返回目录
Mean, Median, Mode
成绩单
现在，我们可以开始谈论统计数据。在大局中，统计数据是一个相当广泛的主题，
但我们只需要知道考试的一些事情。出于测试的目的，统计数据包括工具
making sense of data. So we'll get some data and we'll just be asked about this data.
那么，我们可以询问的最重要的问题之一是数据集的，单个号码是整个组的最多代表？ 这结果在现实世界统计中非常重要。例如，如果我们有一个家庭收入列表 everyone in the United States, we'd wanna know what's one single number that would be most representative of that entire list.
嗯，这些数字被称为，中心的衡量标准。这就是我们在这个视频中开始谈论的东西。 最重要的中心措施是平均值和媒介。平均值只是普通平均水平。 To find the mean, say of, seven numbers on a list, you would add up the seven numbers, and then divide this sum by seven.
In general, on a list with N entries, we add up all the entries, and then divide by N. That's the mean. Just an ordinary average. We can write the formula as mean equals the sum of N entries divided by N. That's the formula for the mean.
Notice that this formula is also quite useful in the following form. If we just multiply both sides by N, we get N times the mean. In other words, the number of people on the list times the mean has to equal the sum of the entries. 思考总和通常是许多关于平均值或卑鄙的问题的关键。那么第二种形式， 它真的被低估了这种表格的强大。
例如，这是一个练习问题。暂停视频，然后我们会谈谈这个。 好吧,在一个类,18个学生参加了一个考试an average of 70. Alicia and Burt then took the test, and the average of all 20 students was 71. 如果艾丽西亚有77岁，那么Burt的成绩是什么？好的，所以，很多人会发现这是一个非常难的问题。
这个问题的关键只是考虑总和。所以，首先，旧金，那18人。 我只是花了70岁的时间18倍。繁殖，这是1260。 好吧，现在所有20名学生的新总和，这将是71的20倍。繁殖，那是1420。
好吧，想一想。旧的总和和新的总和，它们之间有什么区别？ 唯一的区别是，我们将艾丽西亚的得分和Burt的分数添加到其他18分。 So the difference between them should be the sum of the score of Alicia and the score of Burt.
So, we subtract them. That sum. 所以艾丽西亚的分数和Burt的分数必须增加160.好吧，如果我们减去阿利西亚的77分， 然后我们得到83，当然，这必须是Burt的成绩。所以这就是我们如何回答这个问题，纯粹考虑总和。
Now the median. The median is the middle number on a list. 我们必须在这里有点小心。我们必须先将列表以升序命令汇总，即从最小到最大。 Technically, the median is the middle number on an ordered list. So, we can't just write them in any jumbled order and then say, okay, 中位数是碰巧坐在混乱的中间的中位数。
No, we have to put the list in order, from smallest to biggest. And often, incidentally, the test will not do that. They'll give you the numbers in a jumbled order, and then you, yourself, have to put them in order and then find the median. 所以，你必须小心。因此，如果我们有这个列表，中间中间的正确Smack Dab是4。
所以，这是中位数。它下面有三个数字，在它上面的三个数字，清楚地在中间。 现在这个列表有点兴趣，因为列表上的偶数数量不是奇数，所以中间没有一个数字。 在中间，我们有这两个数字，4和5，他们不能成为中位数。所以如果列表上有偶数，我们会做什么，我们平均是两个中间数字。
And so the median is between 4 and 5. We take the average of four and five, which is 4.5. 那是中位数。那些被赋予了这个清单， 好吧，我们要做的第一件事就是按顺序排列。所以现在它是为了。
所以现在中位数，这个名单上有两三四有四有四六七八八个数字。 And so the median has to be between 8 and 13. So we're gonna average 8 and 13. 8 + 13 divided by 2, 21 divided by 2. 10.5, 10.5 is the median of list K.
Notice that the median only takes into account the number or numbers at the very center. 我们可以在列表中的任何一端更改数字，此更改根本不会影响中位数。 所以，在这个特定的清单中，假设我们改变了55到55,000，中位数根本不会改变，意思会改变，但不是中位数。
我们将在下一个视频中讨论这一点。这是一个练习问题。 暂停视频，然后我们会谈谈这个。好的，列表B的中位数恰好4高于列表A的中位数。 All right, so the very first thing we need to do is find the median of List A. We put the terms in order.
