5.4 中中度 -- -- 概率和统计

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• 选择节中中度 - 概率和统计

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  中中度 - 概率和统计

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  What is the median value of a data set ? Why would you need to know how to find it if you already know how to find the mean? Is there really any difference? Can you think of a situation where the mean and median are exactly the same number?
  ::数据集的中位值是多少? 如果你已经知道如何找到中位数,为什么你还需要知道如何找到中位数?真的有什么区别吗?你能想到中位数和中位数完全相同的情况吗?

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  In the lesson below we will discuss all of these concepts.
  ::在下面的教训中,我们将讨论所有这些概念。

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  The Median 
  ::中 中 音

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  The median is defined as the value representing the “middle number” of a data set that has been ordered by increasing value, meaning that exactly  $\begin{array}{r}\frac{1}{2}\end{array}$ of the data is greater than the median, and $\begin{array}{r}\frac{1}{2}\end{array}$ is less. Finding the median value of a set is a very simple process, requiring very little or no actual calculation at all. Regardless, it can be an excellent representation of the central tendency of a set. The median is particularly useful when evaluating sets with multiple non-representative outliers , since the median is not sensitive to very large or very small values at the extremes of a data set.
  ::中位数被定义为代表一个数据集的“ 中位数 ” 的值, 该数据集是按增加值排序的, 也就是说, 精确地说, 12 个数据大于中位数, 12 个更小 。 发现一个数据集的中位值是一个非常简单的过程, 只需要很少或完全没有实际计算 。 尽管如此, 中位数可以很好地代表一个数据集的中心趋势 。 该中位数在用多个无代表性的异常值来评价数据集时特别有用, 因为中位数对数据集最极端的非常大或极小的值并不敏感 。

  

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  To identify the median:
  ::为确定中位数:

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  • First, organized your set in ascending numerical order and count the values:
    ::首先,按照向上数字顺序排列您的组合,并计算数值:
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  • Second:
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    • If there are an odd number of values, the median is the middle number in the series.
      ::如果数值有奇数,中位数是序列中的中位数。
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    • If there is an even number of values, the median is the arithmetic mean of the two middle numbers in the series.
      ::如果数值偶数,中位数是序列中两个中间数字的算术平均值。
    
    ::第二:如果有奇数值,中位数是序列中的中位数。如果有偶数值,中位数是序列中两个中位数的算术平均值。

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  Finding the Median
  ::查找中位数

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  1. Find the median of $\begin{array}{r}z\end{array}$ .
  ::1. 找出z的中位数。

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  $$\begin{array}{r}z=\{1,10,3,8,5,6,7,4,9,2,13,12,15,19\}\end{array}$$

  
  ::z1,10,3,8,8,5,6,7,4,4,9,2,13,12,15,19}

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  First, organize the numbers in ascending numerical order and count the values:
  ::首先,将数字按向上数字顺序排列,并计算数值:

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  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 19: total of 14 values
  ::1、2、3、4、5、6、7、8、9、10、12、13、15、19:共计14个数值

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  Since there is an even number of values (14 of them), the median is the arithmetic mean of the 7 ^th   and 8 ^th numbers in the series: $\begin{array}{r}\frac{7+8}{2}=7.5\end{array}$
  ::由于数值偶数(14个),中位数是系列中第7和第8位数的算术平均数:7+82=7.5。

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  $$\begin{array}{r}\therefore 7.5\text{is the median of set}z\end{array}$$

  
  ::$7.5 是 z 集的中位值

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  2. Find the median of set $\begin{array}{r}x\end{array}$ .
  ::2. 找到一组的中位数x。

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  $$\begin{array}{r}x=\left\{\frac{1}{8},\frac{3}{4},\frac{4}{5},\frac{19}{21},\frac{17}{20},\frac{31}{42},\frac{2}{3},\frac{5}{8},\frac{3}{8},\right\}\end{array}$$

  
  ::x18,34,45,1921,1720,3142,23,58,38,}

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  First, organize the values in ascending numerical order and count them:
  ::首先,将数值按向上数字顺序排列并计数:

