3.6.1 Position average: a robust expression of central tendency 1 位置平均数：集中趋势之稳健表达1在线视频

位置平均数
Location average

就是有我们有四分位数
includes quartile

中位数 众数
median and mode

我们先讲中位数
Let us first talk about median

中位数就是变量值
Median is the value of variables

由大到小排列
sorted in a descending order

当然你也可以
Of course, it can also be

由小到大排列
in an ascending order

排列处于中间位置的
The middle number

那个变量值
in this list

就属于中位数
is the median

中位数在整个变量数列中
There is only one median

只有一个中位数
in the list of the values

所以它在计算的时候
Therefore, the determination of

我们要视情况而定
the median depends

如果是我们讲的
If the data has not been

未分组资料
grouped yet

首先第一步先确定位置
we should first determine its location

位置就是n+1/2
by n+1/2

n就是个数
n is the number

变量个数
the number of variable values

n+1/2那就是中间位置
n+1/2 shows the middle position

中间位置的那个
or the variable value XI

对应的那个变量值XI
at the middle position

它就属于中位数
It is the median

用M{\fs10}e{\r}表示
expressed as M{\fs10}e{\r}

当然它也有分组资料
If the information is grouped

分组资料你要看
in grouped data, we need to know

是单项数列还是组距数列
whether it is a single-valued series or a class interval series

如果是单项数列的话
In a single-valued series

也是用这个n+1/2
n+1/2 also applies

处于中间位置的累计
to find whether the accumulative in the middle

向上累计还是向下累计
is a cumulative up or cumulative down

把它算出来 中间在哪里
Calculate it and find where the middle is

就是对应的那个变量值属于中间的
The corresponding variable value is the one in middle position

组距数列的话就复杂一点
It is a little more complicated with class interval series

因为组距数列第一步
Because we should start with

确定到的是中位数所在组
finding the group the median is in

在这个组里面
This group

它也存在着下限和上限
also has its upper limit and lower limit

下限和上限中间的哪一个地方
The middle between the upper limit and lower limit

是中位数
is the median

我们就要用上限公式
We can use the upper limit formula

或者下限公式
or the lower limit formula

大家看一下那两个公式
Let us look at the two formulas

比如说大妞 二妞
Suppose the test scores of

她们那个考试的分数
Da Niu and Er Niu

在班上中位数里面
is the median of the class

她们的分数属于在哪个位置
Where are their grades in the list

这个可以看得到
We can see it directly

下面我们运用中位数
Now let’s use the concept of median

来看看二妞这次统计学考试
to analyze the standing list of the

在班上的排名情况
Statistics test scores in Er Niu’s class

通过前面所形成的频数分布表
According to the frequency distribution table

我们可以首先计算这一组数据的
we can first calculate the middle position

中点位置
of this list of data

是Σf/2
It is Σf/2

也就是50/2=25
or 50/2=25

计算出了中点位置为25
After calculating the middle position, 25

那么根据这张频数分布表
according to the frequency distribution table

我们就可以得到25
we can see 25 is in the group

所对应的这一组是70到80这一组
from 70 to 80

那么我们下一步就要找到
Then we need to find

这个中点位置所对应的中位数
the corresponding median at the middle position

接下来我们来看一看
Now let us look at the

中位数的推算过程
calculation of the median

首先我们将二妞班上
First, let us set an abscissa

这次统计学考试的成绩
representing the Statistics test scores

和相对应的人数
and the corresponding

用一个横坐标表示如下
number of test takers

通过这张横坐标
From the abscissa

我们可以看到
we can see

在横坐标的上方
that above it

显示的是这次统计学考试的成绩
are the Statistics test scores

在这个横坐标的下面
and below it

显示的是其相对应的人数
are the number of test takers.

