5.8 计算标准偏差

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  计算标准偏差

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  What is standard deviation ? How is the standard deviation of a set related to variance ? Is the standard deviation of a sample different from that of a population , the way it is with variation?
  ::标准偏差是什么?一组标准偏差与差异有何关系?抽样的标准偏差是否与人口的标准偏差不同,变化如何?

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  This lesson details the process of calculating standard deviation, and introduces a few examples of its use. After the lesson we’ll review the questions above, using the knowledge we have gained.
  ::这一教训详细说明了计算标准偏差的过程,并举了一些使用它的例子。 在吸取了教训之后,我们将利用我们所获得的知识来审查上述问题。

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  Standard Deviation 
  ::标准偏离

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  Standard deviation   $\begin{array}{r}(\sigma )\end{array}$ is very common in statistics - it   is sort of a “reference difference from the mean” that you can use to evaluate the spread of the data in a set.  It is easy to calculate if you have already identified the variance , since the standard deviation is just the square root of the variance.
  ::标准偏差 (- 12) 在统计中非常常见, 这是一种“ 参考值差异 ” , 您可以用它来评估数据集中数据分布的“ 参考值差异 ” 。 如果您已经确定了差异, 很容易计算, 因为标准偏差只是差异的平方根 。

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  To illustrate how standard deviation is used , assume the mean of a particular set is 6 and the standard deviation is 4.
  ::为了说明如何使用标准差,假设某一组的平均值为6,标准差为4。

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  • In this set, a value of 18  is very uncommon (less than 0.3%) , since  18 is  3 standard deviations away from the mean.  6 + (3  $\begin{array}{r}\times \end{array}$  4) = 18
    ::在这一组中,18的值非常罕见(不到0.3%),因为18是3个标准差,与平均值相差。 6+(3×4)=18
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  • A value of 8 is much more likely (about 40%), given that it is only  $\begin{array}{r}\frac{1}{2}\end{array}$ of a standard deviation (SD) away from the mean.  6 + ( $\begin{array}{r}\frac{1}{2}\end{array}$ $\begin{array}{r}\times \end{array}$  4) = 8
    ::8值的可能性大得多(约40%),因为只有12个标准差(SD)偏离平均值。 6+(12x4)=8

  ---

  Standard deviation is simply the square root of the variance

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  To calculate standard deviation, start by determining the variance.
  ::要计算标准偏差,首先确定差异。

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  Recall from the lesson Calculating Variance that variance is a measure of the relative amount of ‘scattering’  in a given set. The greater the variance, the more the data values in the set are spread out away from the mean.
  ::从计算差异的教训中回顾,差异是衡量某个集中“断层”相对数量的尺度。 差异越大,数据集中的数据值越远离平均值。

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  To calculate the variance of a population:
  ::计算人口差异:

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  • First, identify the arithmetic mean (the average) of your data by finding the sum of the values and dividing it by the number of values.
    ::首先,通过找到数值的总和并将其除以数值数,来标明数据中的算术平均值(平均值)。
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  • Next, subtract each value from the mean and record the result. This value is called the deviation of each score from the mean.
    ::接下来,从平均值中减去每个值并记录结果。这个值被称为每个得分与平均值的偏差。
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  • For each value, square the deviation.
    ::每个值的偏差平方。
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  • Finally,  find the mean of the squared deviations . This is  the variance of the set.
    ::最后,找到正方形偏差的平均值。 这是一组偏差的平均值 。

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  • To find the SD, simply take the square root of the variance .
    ::为了找到SD, 只需选择差异的平方根。

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  One important difference between the variance and the standard deviation is that the units associated with variance are the square of the units of the original values, but the units associated with the standard deviation are the same as the units in the original set.
  ::差异和标准差之间的一个重要区别是,与差异有关的单位是原值单位的平方,但与标准差有关的单位与原数据集中的单位相同。

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  We will return to SD in our chapter on “Normal Distribution”, when we will further discuss the uses of the SD of both samples and populations.
  ::我们将在关于“正常分布”的一章中回到自毁问题,届时我们将进一步讨论样品和人口对自毁的用途。

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  Calculating Standard Deviation Given Variance 
  ::计算标准差 差 差

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  What is the standard deviation of a set with  $\begin{array}{r}{\sigma }^2\end{array}$ of 14.6?
  ::0.12乘以14.6的标准偏差是多少?

