入两个数据点来找到等式:

\begin{aligned} m & = \frac{22.8 - 24.3}{1950 - 1940} = \frac{- 1.5}{10} = - 0.15 \\ y & = - 0.15 x + b \\ 24.3 & = - 0.15 (1940) + b \\ b & = 315.3 \end{aligned}

::m=22.8-24.31950-1940}1.5100.15y0.15x+b24.30.15(1940)+bb=3153

Our equation is $\begin{aligned} y = - 0.15 x + 315.3 \end{aligned}$ .
::我们的方程式是YO0.15x+315.3

To estimate the median age of marriage of males in the year 1946, we plug $\begin{aligned} x = 1946 \end{aligned}$ into the equation we just found:
::为了估计1946年男性结婚的中位年龄, 我们将x=1946 插入我们刚刚发现的方程式中:

$\begin{aligned} y = - 0.15 (1946) + 315.3 = 23.4 \end{aligned}$ years old
::y0.15(1946)+315.3=23.4岁

For non-linear data, linear interpolation is often not accurate enough for our purposes. If the points in the data set change by a large amount in the interval you’re interested in, then linear interpolation may not give a good estimate. In that case, it can be replaced by polynomial interpolation, which uses a curve instead of a straight line to estimate values between points. But that’s beyond the scope of this lesson.
::对于非线性数据来说,线性内插往往不足以为我们的目的提供准确性。如果数据集中的点在您感兴趣的时间间隔内变化很大,那么线性内插可能不会给出一个良好的估计。在这种情况下,它可以被多线性内插所取代,多线性内插用曲线而不是直线来估计点之间的值。但这超出了这一教训的范围。

Linear Extrapolation
::线外推法

Linear extrapolation can help us estimate values that are outside the range of our data set. The strategy is similar to linear interpolation: we pick the two data points that are closest to the one we’re looking for, find the equation of the line between them, and use that equation to estimate the coordinates of the missing point.
::线性外推法可以帮助我们估算数据组范围以外的数值。策略类似于线性内推法:我们选择最接近我们所寻找的数据点的两个数据点,找到它们之间的线条方程式,并用这个方程式估计缺失点的坐标。

The winning times for the women’s 100 meter race are given in the following table. Estimate the winning time in the year 2010. Is this a good estimate?
::下表给出了女性100米比赛的胜选时间。估计2010年的胜选时间。这是否是一个良好的估计?

Winner	Country	Year	Time (seconds)
Mary Lines	UK	1922	12.8
Leni Schmidt	Germany	1925	12.4
Gerturd Glasitsch	Germany	1927	12.1
Tollien Schuurman	Netherlands	1930	12.0
Helen Stephens	USA	1935	11.8
Lulu Mae Hymes	USA	1939	11.5
Fanny Blankers-Koen	Netherlands	1943	11.5
Marjorie Jackson	Australia	1952	11.4
Vera Krepkina	Soviet Union	1958	11.3
Wyomia Tyus	USA	1964	11.2
Barbara Ferrell	USA	1968	11.1
Ellen Strophal	East Germany	1972	11.0
Inge Helten	West Germany	1976	11.0
Marlies Gohr	East Germany	1982	10.9
Florence Griffith Joyner	USA	1988	10.5

We start by making a scatter plot of the data; then we connect the last two points on the graph and find the equation of the line.
::我们首先绘制数据散射图; 然后连接图形上最后两点, 找到线的方程式。

\begin{aligned} m & = \frac{10.5 - 10.9}{1988 - 1982} = \frac{- 0.4}{6} = - 0.067 \\ y & = - 0.067 x + b \\ 10.5 & = - 0.067 (1988) + b \\ b & = 143.7 \end{aligned}

::m=10.5-10.91988-1982_0.460.067y0.067x+b10.50.067(1988)+bb=143.7

Our equation is $\begin{aligned} y = - 0.067 x + 143.7 \end{aligned}$ .
::我们的方程式是Y 0.067x+143.7

The winning time in year 2010 is estimated to be:
::2010年的获胜时间估计为:

$\begin{aligned} y = - 0.067 (2010) + 143.7 = 9.03 \end{aligned}$ seconds.
::y0.067(2010)+143.7=9.03秒。