下一步我们就要找到
Next, we need to find

我们刚刚所计算到的
the variable value

这个中点位置25所对应的
of 25 in the middle position

变量值为多少
that we just calculated

通过这个横坐标
From the abscissa

我们大致可以找到
we can find

第25个人所对应的变量值
the corresponding the variable value

大概在这个位置
of the 25th person in this position

也就是靠近70的位置
close to 70

但是具体为多少呢
But what is the specific figure

我们肯定还要进一步通过计算得到
We must do some further calculations

这里我们假定中位数所在的组
Suppose the variable values of group

它的变量值是呈现均匀分布的
the median is in is evenly distributed

那么下一步呢
What’s next

我们就可以采用比例插值法
We can us interpolation by proportional parts

可以得到下面的公式
to get this formula

也即（公式如上）
namely (see the formula above)

那么这个恒等式是怎么得到呢
How are the identities achieved

接下来我们来看一下
Let us look at them

在这个恒等式的左边
On the left of the identities

10:16它表示是什么意思呢
what does “10:16” mean

我们可以看到10表示的
We can see the “10”

是70到80分
is the class interval of

这一组的组距
the group from 70 to 80

16表示的是其对应的人数
16 is the corresponding number of test takers

10:16实际上得到的
10:16 is actually showing

就是这一组
the average grade of test takers

平均每人得到了多少分
in this group

右边的恒等式那么当然
The right side of the identities, of course

就等于25-23人
show the corresponding grade

这么多人 它所对应的分数为多少
of 25 to 23 test takers

根据刚刚所建立的恒等式
If this established identities

我们可以进一步将其转化为
are further converted into

数学符号
mathematical notations

也就是等于i:f{\fs10}m{\r}
as i:f{\fs10}m{\r}

等于（计算如上）
equal to (see from the above calculation)

进一步推导得到
Further derivation shows

(计算如上)
(see from the above calculation)

那么这个x在这个横坐标上
x on the abscissa

它表示的是实际上就是
is actually

第25个人所在的位置
the position of the 25th test taker

它距离这一组的下限
Its interval from the lower limit

也就是70这个点的距离为x
70, of this group is x

由此我们可以得到它的中位数
So, we can see its median

等于L+x
is L+x

进一步代入可以得到M{\fs10}e{\r}
After substitution, we get M{\fs10}e{\r}

等于（计算如上）
equal to (see from the above calculation)

由于x=1.25
because x=1.25

因此中位数M{\fs10}e{\r}就等于L+x
the median, M{\fs10}e{\r}, equal to L+x

等于70+1.25
is equal to 70+1.25

等于71.25分
is equal to 71.25

有了中位数
Now we have the median

中位数其实就是
The median divides

把整个数列区分为两段
the list into two sections

就是中间把它分成两段
breaking off at the middle position

一部分比它少
One section has more

一部分比它多
than the other

才属于中间位置
so there is a middle position

你按照这个思路
Follow this way of thinking

那你可以把这个数
you can divide the list of

这一段数列 这个整个数列
variables, the whole list

分成四段
into four groups

就中间
cutting from the middle

就分两下
twice

就分成了四段
to make four segments

那四段里面
Among the four groups

这个就形成了
the quartile steps in

那个分开的那个界限
as the division point

就是属于四分位数
that breaks the list

它的计算原理跟中位数一样
It has the same calculation method as the median

这里我们仍然以二妞班上
Again we use the grouped data of

这次统计学考试的分组数据为例
Statistics test scores from Er Niu’s class

来确定班上50名同学
to determine the quartile

这次统计学考试成绩的四分位数
among the 50 test takers

和应用中位数的方法一样
In the same way as calculating the median

首先我们要确定每一个
first we need to determine

四分位数的位置
the position of the quartile

也就是说我们首先要分别确定
That is to say, we should find the respective

第一四分位数 第二四分位数
positions of first quartile, the second quartile

和第三四分位数的具体位置
and the third, fourth quartiles

和计算中位数位置的公式类似
Similar to how we calculate the position of the median

我们这里四分位数的位置
the formula to determine the quartiles

等于（公式如上）
are equal to (the above formula)

