课程： 10.2 理解平均值

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理解平均值

The school track team is heading to regionals. Mackenzie wants to figure out the team standings in order to set goals to improve. Mackenzie began to figure out his standing. He is a high-jumper at his school. He has 8 teammates whose records are 172 cm, 174 cm, 175 cm, 179 cm, 181 cm, 181 cm, 182 cm, and 185 cm. If Mackenzie’s record is 176 cm, how does he compare to the rest of the team?
::学校轨道队正在前往地区。Mackenzie想弄清楚球队的站姿,以设定改进的目标。Mackenzie开始弄清楚他的站势。他在学校里是一名高音官。他有8名队友,他们的记录是172厘米、174厘米、175厘米、179厘米、181厘米、181厘米、182厘米和185厘米。如果Mackenzie的成绩是176厘米,那么他与球队其他人相比如何?

In this concept, you will learn to understand the mean.
::在这个概念中,你会学会理解这个意思。

Mean
::平均平均

Statistical measures help to generalize a group of data , make inferences about it, and compare it with other groups of data. Depending on the situation, certain measures may be more helpful than others in interpreting data.
::统计措施有助于概括一组数据,对其作出推论,并与其他一组数据进行比较。根据情况,某些措施在解释数据方面可能比其他措施更有帮助。

The mean, , and mode are three common measures of central tendency ; they are three mathematical tools frequently used to analyze data.
::中值、中值和模式是三个常见的中心趋势衡量标准;它们是三个经常用来分析数据的数学工具。

The mean is the sum of all the data items divided by the number of data items.
::平均值是所有数据项目除以数据项目数的总和。

The median is the middle number in a set of data that is ordered from lowest to highest. If there is an even number of data, you take the average of the middle two numbers to find the median.
::中位数是一组数据中的中位数,该组数据从最低到最高排序。如果数据数量相等,则用中间两个数字的平均值来找到中位数。

The mode is the number that occurs most often.
::模式是最常见的数字。

When you look for the deviation from the mean , you are looking for the difference between an average and a value.
::当您寻找与平均值的偏差时,您正在寻找平均值和值之间的差。

Let’s look at an example that shows what happens when the difference between the mean and another value is determined.
::让我们看看一个示例, 说明在确定平均值与其它值之间的差值时会发生什么。

At a restaurant, the food servers report how much money in tips they earn each night. One Saturday, the food servers reported the following tips: $45, $37.50, $51, $69, $47, and $55. Greg is the newest food server and wants to know how the tips compare to each other. One way to do this is to find the deviation from the mean. This tells you how far away from the mean, or average, each food server was.
::在一家餐馆,食品服务器报告他们每晚赚到多少小费。一个星期六,食品服务器报告以下小费:45美元、37.50美元、51美元、69美元、47美元和55美元。Greg是最新的食品服务器,想知道这些小费是如何相互比较的。其中一个方法就是找到偏离平均值的方法。这说明每个食品服务器离平均值或平均值有多远。

First, find the mean.
::首先,找到那个刻薄的。

\begin{aligned} \begin{array}{rcl} mean & = & \frac{45 + 37.50 + 51 + 69 + 47 + 55}{6} \\ = & \frac{304.50}{6} \\ = & 50.75 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} \text{mean} &=& \frac{45+37.50+51+69+47+55}{6} \\ &=& \frac{304.50}{6} \\ &=& 50.75 \\ \end{array}\end{align*}$
::=45+37.50+51+69+47+556=304.506=50.75

Next, find the difference of each food server tip from the mean. This is the deviation from the mean. Notice that Greg’s value is always used because you are looking for the deviation between the tips of the other waiters and Greg.
::下一步, 找到每个食物服务器提示与平均值的差别。这是与平均值的偏差。注意Greg 的价值总是被使用, 因为您正在寻找其他服务员和 Greg 的提示之间的偏差。

When subtracting, place the larger of the two numbers first so that the difference is positive.
::减法时,先将两个数字中的较大数字放在首位,这样差数是正数。

Difference from mean	Deviation from the mean
50.75 - 37.5	13.25
50.75 - 45	5.75
50.75 - 47	3.75
51 - 50.75	0.25
55 - 50.75	4.25
69 - 50.75	18.25