Unfortunately, this estimate actually isn’t very accurate. This example demonstrates the weakness of linear extrapolation; it uses only a couple of points, instead of using all the points like the best fit line method, so it doesn’t give as accurate results when the data points follow a linear pattern . In this particular example, the last data point clearly doesn’t fit in with the general trend of the data, so the of the extrapolation line is much steeper than it would be if we’d used a line of best fit. (As a historical note, the last data point corresponds to the winning time for Florence Griffith Joyner in 1988. After her race she was accused of using performance-enhancing drugs, but this fact was never proven. In addition , there was a question about the accuracy of the timing: some officials said that tail-wind was not accounted for in this race, even though all the other races of the day were affected by a strong wind.)
::不幸的是,这一估计实际上并不十分准确。这个例子显示了线性外推的弱点;它只使用了几个点,而不是使用最合适的线性方法等所有点,因此当数据点遵循线性模式时,它不会给出准确的结果。在这个具体的例子中,最后一个数据点显然不符合数据的总趋势,因此外推线比我们使用最合适的线性线要陡峭得多。 (作为历史的注解,最后一个数据点与1988年佛罗伦萨·格里菲斯·乔伊纳的胜利时间相对应。在她被指责使用提高性能药物的赛跑后,她被指控使用提高性能药物,但这一事实从未被证明。此外,还有一个关于时间准确性的问题:一些官员说,这场比赛没有计算尾风,尽管当天所有其他种族都受到强风的影响。 )

Here's an example of a problem where linear extrapolation does work better than the line of best fit method.
::这里有一个例子线性外推法确实比最合适方法的线性效果更好

A cylinder is filled with water to a height of 73 centimeters. The water is drained through a hole in the bottom of the cylinder and measurements are taken at 2 second intervals. The following table shows the height of the water level in the cylinder at different times.
::气瓶装满的水高度为73厘米,水通过气瓶底部的一个洞排干,测量间隔为2秒,下表显示气瓶中水位在不同时间的高度。

Time (seconds)	Water level (cm)
0.0	73
2.0	63.9
4.0	55.5
6.0	47.2
8.0	40.0
10.0	33.4
12.0	27.4
14.0	21.9
16.0	17.1
18.0	12.9
20.0	9.4
22.0	6.3
24.0	3.9
26.0	2.0
28.0	0.7
30.0	0.1

Here’s what the line of best fit would look like for this data set:
::以下是该数据集最合适的直线:

Notice that the data points don’t really make a line, and so the line of best fit still isn’t a terribly good fit. Just a glance tells us that we’d estimate the water level at 15 seconds to be about 27 cm, which is more than the water level at 14 seconds. That’s clearly not possible! Similarly, at 27 seconds we’d estimate the water to have all drained out, which it clearly hasn’t yet.
::注意数据点并没有真正划出一条线,因此最合适的水线还是不太合适。只要一眼就能看出,我们估计水位为15秒,大约为27厘米,比14秒的水位高出27厘米。这显然是不可能的。同样,在27秒,我们估计水位已经全部排干,但显然还没有排干。

So let’s see what happens if we use linear extrapolation and interpolation instead. First, here are the lines we’d use to interpolate between 14 and 16 seconds, and between 26 and 28 seconds.
::让我们来看看如果我们使用线性外推法和内插法来代替的话会发生什么。首先,我们使用的是14至16秒之间的内插线,26至28秒之间的内插线。

a) Find the water level at time 15 second.
:a) 在15秒时找到水位。

The slope of the line between points (14, 21.9) and (16, 17.1) is $\begin{aligned} m = \frac{17.1 - 21.9}{16 - 14} = \frac{- 4.8}{2} = - 2.4 \end{aligned}$ . So $\begin{aligned} y = - 2.4 x + b \Rightarrow 21.9 = - 2.4 (14) + b \Rightarrow b = 55.5, \end{aligned}$ and the equation is $\begin{aligned} y = - 2.4 x + 55.5 \end{aligned}$ .
::点(14,21.9)和点(16,17.1)之间的斜坡为m=17.1-21.916-144.822.4。y2.4x+21.92.4(14)+bb=55.5,等式为y2.4x+55.5。

Plugging in $\begin{aligned} x = 15 \end{aligned}$ gives us $\begin{aligned} y = - 2.4 (15) + 55.5 = 19.5 c m \end{aligned}$ .
::插入 x=15 中显示 y2.4 (15)+55.5=19.5 cm。

b) Find the water level at time 27 second.
:b) 在27秒的时间找到水位。

The slope of the line between points (26, 2) and (28, 0.7) is $\begin{aligned} m = \frac{0.7 - 2}{28 - 26} = \frac{- 1.3}{2} = - .65 \end{aligned}$ , so $\begin{aligned} y = - .65 x + b \Rightarrow 2 = - .65 (26) + b \Rightarrow b = 18.9, \end{aligned}$ and the equation is $\begin{aligned} y = - .65 x + 18.9 \end{aligned}$ .
::点(26, 2)和点(28, 0.7)之间的斜坡为m=0.7-228-2261.32.65, 所以y. 65x+b22.65(26)+bb=18.9, 方程为y65x+18.9。