由此根据公式
According to the formula

我们可以分别求得
we can find the respective

Q{\fs10}1{\r}位置=50/4=12.5
position of Q{\fs10}1{\r} =50/4=12.5

Q{\fs10}2{\r}位置=（2×50）/4=25
of Q{\fs10}2{\r} =（2×50）/4=25

Q{\fs10}3{\r}位置=（3×50）/4=37.5
and of Q{\fs10}3{\r} =（3×50）/4=37.5

然后我们就可以根据累计频数
Then according to cumulative frequency

找出四分位数所在的组
we can find the individual groups of the quartiles

最后用差值法
At last we use the differential technique

按比例计算四分位数的近似值
to calculate the approximate value of the quartiles

如Q{\fs10}1{\r}位置=12.5
such as the position of Q{\fs10}1{\r} =12.5

那么根据累计频数
And cumulative frequency shows

我们可以看到
us that

它应该属于2到23这一组
it should belong to the group from 2 to 23

也就是它对应的是60到70这一组
or the corresponding group from 60 to 70

找到了这一四分位数所在的组
After finding the groups of the quartiles

下一步我们就可以运用
we can further apply

和中位数类似的方法
the same way to calculate median

来计算四分位数的近似值
to find their approximate values

在这里我使用的是
Here is how I use

计算四分位数的下限公式
the lower limit formula to calculate the quartile

（公式如上）
(see the above formula)

（公式如上）
(see the above formula)

（公式如上）
(see the above formula)

（公式如上）
(see the above formula)

如我们刚刚已经通过
As we calculated previously

计算得到Q{\fs10}1{\r}的位置等于12.5
the position of Q{\fs10}1{\r} was 12.5

同时也确定了它属于60到70这一组
and it belonged to the group from 60 to 70

这样我们就可以根据刚刚的
we can use the lower limit formula

下限公式来计算Q{\fs10}1{\r}的近似值
mentioned above to calculate the approximate value of Q{\fs10}1{\r}

L{\fs10}Q{\r}{\fs7}1{\r}实际上就是60到70
L{\fs10}Q{\r}{\fs7}1{\r} is from 60 to 70

这一组的下限值
the lower limit of this group

iΣf/4实际上就是
iΣf/4 is actually

1/4乘以50
1/4 multiplies 50

S{\fs10}Q{\r}{\fs7}i{\r}{\fs10}-1{\r}为小于这一四分位数的
S{\fs10}Q{\r}{\fs7}i{\r}{\fs10}-1{\r} is the sum of frequencies

各组次数之和
smaller than the quartiles in each group

根据向上累计频数
According to cumulative up

我们可以看到S{\fs10}Q{\r}{\fs7}i{\r}{\fs10}{\fs7}-1{\r\fs10}{\r}是2
we know S{\fs10}Q{\r}{\fs7}i{\r}{\fs10}{\fs7}-1{\r\fs10}{\r} is 2

f{\fs10}Q{\r}{\fs7}i{\r}则为这一四分位数
and f{\fs10}Q{\r}{\fs7}i{\r} is the frequency of

所在组的次数
the group of this quartile

通过频数分布表
According to frequency distribution table

我们可以看到
we can see

60到70这一组所对应的频数为21
the frequency of the group from 60 to 70 is 21

而d{\fs10}Q{\r}{\fs7}i{\r}周围这一组组距
and the class interval around d{\fs10}Q{\r}{\fs7}i{\r}

也就是70-60=10代入
is substituted by 70-60=10

最后得到Q{\fs10}1{\r}=65
and at last we get Q{\fs10}1{\r}=65

最后我们可以用同样的方法
Consequently, we can use the same method to

来分别计算Q{\fs10}2{\r}和Q{\fs10}3{\r}的近似值
calculate the respective approximate values of Q{\fs10}2{\r} and Q{\fs10}3{\r},

计算结果分别为Q{\fs10}2{\r}=71.25
with the output being Q{\fs10}2{\r}=71.25

Q{\fs10}3{\r}=79.06
and Q{\fs10}3{\r}=79.06.

当然如果有兴趣的同学
Of course if you are intrigued

还可以使用上限公式
you can also use the upper limit formula

来计算一下
to calculate

这一四分位数的近似值
the approximate value of this quartile

看看结果是不是一样的
to if the output is the same.