Then, find the mean of these deviations. The deviation from the mean can let each food server know how far he or she was from the average tips that night.
::然后,找到这些偏差的平均值。与平均值的偏差可以让每个食物服务器知道,他/她离当晚的平均提示有多远。

\begin{aligned} \begin{array}{rcl} mean & = & \frac{13.25 + 5.75 + 3.75 + 0.25 + 4.25 + 18.25}{6} \\ = & \frac{45.5}{6} \\ = & 7.58 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} \text{mean} &=& \frac{13.25+5.75+3.75+0.25+4.25+18.25}{6} \\ &=& \frac{45.5}{6} \\ &=& 7.58 \end{array}\end{align*}$
::=13.25+5.75+3.75+0.25+4.25+18.25=18.256=45.56=7.58

The answer is 7.58. Therefore the tips received by each of the servers are on average $7.58 from the mean amount of tips ($50.75).
::答案是7.58,因此,每个服务器收到的提示平均为7.58美元,来自平均提示数(50.75美元)。

Another statistical measure that can be useful is the mean absolute deviation . The deviation from the mean, which is how far an individual item is from the mean, has already been calculated. When only the positive difference is found, the mean absolute deviation is found.
::另一个有用的统计尺度是平均绝对偏差。与平均值的偏差,即单个项目与平均值的距离,已经计算过了。当只发现正差时,就会发现平均绝对偏差。

The range is found by subtracting the smallest number from the largest number. This gives an idea of the span or the breadth of the data.
::通过从最大数字中减去最小数字来发现该范围。这样就可以了解数据的宽度或宽度。

For instance, if a new car buyer has just entered the work force, it may help for the buyer to know that the mean price of new cars is $22,300. This may be too high, but the mean doesn’t really supply enough information. If the buyer knows that new car prices vary from $10,500 to $89,900, then he may have a better idea about what a car might cost at the lower end, easier for his income level. The range of car prices, in this example is $\begin{align*}\$89, 900 - \$10, 500 = \$79, 400\end{align*}$ . It is a broad range that should allow him to purchase a car.
::例如,如果新买车者刚刚加入劳动力大军,买车者也许知道新车的平均价格是22 300美元,这也许太高了,但实际上并没有提供足够的信息。如果买车者知道新车价格从10 500美元到89 900美元不等,那么他可能更清楚地知道汽车在较低端的成本会如何,对收入水平来说比较容易。汽车价格的范围,例如89 900美元-10 500美元=79 400美元。购买汽车的范围应该很广,应该允许他购买汽车。

Let’s look at another example.
::让我们再看看另一个例子。

A city surveyor took elevation measurements around a coastal city that has a reported mean elevation of 35 feet above sea level. He went to various homes and gathered the following data: 152, 316, 26, 64, 20, 506, 210, and 89. Find the range and mean absolute deviation.
::一名城市测量员在一个沿海城市周围进行了海拔测量,据报该城市平均海拔为35英尺,他前往多个家庭,收集了以下数据:152、316、26、64、20、2506、210和89。

First, find the range. Subtract the smallest number from the largest.
::首先,找到范围。从最大的中减去最小的数字。

$\begin{align*}506 - 20 =486\end{align*}$

Next, knowing the mean is 35 feet (given); find the deviations from the mean.
::其次,知道平均值是35英尺(准);找到与平均值的偏差。

Difference from mean	Deviation from the mean ::偏离平均值
35 - 20	15
35 - 26	9
64 - 35	29
89 - 35	54
152 - 35	117
210 - 35	175
316 - 35	281
506 - 35	471

Then, find the mean of these deviations.
::然后,找到这些偏差的平均值。

\begin{aligned} \begin{array}{rcl} mean & = & \frac{15 + 9 + 29 + 54 + 117 + 175 + 281 + 471}{8} \\ = & \frac{1151}{8} \\ = & 143.875 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} \text{mean} &=& \frac{15+9+29+54+117+175+281+471}{8} \\ &=& \frac{1151}{8} \\ &=& 143.875 \end{array}\end{align*}$
::=15+9+29+54+117+175+281+4718=11518=143.875