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  $\begin{array}{r}\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{2}{3},\frac{31}{42},\frac{3}{4},\frac{4}{5},\frac{17}{20},\frac{19}{21}\end{array}$  : Total of 9 values
  ::18、38、58、23、3142、34、45、1720、1921:总计9个数值

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  • Since there are an odd number of values (9 of them), the median is the 5 ^th value: $\begin{array}{r}\frac{31}{42}\end{array}$ .
    ::由于有奇数的值(9个),中位数是第5位值:3142。

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  $$\begin{array}{r}\therefore \frac{31}{42}\text{is the median of set}x\end{array}$$

  
  ::3142是集的中位数 x

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  Real-World Application: Job Searching
  ::Real- World 应用程序: 寻找工作

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  Brian is looking for an entry-level job at a retail store, and is debating between two options. If he wants to apply for the highest paying job, which of the two employers described below should he choose? Why?
  ::布赖恩正在零售商店寻找初级工作,并且正在两个选项之间进行争论。 如果他想申请最高报酬的工作,他应该选择下面描述的两个雇主中的哪一个? 为什么?

  

  Employer #1	Employer #2
  Company’s mean salary: $48,000	Company’s mean salary: $42,000
  Company’s median salary $22,000	Company’s median salary: $28,000

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  Since Brian is looking for an entry-level position, he would likely earn more income with Employer #2.
  ::由于Brian正在寻找一个起职职位,他很可能在雇主2号那里赚取更多的收入。

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  Recall that the mean salary is the average of all the salaries in the company, including any managers or owners on the payroll. Since those positions are undoubtedly higher paid, they would artificially raise the mean income well above that of most employees.
  ::回顾平均工资是公司所有工资的平均数,包括工资单上的任何管理人员或所有人,由于这些职位的工资无疑较高,它们会人为地使平均收入大大高于大多数雇员。

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  By contrast, the median income is probably closer to the salary of a typical salesman, since there would be many more salesmen than managers or owners.
  ::相比之下,中位收入可能更接近典型销售人员的工资,因为销售人员比经理或业主多得多。

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  Earlier  Problem Revisited
  ::重审先前的问题

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  What is the median value of a set? Why would you need to know how to find it if you already know how to find the mean? Is there really any difference? Can you think of a situation where the mean and median are exactly the same number?
  ::一组值的中值是什么? 如果你已经知道如何找到中值, 为什么你需要知道如何找到中值? 是否真的有区别? 你能想到中值和中值完全相同的情况吗 ?

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  The median value is the middle number when the set is organized in ascending numerical order. Depending on the composition of the set, there can actually be quite a large difference between the mean and the median, the mean being much more influenced by outliers. In a normally distributed set, the mean, median, and mode are all the same.
  ::中位值是按升序数字排列集时的中间数字。根据集的构成,中位数和中位数之间实际上可能有相当大的差别,平均值受外部线的影响要大得多。在通常分布的集中,中位数、中位数和模式都是一样的。

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  Examples 
  ::实例

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  Identify the median of each set.
  ::确定每组的中位数。

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  Example 1
  ::例1

  $\begin{array}{r}\{12,17,32,63,85,12,54,23,39\}\end{array}$

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  Put the values in numerical order first: 12, 12, 17, 23, 32, 39, 54, 63, 85. There are nine values, an odd count, so we just take the middle number. 32 is the median.
  ::先按数值顺序排列数值: 12,12,17,23,32,39,54,63,85。 有9个数值,一个奇数,所以我们选中间数字。32是中位数。

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  Example 2
  ::例2

  $\begin{array}{r}\begin{array}{cccc}6& 7& 7& 9\\ 7& 3& 4& 4& 6& 8& 9\\ 8& 0& 1& 2& 3& 3& 3& 3& 4& 4& 6& 7& 7\\ 9& 0\end{array}\end{array}$