Plugging in $\begin{aligned} x = 27 \end{aligned}$ , we get $\begin{aligned} y = - .65 (27) + 18.9 = 1.35 c m \end{aligned}$ .
::插入x=27,我们得到y.65(27)+18.9=1.35厘米。

c) What would be the original height of the water in the cylinder if the water takes 5 extra seconds to drain? (Find the height at time of -5 seconds)
:c) 如果水需要额外5秒钟来排水,圆柱体中水的最初高度是什么? (在5秒时找到高度)

Finally, we can use extrapolation to estimate the height of the water at -5 seconds. The slope of the line between points (0, 73) and (2, 63.9) is $\begin{aligned} m = \frac{63.9 - 73}{2 - 0} = \frac{- 9.1}{2} = - 4.55 \end{aligned}$ , so the equation of the line is $\begin{aligned} y = - 4.55 x + 73 \end{aligned}$ .
::最后,我们可以使用外推法估计水的高度为-5秒。点(0,73)和点(2,63.9)之间的斜度是m=63.9-732-09.124.55,所以线的方程式是y4.55x+73。

Plugging in $\begin{aligned} x = - 5 \end{aligned}$ gives us $\begin{aligned} y = - 4.55 (- 5) + 73 = 95.75 c m \end{aligned}$ .
::插入 x5 显示 y4.55(-5)+73=95.75厘米。

Example
::示例示例示例示例

The Center for Disease Control collects information about the health of the American people and behaviors that might lead to bad health. The following table shows the percent of women who smoke during pregnancy.
::疾病控制中心收集了有关美国人民健康和可能导致健康不良的行为的信息。下表显示了怀孕期间吸烟妇女的比例。

Year	Percent of pregnant women smokers
1990	18.4
1991	17.7
1992	16.9
1993	15.8
1994	14.6
1995	13.9
1996	13.6
2000	12.2
2002	11.4
2003	10.4
2004	10.2

Example 1
::例1

Estimate the percentage of pregnant women that were smoking in the year 1998.
::估计1998年吸烟的孕妇百分比。

We connect the two points on either side of 1998 with a straight line and find its equation. Here’s how that looks on a scatter plot:
::我们把1998年两边的两点与一条直线连接起来,找到它的等式。

We find the equation by plugging in the two data points:
::我们通过插入两个数据点来找到等式:

\begin{aligned} m & = \frac{12.2 - 13.6}{2000 - 1996} = \frac{- 1.4}{4} = - 0.35 \\ y & = - 0.35 x + b \\ 12.2 & = - 0.35 (2000) + b \\ b & = 712.2 \end{aligned}

::m=12.2-13.62000-1996_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Our equation is $\begin{aligned} y = - 0.35 x + 712.2 \end{aligned}$ .
::我们的方程式是y=0.35x+712.2

To estimate the percentage of pregnant women who smoked in the year 1998, we plug $\begin{aligned} x = 1998 \end{aligned}$ into the equation we just found:
::为了估计1998年吸烟的孕妇百分比,我们把x=1998插进我们刚刚发现的方程式中:

\begin{aligned} y = - 0.35 (1998) + 712.2 = 12.9 % \end{aligned}

::y0.35(1998年)+712.2=12.9%

Review
::回顾

Use the data from Example 1 ( Median age at first marriage ) to estimate the age at marriage for females in 1946. Fit a line, by hand, to the data before 1970.
::使用例1(初婚时的中间年龄)的数据来估计1946年女性的结婚年龄。

Use the data from Example 1 ( Median age at first marriage ) to estimate the age at marriage for females in 1984. Fit a line, by hand, to the data from 1970 on in order to estimate this accurately.
::利用例1(初婚时的中间年龄)的数据来估计1984年女性的结婚年龄。为了准确估计这一年龄,用手对1970年的数据进行直线调整。

Use the data from Example 1 ( Median age at first marriage ) to estimate the age at marriage for males in 1995. Use linear interpolation between the 1990 and 2000 data points.
::使用例1(初婚时的中间年龄)的数据估计1995年男性的结婚年龄,使用1990年至2000年的线性内插数据点。

Use the data from Example 2 ( Pregnant women and smoking ) to estimate the percentage of pregnant smokers in 1997. Use linear interpolation between the 1996 and 2000 data points.
::使用例2(孕妇和吸烟)的数据估算1997年怀孕吸烟者的百分比,使用1996年和2000年数据点之间的线性内插。

Use the data from Example 2 ( Pregnant women and smoking ) to estimate the percentage of pregnant smokers in 2006. Use linear extrapolation with the final two data points.
::使用例2(孕妇和吸烟)的数据估算2006年怀孕吸烟者的百分比,使用最后两个数据点的线性外推法。