The answer is 143.875.
::答案是143.875。

Therefore, the mean absolute deviation is 143.9 feet.
::因此,平均绝对偏差为143.9英尺。

Examples
::实例

Example 1
::例1

Earlier , you were given a problem about Mackenzie the high-jumper at his school. He has 8 teammates whose records are 172 cm, 174 cm, 175 cm, 179 cm, 181 cm, 181 cm, 182 cm, and 185 cm.
::早些时候,有人给了你一个问题关于麦肯锡是他的学校的高中生他有8个队友记录是172厘米 174厘米 175厘米 179厘米 181厘米 181厘米 181厘米 182厘米 185厘米

If Mackenzie’s record is 176 cm, how does he compare to the rest of the team?
::如果Mackenzie的记录是176厘米,

In this case, finding the deviation from the mean would be most useful in answering the question because he wants to compare his individual record to that of the team.
::在本案中,发现偏离平均值的情况对回答问题最为有用,因为他想将个人记录与团队记录进行比较。

First, find the mean.
::首先,找到那个刻薄的。

\begin{aligned} \begin{array}{rcl} mean & = & \frac{172 + 174 + 179 + 181 + 181 + 182 + 185 + 176}{8} \\ = & \frac{1430}{8} \\ = & 178.75 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} \text{mean} &=& \frac{172+174+179+181+181+182+185+176}{8}\\ &=& \frac{1430}{8} \\ &=& 178.75 \end{array}\end{align*}$
::= 172+174+179+179+181+181+182+185+1768=14308=178.75

Next, calculate Mackenzie’s deviation from the mean.
::计算Mackenzie与平均值的偏差。

\begin{aligned} 178.75 - 176 = 2.75 \end{aligned}

$\begin{align*}178.75 - 176 = 2.75\end{align*}$

The answer is 2.75.
::答案是2.75。

Therefore Mackenzie’s record is 2.75 cm lower than the mean.
::因此Mackenzie的记录比平均值低2.75厘米。

Example 2
::例2

The town keeps statistics on its local races. Here are the times from a recent 5 km race. The average time was 23 minutes. The times are: 21, 21, 22, 18, 19, 23, 25, 27, 30.
::该镇保存当地种族的统计数字,这是最近5公里种族的间隔时间,平均时间为23分钟,21、21、22、18、19、23、25、27、30。

What are the range and the mean absolute deviation?
::绝对偏差的幅度和平均值是多少?

First, find the range. Subtract the smallest number from the largest.
::首先,找到范围。从最大的中减去最小的数字。

$\begin{align*}30 - 18 = 12\end{align*}$

Next, knowing the mean is 23 minutes (given); find the deviations from the mean.
::其次,了解平均值为23分钟(给定);发现与平均值的偏差。

Difference from mean	Deviation from the mean
23 - 18	5
23 - 19	4
23 - 21	2
23 - 21	2
23 - 22	1
23 - 23	0
25 - 23	2
27 - 23	4
30 - 23	7

Then, find the mean of these deviations.
::然后,找到这些偏差的平均值。

\begin{aligned} \begin{array}{rcl} mean & = & \frac{5 + 4 + 2 + 2 + 1 + 0 + 2 + 4 + 7}{9} \\ = & \frac{27}{9} \\ = & 3 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} \text{mean} &=& \frac{5+4+2+2+1+0+2+4+7}{9} \\ &=& \frac{27}{9} \\ &=& 3 \end{array}\end{align*}$
::=5+4+2+2+2+1+0+2+4+79=279=3

The answer is 3.
::答案是3

The mean absolute deviation of the race times is 3 minutes.
::比赛时间的绝对偏差是3分钟

The following scores were earned on a math test: 65, 70, 82, 83, 50, 90 and 88. Use this information to answer each question below.
::数学测验得分如下:65、70、82、83、50、90和88。使用这一信息回答以下每个问题。

Example 3
::例3

What is the mean of the scores?
::分数是什么意思?