那现在我们看众数
Now let us look at mode

众数就是出现次数最多的变量值
Mode is the variable value that appears the most frequently

众数的使用也非常广
Mode is also widely used

刚才我们讲的衣服 帽 鞋等等
In the previous example of clothing, hat and shoes

它的尺寸确定的时候
when we need to determine the size

是用众数
we use mode

还有我们那个物价指数的计算CPI
And in the example to calculate CPI, Consumer Price Index

在那个单独采购那个
the commodity price

商品物价的时候
in separate purchase

也用众数价格
is also mode price

等等都是使用了众数
These are how mode is used on various occasions

众数就是在一组数列里面
Mode is the variable value that appears

出现次数最多的那个变量值
the most frequently in a set

它就是众数
This is mode

当然 在一个数列里面
Of course, there might be more than one

可能有n个众数
or n modes in a list

比如我们看看例子里面
Let me cite a few examples

比如我们穿鞋
Take the shoe size for example

在南方的男士穿鞋
the mode size of men’s shoes

可能众数是40码
in south China might be 40

也可能是41码
or 41

这两个都是属于众数
Both are modes

所以它这两个鞋码做的就特别多
And shoes of these two sizes are produced in larger quantities

这个是两个众数
There can be two modes

它不像中位数
unlike median

中位数只有一个
There is only one median in a list

众数有两个
But there can be two or more modes

当然众数的确定
The determination of mode

也可以像中位数一样
can be the same as of median

它也有未分组的情况下
When the data is ungrouped

单项数列的情况下
in single-valued series

和组距数列的情况下
and in class interval series

它怎么计算
the calculation varies

组距数列相对复杂一点
The calculation in class interval series is little more complicated

组距数列计算的时候
In class interval series

它按照先确定众数所在组
we should first find the group of the mode

再在组里面找众数
before we calculate the mode

下面我们来以二妞班上
Now let us see in the case of Statistic test scores

这次统计学考试成绩为例
of Er Niu’s class

来计算他们的众数
how to calculate the mode

通过我们之前所形成的频数分布表
According to the previous frequency distribution table

可以看到
we can see

这里显示的是组距式数列
this is a class interval distribution series

既然是组距式数列
In a class interval distribution series

我们第一步就要确定它的众数组
we should first determine the modal class

大家可以很清楚的看到
We can see clearly

21为这里数值最大的
21 is the maximum value

也就是21所对应的这个组为众数组
and the group of 21 is the modal class

就是60到70这一组
namely, the group from 60 to 70

这里我们同样通过画图的形式
Again we use a chart

来找到众数组所对应的众数
to locate the corresponding the modal class

通过这张图可以看到
From this chart, we can see

在众数组相邻左右两边的组
the neighbor sets of the modal class

其所对应的次数分别为2和16
have their respective frequency of 2 and 16

而众数又是受到次数影响的
Mode is an indicator that is related to frequency

一个指标
related to frequency

因此我们可以想像
So we can presume

在60到70这一众数组的众数
the modal class from 60 to 70

它应该是偏向70到80这个方向
is deviated to from 70 to 80

那具体的位置会在哪里呢
But where exactly is it

一般我们可以通过做图的形式
Generally, we locate the position of mode

来找到这一众数
by chart-drawing

具体来说就是将众数组
Specifically speaking, we connect

所对应的直方图的顶点AD
the modal class’s histogram vertex AD

和与其相邻两个组的
with the corresponding histogram intersection CF

对应直方图的交点CF相连接
of its neighbor sets

画出AF和CD这两条直线
Draw straight lines AF and CD

在这里我们可以看到
and we can see

这两条直线相交
the two straight lines cross

形成一个交点为O点
at the intersection O

沿O点做一条垂线
draw a perpendicular line through O

与X轴的交点被认为是
its intersection on X-axis is seen

众数所在的位置
to be the position of mode

原因在于O点位置的确定
That is because the position of O is determined

是受到众数所在的组
by both the group of the mode

和其相邻组次数的影响
and frequency of the neighbor sets

因为在图中AC所表示的
As shown in the chart, AC represents

就是众数所在的组