The median is the average of 7 and 10, which is 8.5. That's the median of List A. 列表B的中位数必须高于4，因此它必须是12.5，即8.5加4。 好吧，所以现在让我们想一想。有三个数字小于中位数，4,7和 10 are all less than the median.
18,25和X必须大于中位数，实际上，12.5必须是10和x的平均值。 So 12.5 is the average of 10 and x. We'll multiply by 2. 我们得到的是12.5次2是25减去10，我们得到x = 15.这就是x的值。
如果x具有15的值，则设定B将具有12.5的中位数，这正是4高于8.5。 一个最终中心的尺寸是模式。你经常听到这三个说在一起，平均值，中位数和模式。 模式是什么？该模式是列表中最常出现的数字。
The number that makes the greatest number of appearances. This is far less important than either the mean or the median for 各种原因。一些列表具有单个模式，例如有很多 在此列表中的发行有三个三个，每个其他号码只出现一次。所以模式显然是3。
Some lists have two modes, for example in this list we have a pair of twos and also a pair of fives, so the modes are 2 and 5. If all the numbers on the list are different from one another, as is usually the case ,then there simply is no mode. So sometimes there is a mode, sometimes there's more than 1 and often there's just no mode at all.
所以考虑一下这个，行星上的每一个列表都有一个平均值，地球上的每个单个列表都有一个中位数。 But, only some lists have modes. Some may have more than one mode and many have no mode at all and so this is one of 该模式与平均值或中位数几乎同样重要的大原因。总之，平均值平均值， 经常有关卑鄙的问题，在总和方面有助于思考。
So the sum of the entry equals the number of entries times the mean. We just rewrite that formula on the top to get this. 中位数是有序列表的中间数量。如果中间有两个数字，那么我们只是平均平均这些数字。 And the mode, the most frequently appearing number is just less important. Some lists have a mode, some have more than one, and many have no mode at all.
Theoretically the test could ask you about a mode, but it's much, much less common than questions about mean or median.
阅读完整成绩单
那么，我们可以询问的最重要的问题之一是数据集的，单个号码是整个组的最多代表？ 这结果在现实世界统计中非常重要。例如，如果我们有一个家庭收入列表 everyone in the United States, we'd wanna know what's one single number that would be most representative of that entire list.
嗯，这些数字被称为，中心的衡量标准。这就是我们在这个视频中开始谈论的东西。 最重要的中心措施是平均值和媒介。平均值只是普通平均水平。 To find the mean, say of, seven numbers on a list, you would add up the seven numbers, and then divide this sum by seven.
In general, on a list with N entries, we add up all the entries, and then divide by N. That's the mean. Just an ordinary average. We can write the formula as mean equals the sum of N entries divided by N. That's the formula for the mean.
Notice that this formula is also quite useful in the following form. If we just multiply both sides by N, we get N times the mean. In other words, the number of people on the list times the mean has to equal the sum of the entries. 思考总和通常是许多关于平均值或卑鄙的问题的关键。那么第二种形式， 它真的被低估了这种表格的强大。
例如，这是一个练习问题。暂停视频，然后我们会谈谈这个。 好吧,在一个类,18个学生参加了一个考试an average of 70. Alicia and Burt then took the test, and the average of all 20 students was 71. 如果艾丽西亚有77岁，那么Burt的成绩是什么？好的，所以，很多人会发现这是一个非常难的问题。
这个问题的关键只是考虑总和。所以，首先，旧金，那18人。 我只是花了70岁的时间18倍。繁殖，这是1260。 好吧，现在所有20名学生的新总和，这将是71的20倍。繁殖，那是1420。
好吧，想一想。旧的总和和新的总和，它们之间有什么区别？ 唯一的区别是，我们将艾丽西亚的得分和Burt的分数添加到其他18分。 So the difference between them should be the sum of the score of Alicia and the score of Burt.