\begin{aligned} \begin{array}{rcl} mean & = & \frac{65 + 70 + 82 + 83 + 50 + 90 + 88}{7} \\ = & \frac{528}{7} \\ = & 75.43 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} \text{mean} &=& \frac{65+70+82+83+50+90+88}{7} \\ &=& \frac{528}{7} \\ &=& 75.43 \end{array}\end{align*}$
::=65+70+82+83+83+50+90+887=5287=75.43

The answer is 75.43.
::答案是75.43

The mean score on the math test is 75.4.
::数学测验的平均分数是75.4

Example 4
::例4

Find the absolute mean deviation from the mean for each score.
::找出每个得分的绝对平均偏差。

Difference from mean	Deviation from the mean
75.4 - 50	25.4
75.4 - 65	10.4
75.4 - 70	5.4
82 - 75.4	6.6
83 - 75.4	7.6
88 - 75.4	12.6
90 - 75.4	14.6

\begin{aligned} \begin{array}{rcl} mean & = & \frac{25.4 + 10.4 + 5.4 + 6.6 + 7.6 + 12.6 + 14.6}{7} \\ = & \frac{82.6}{7} \\ = & 11.8 \end{array} \end{aligned}

$\begin{align*}\begin{array}{rcl} \text{mean} &=& \frac{25.4+10.4+5.4+6.6+7.6+12.6+14.6}{7} \\ &=& \frac{82.6}{7} \\ &=& 11.8 \end{array}\end{align*}$
::=25.4+10.4+5.4+6.6+7.6+12.6+14.67=82.67=11.8

The answer is 11.8.
::答案是11.8。

The absolute mean deviation is 11.8.
::绝对平均偏差为11.8。

Example 5
::例5

How did Kara’s score of 82 compare to the others?
::卡拉的82分得分与其他得分相比如何?

Kara’s score is 82%.
::卡拉的得分是82%。

The mean score on the math test is 75.4%.
::数学测试的平均分数是75.4%。

Therefore, Kara’s score is 6.6 points above the mean.
::因此,卡拉的得分是高于平均值的6.6分。

Review
::回顾

Define each term.
::界定每个术语。

1. Mean
::1. 平均值

2. Median
::2. 中中

3. Mode
::3. 模式

4. Deviation from the mean
::4. 偏离平均值

5. Range
::5. 范围

Use this data to complete the following questions.
::使用此数据完成下列问题。

Two groups of adult female harbor seals were weighed from different parts of the globe, one from the Pacific Ocean and one from the Atlantic Ocean.
::对全球不同地区的两组成年女性港口海豹进行了加权,其中一组来自太平洋,另一组来自大西洋。

The Pacific Ocean group had weights of: 126 kg, 130 kg, 135 kg, 136 kg, 137 kg, 140 kg, 148 kg, and 150 kg. The Atlantic Ocean group had weights of 117 kg, 119 kg, 122 kg, 123 kg, 130 kg, 131 kg, 141 kg, 149 kg, and 152 kg. A marine biologist decided to gather data to compare the two groups.
::太平洋组的重量为126公斤、130公斤、135公斤、136公斤、137公斤、140公斤、148公斤和150公斤,大西洋组的重量为117公斤、119公斤、122公斤、123公斤、130公斤、131公斤、141公斤、149公斤和152公斤。

Beginning with the Pacific Ocean Group.
::从太平洋集团开始。

6. What is the mean of the data set?
::6. 数据集的含义是什么?

7. What is the median?
::7. 中位数是多少?

8. What is the mode?
::8. 模式是什么?

9. What is the range?
::9. 范围是什么?

10. If a new seal was weighed with a weight of 137 kg, what would be the deviation from the mean?
::10. 如果新封条的重量重为137公斤,那么与平均值的偏差是多少?

Now with the Atlantic Ocean Group.
::现在,大西洋集团。

11. What is the mean of the data set?
::11. 数据集的含义是什么?

12. What is the median?
::12. 中位数是多少?

13. What is the mode?
::13. 模式是什么?

14. What is the range?
::14. 范围是什么?

15. If a new seal was weighed with a weight of 137 kgs, what would be the deviation from the mean?
::15. 如果用137公斤的重量对新封条称重,那么偏离平均值是多少?

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。

章节大纲

常规

理解平均值

Mean ::平均平均

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Resources ::资源