the frequency difference between

和其相邻组
the group of the mode and its neighbor set

也就是左边的组的次数差
or the set on its left

通常我们假设AC等于△{\fs10}1{\r}
Normally we assume AC is equal to △{\fs10}1{\r}

而DF表示的值是
then the value of DF

众数所在此组的次数
is the difference of frequency between

和其相邻右边组
the group of mode

所在的次数的差
and its neighbor set on the right

通常我们将DF等于△{\fs10}2{\r}
Normally we assume DF is equal to △{\fs10}2{\r}

而AC和DF的程度
The dimension of AC and DF

将会决定O点投影在X轴上的
decides the specific position of

具体位置
point O projected on X axis.

由此可以进一步得到
So we can further infer

众数距离60所在的
the distance of mode from 60,

L点位置为x 距离70所在的
point L is x, and from 70

U点位置为y
point U is y

由此进一步可以建立
to establish

(公式如上)
(see the formula above)

又由于图中显示的三角形ACO
Again the triangle ACO in the chart

类似于三角形ODF
is similar to triable ODF

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

又由于三角形ABO类似于
and is also similar to triable ABO

三角形OFE
and triable OFE

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation).

通过前面两个三角形
We can get the identities

所得到的恒等式
based on the two triangles above

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

同时我们前面已知
Meanwhile we know

（公式如上）
(see formula above)

（公式如上）
(see formula above)

把刚刚我们得到的x代入
substitute x from previous calculation

（计算如上）
(see from the calculation above)

（计算如上）
(see from the calculation above)

（计算如上）
(see from the calculation above)

这里我们使用的是下限公式
we have used the lower limit formula

同理我们可以证得
Similarly, we can prove

（计算如上）
(see from the calculation above)

（计算如上）
(see from the calculation above)

（计算如上）
(see from the calculation above)

在这里我们还要进一步了解
At last, we need to learn

这么一个符号的含义
the meaning of the symbols here

L表示的是众数组的下限
L represents the lower limit of the modal class

U表示的是上限
U，the upper limit

i为众数组的组距
i, the interval

△{\fs10}1{\r}=f{\fs10}m{\r}-f{\fs10}m{\r}{\fs10}-1{\r}
△{\fs10}1{\r}=f{\fs10}m{\r}-f{\fs10}m{\r}{\fs10}-1{\r}

也就是表示众数组的次数
represents the frequency difference between modal class

与前一组次数之差
and its neighbor set in front

△{\fs10}2{\r}=f{\fs10}m{\r}-f{\fs10}m{\r}{\fs10}+1{\r}
△{\fs10}2{\r}=f{\fs10}m{\r}-f{\fs10}m{\r}{\fs10}+1{\r}

表示的是众数组的次数
represents the difference between modal class

与后一组次数之差
and its neighbor set in the back

最后我们假设使用下限公式
At last, we assume if we use lower limit formula

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

（计算如上）
(see from the above calculation)

在这里大家还要注意的是
There is one more thing that we should pay attention to

我们刚刚所介绍的是
In what have previously mentioned

针对组距式数列
the process to calculate mode

来计算众数的过程
in class interval distribution series

对于未分组资料和单项式数列的
the mode in ungrouped information and single-valued distribution series

众数是非常容易找的
is easy to find

实际上呢就是数数
In fact, it can be done by counting

这里我们就不举例说明了
I will not further illustrate on this

如果有兴趣的同学
If you are interested

还可以进一步思考
you can continue to think

如果两个相邻组的次数是相等的
if the frequency of neighbor sets are the same

那么众数怎么计算呢
how can we calculate the mode

大家可以算一算
You can try to calculate

这个时候众数实际上
At that time, actually

就等于众数组的组中值
the mode is the same as its class mid-point

当然在这里面大家可以讨论
You can think about it

有一些同学可能会认为
Some of you may think

众数不一定在众数组
mode may not always be found in modal class

如果不在众数组
If it is not in modal class

你这个公式就要进行
you may need to adjust the formula

就值得商讨了
in use.

大家可以进行讨论
Welcome to express your views