So, we subtract them. That sum. 所以艾丽西亚的分数和Burt的分数必须增加160.好吧，如果我们减去阿利西亚的77分， 然后我们得到83，当然，这必须是Burt的成绩。所以这就是我们如何回答这个问题，纯粹考虑总和。
Now the median. The median is the middle number on a list. 我们必须在这里有点小心。我们必须先将列表以升序命令汇总，即从最小到最大。 Technically, the median is the middle number on an ordered list. So, we can't just write them in any jumbled order and then say, okay, 中位数是碰巧坐在混乱的中间的中位数。
No, we have to put the list in order, from smallest to biggest. And often, incidentally, the test will not do that. They'll give you the numbers in a jumbled order, and then you, yourself, have to put them in order and then find the median. 所以，你必须小心。因此，如果我们有这个列表，中间中间的正确Smack Dab是4。
所以，这是中位数。它下面有三个数字，在它上面的三个数字，清楚地在中间。 现在这个列表有点兴趣，因为列表上的偶数数量不是奇数，所以中间没有一个数字。 在中间，我们有这两个数字，4和5，他们不能成为中位数。所以如果列表上有偶数，我们会做什么，我们平均是两个中间数字。
And so the median is between 4 and 5. We take the average of four and five, which is 4.5. 那是中位数。那些被赋予了这个清单， 好吧，我们要做的第一件事就是按顺序排列。所以现在它是为了。
所以现在中位数，这个名单上有两三四有四有四六七八八个数字。 And so the median has to be between 8 and 13. So we're gonna average 8 and 13. 8 + 13 divided by 2, 21 divided by 2. 10.5, 10.5 is the median of list K.
Notice that the median only takes into account the number or numbers at the very center. 我们可以在列表中的任何一端更改数字，此更改根本不会影响中位数。 所以，在这个特定的清单中，假设我们改变了55到55,000，中位数根本不会改变，意思会改变，但不是中位数。
我们将在下一个视频中讨论这一点。这是一个练习问题。 暂停视频，然后我们会谈谈这个。好的，列表B的中位数恰好4高于列表A的中位数。 All right, so the very first thing we need to do is find the median of List A. We put the terms in order.
The median is the average of 7 and 10, which is 8.5. That's the median of List A. 列表B的中位数必须高于4，因此它必须是12.5，即8.5加4。 好吧，所以现在让我们想一想。有三个数字小于中位数，4,7和 10 are all less than the median.
18,25和X必须大于中位数，实际上，12.5必须是10和x的平均值。 So 12.5 is the average of 10 and x. We'll multiply by 2. 我们得到的是12.5次2是25减去10，我们得到x = 15.这就是x的值。
如果x具有15的值，则设定B将具有12.5的中位数，这正是4高于8.5。 一个最终中心的尺寸是模式。你经常听到这三个说在一起，平均值，中位数和模式。 模式是什么？该模式是列表中最常出现的数字。
The number that makes the greatest number of appearances. This is far less important than either the mean or the median for 各种原因。一些列表具有单个模式，例如有很多 在此列表中的发行有三个三个，每个其他号码只出现一次。所以模式显然是3。
Some lists have two modes, for example in this list we have a pair of twos and also a pair of fives, so the modes are 2 and 5. If all the numbers on the list are different from one another, as is usually the case ,then there simply is no mode. So sometimes there is a mode, sometimes there's more than 1 and often there's just no mode at all.
所以考虑一下这个，行星上的每一个列表都有一个平均值，地球上的每个单个列表都有一个中位数。 But, only some lists have modes. Some may have more than one mode and many have no mode at all and so this is one of 该模式与平均值或中位数几乎同样重要的大原因。总之，平均值平均值， 经常有关卑鄙的问题，在总和方面有助于思考。
So the sum of the entry equals the number of entries times the mean. We just rewrite that formula on the top to get this. 中位数是有序列表的中间数量。如果中间有两个数字，那么我们只是平均平均这些数字。 And the mode, the most frequently appearing number is just less important. Some lists have a mode, some have more than one, and many have no mode at all.
Theoretically the test could ask you about a mode, but it's much, much less common than questions about mean or median.