Use the data from Example 3 ( Winning times ) to estimate the winning time for the female 100-meter race in 1920. Use linear extrapolation because the first two or three data points have a different slope than the rest of the data.
::使用例3(Winning times)中的数据来估计1920年女性100米赛跑的获胜时间。使用线性外推法是因为前两个或三个数据点的斜度不同于其他数据点。

The table below shows the highest temperature vs. the hours of daylight for the $\begin{aligned} 15^{t h} \end{aligned}$ day of each month in the year 2006 in San Diego, California.
::下表显示2006年加利福尼亚州圣地亚哥最高气温与每月15天的日光时数之比。

Hours of daylight	High temperature (F)
10.25	60
11.0	62
12	62
13	66
13.8	68
14.3	73
14	86
13.4	75
12.4	71
11.4	66
10.5	73
10	61

(a) What would be a better way to organize this table if you want to make the relationship between daylight hours and temperature easier to see?
:a) 如果你想使日光时间和温度之间的关系更容易看到,什么是安排这个表格的更好办法?

(b) Estimate the high temperature for a day with 13.2 hours of daylight using linear interpolation.
:b) 使用线性内插法估计一天高温13.2小时的日光。

(c) Estimate the high temperature for a day with 9 hours of daylight using linear extrapolation. Is the prediction accurate?
:c) 使用线性外推法估计一天9小时白天的高温。预测是否准确?

(d) Estimate the high temperature for a day with 9 hours of daylight using a line of best fit.
:d) 使用最合适的线估计一天9小时白天的高温。

The table below lists expected life expectancies based on year of birth (US Census Bureau). Use it to answer questions 8-15.
::下表按出生年份列出预期预期寿命(美国人口普查局),用来回答问题8-15。

Birth year	Life expectancy in years
1930	59.7
1940	62.9
1950	68.2
1960	69.7
1970	70.8
1980	73.7
1990	75.4
2000	77

Make a scatter plot of the data.
::绘制数据分布图。

Use a line of best fit to estimate the life expectancy of a person born in 1955.
::使用最合适的线估计1955年出生的人的预期寿命。

Use linear interpolation to estimate the life expectancy of a person born in 1955.
::利用线性内插估计1955年出生的人的预期寿命。

Use a line of best fit to estimate the life expectancy of a person born in 1976.
::使用最合适的线估计1976年出生的人的预期寿命。

Use linear interpolation to estimate the life expectancy of a person born in 1976.
::使用线性内插估计1976年出生的人的预期寿命。

Use a line of best fit to estimate the life expectancy of a person born in 2012.
::使用最合适的线估计2012年出生者的预期寿命。

Use linear extrapolation to estimate the life expectancy of a person born in 2012.
::使用线性外推法估计2012年出生者的预期寿命。

Which method gives better estimates for this data set? Why?
::哪种方法为这一数据集提供更好的估计?为什么?

The table below lists the high temperature for the fist day of the month for the year 2006 in San Diego, California (Weather Underground). Use it to answer questions 16-21.
::下表列出了2006年加利福尼亚州圣地亚哥(地铁地下)本月拳击日的高温,用于回答问题16-21。

Month number	Temperature (F)
1	63
2	66
3	61
4	64
5	71
6	78
7	88
8	78
9	81
10	75
11	68
12	69

Draw a scatter plot of the data.
::绘制数据散射图。

Use a line of best fit to estimate the temperature in the middle of the $\begin{aligned} 4^{t h} \end{aligned}$ month (month 4.5).
::使用最合适的线估计4个月中(4.5个月)的温度。

Use linear interpolation to estimate the temperature in the middle of the $\begin{aligned} 4^{t h} \end{aligned}$ month (month 4.5).
::使用线性内插法估计4个月中(4.5个月)的温度。

Use a line of best fit to estimate the temperature for month 13 (January 2007).
::使用最合适的线估计第13个月(2007年1月)的温度。

Use linear extrapolation to estimate the temperature for month 13 (January 2007).
::使用线性外推法估计第13个月(2007年1月)的温度。

Which method gives better estimates for this data set? Why?
::哪种方法为这一数据集提供更好的估计?为什么?

Name a real-world situation where you might want to make predictions based on available data. Would linear extrapolation/interpolation or the best fit method be better to use in that situation? Why?
::列出您可能希望根据现有数据作出预测的实际情况。线性外推法/ 内插法或最合适的方法是否更适合在这种情况下使用? 为什么?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Texas Instruments Resources
::得克萨斯州工具资源

In the CK-12 Texas Instruments Algebra I FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See .
::在CK-12得克萨斯州仪器代数I FlexBook资源中,有图表计算活动,旨在补充本章某些经验教训的目标。

章节大纲

常规

线性内插和外推

Linear Interpolation and Extrapolation ::线性内插和外推

Linear Interpolation ::线线内插

Linear Extrapolation ::线外推法

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Texas Instruments Resources ::得克萨斯州工具资源