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  First, put the values in numerical order, remembering that, as a stem plot, this data is listed with the tens place to the left of the vertical, and each separate ones place to the right. That means our values are: 67, 67, 69, 73, 74, 74, 76, 78, 79, 80, 81, 82, 83, 83, 83, 83, 84, 84, 86, 87, 87, 90. Since there are 22 values, an even number, we find the arithmetic mean of the middle two: 81 and 82, to get the median. The median is 81.5.
  ::首先,将数值按数字顺序排列,记住,作为一个干图,这一数据与垂直左侧的数十个位置和右侧的每个位置分别列出。这意味着我们的数值是:67,67,69,73,74,74,76,78,78,79,80,81,82,83,83,83,83,83,84,84,86,87,87,90。由于有22个值,一个偶数,我们发现中间两点的算术平均值:81和82,以获得中位数。中位数是81.5。

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  Example 3
  ::例3

  $\begin{array}{r}\{93,91,95,92,92,93,95,94,97,93,86,92,94,89,92,91,92,93,94,100\}\end{array}$

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  Put the values in numerical order: 86, 89, 91, 91, 92, 92, 92, 92, 92, 93, 93, 93, 93, 94, 94, 94, 95, 95, 97, 100. There are 20 numbers, an even count, so we average the middle two, which are both 93. The median is 93.
  ::按数字顺序排列数值:86、89、91、91、92、92、92、92、92、92、92、93、93、93、93、93、94、94、94、94、95、95、97、100。 有20个数字,一个偶数,所以我们平均使用中间两个数字,两者均为93。 中位数为93。

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  Example 4
  ::例4

  $\begin{array}{r}\{275,281,269,280,268,278,279,274,275,281,285,285,278,269,\\ 283,263,277,276,269,281,272,275,276\}\end{array}$

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  Put the values in numerical order: 263, 268, 269, 269, 269, 272, 274, 275, 275, 275, 276, 276, 277, 278, 278, 279, 280, 281, 281, 281, 283, 285, 285. There are 23 numbers, an odd count, so we just take the middle value. The median is 276.
  ::按数值顺序排列:263、268、269、269、269、269、272、274、275、275、276、276、277、278、278、278、279、280、281、281、281、281、283、285、285。有23个数字,一个奇数,所以中值是276。中值是276。

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  Review 
  ::回顾

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  Find the median value for each set in questions 1 – 11.
  ::查找问题1至问题11中每一组的中位值。

  1. $\begin{array}{r}\{326,314,325,315,315,307,318,318,320,322,325,321,322,320,312,325,326,325\}\end{array}$
  2. $\begin{array}{r}\{35,37,30,42,32,42,30,45,34,43,37,43,27,41,27,45,31,44,28,45\}\end{array}$
  3. $\begin{array}{r}\{123,167,150,132,152,128,129,160,140,121\}\end{array}$
  4. $\begin{array}{r}\{2120,3040,2180,1892,923,9231,8231\}\end{array}$
  5. $\begin{array}{r}\{1,23,41,23,61,130,210,109,592,203,12\}\end{array}$
  6. $\begin{array}{r}\{23.43,32.52,23.92,32.25,23.43,29.55,28.30,31.54\}\end{array}$
  7. $\begin{array}{r}\left\{\frac{1}{2},\frac{4}{9},\frac{3}{7},\frac{2}{5},\frac{21}{23},\frac{16}{27},\frac{3}{4}\right\}\end{array}$
  8. $\begin{array}{r}\left\{.57,\frac{23}{100},.05,\frac{17}{100},\frac{52}{100},.42,.44,\frac{45}{100}\right\}\end{array}$
  9. $\begin{array}{r}\{12,1.2,.12,102,120,.012,.120,1202\}\end{array}$
  10. $\begin{array}{r}\left\{123,12.3,\frac{12}{30},\frac{123}{120},\frac{120}{123},\frac{30}{12},\frac{1}{123},1.23,.123\right\}\end{array}$
  11. . .

  $\begin{array}{r}\begin{array}{ccc}2& 2& 2\\ 3& 3& 5& 5& 6& 9& 9& 9\\ 4& 1& 2& 2& 3& 3& 4& 4& 8\\ 5& 2& 4& 5& 5& 7& 7& 7\end{array}\end{array}$

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  Review (Answers)
  ::回顾(答复)

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  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 。