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  Solution: The standard deviation  $\begin{array}{r}(\sigma )\end{array}$ is simply the square root of the variance $\begin{array}{r}({\sigma }^2)\end{array}$ . As a formula: $\begin{array}{r}\sqrt{{\sigma }^2}\end{array}$ .
  ::解决办法:标准偏差(12-12)只是差异(12-12)的平方根。

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  In this case we have:  $\begin{array}{r}\sqrt{14.6}=3.821\therefore \sigma =3.821\end{array}$
  ::在这种情况下,我们有:14.6=3.821 3.821

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  Calculating the Standard Deviation of a Data Set 
  ::计算一套数据集的标准偏离

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  What is the  $\begin{array}{r}\sigma \end{array}$ of set $\begin{array}{r}x\end{array}$ ?
  ::什么是... 数组x?

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  $$\begin{array}{r}x=\left\{3,4,5,6,7,8,9\right\}\end{array}$$

  
  ::x3,4,5,6,7,7,8,9}

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  First find the variation of the set:
  ::首先找到集的变异 :

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  • $\begin{array}{r}\mu (mean)=\frac{3+4+5+6+7+8+9}{42}=6\end{array}$
    ::μ( 平均值) = 3+4+5+6+7+8+942=6
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  • Deviations and squared deviations:
    • $\begin{array}{r}3-6=-3\to (-3{)}^2=9\end{array}$
    • $\begin{array}{r}4-6=-2\to (-2{)}^2=4\end{array}$
    • $\begin{array}{r}5-6=-1\to (-1{)}^2=1\end{array}$
    • $\begin{array}{r}6-6=0\to (0{)}^2=0\end{array}$
    • $\begin{array}{r}7-6=+1\to (+1{)}^2=1\end{array}$
    • $\begin{array}{r}8-6=+2\to (+2{)}^2=4\end{array}$
    • $\begin{array}{r}9-6=-3\to (-3{)}^2=9\end{array}$
    
    ::偏差和正方形偏差: 3-63(-3-3)2=9 4-62(-2)2=4 5-61(-1)2=1 6-6=0.6=0.6(0)2=0 7-61(+1)2=1 8-622(+2)2=4 9-63(-3)2=9
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  • $\begin{array}{r}\text{Sum of squared deviations}=9+4+1+0+1+4+9=28\end{array}$
    ::平方偏差=9+4+1+0+1+4+4+9=28
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  • $\begin{array}{r}\text{Variation}=\frac{28}{7}=4\end{array}$
    ::变化=287=4

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  $$\begin{array}{r}\therefore Standarddeviationofsetx=\sqrt{4}=2\end{array}$$

  
  ::X=4=2 集的标准偏差

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  Finding the Variance and Standard Deviation 
  ::查找差异和标准偏离

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  Katrina wants to use the average scores of the top long jumpers at the 5 schools in her district to predict the average long jumps for top competitors at all schools in her state. Data for her district is below. Find the appropriate of the jumps.
  ::卡特里娜想用她所在地区的5所学校中顶尖长途跳跃者的平均分数来预测她所在州所有学校中顶尖竞争者的平均长途跳跃。 她所在地区的数据在下面。 找到跳跃的适当位置 。

  

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  $$\begin{array}{rl}& \text{School}\mathrm{\#}1{24}^{\mathrm{\prime }}{10.5}^{\mathrm{\prime }\mathrm{\prime }}\\ & \text{School}\mathrm{\#}2{24}^{\mathrm{\prime }}{8.5}^{\mathrm{\prime }\mathrm{\prime }}\\ & \text{School}\mathrm{\#}3{24}^{\mathrm{\prime }}{4.25}^{\mathrm{\prime }\mathrm{\prime }}\\ & \text{School}\mathrm{\#}4{24}^{\mathrm{\prime }}{1.75}^{\mathrm{\prime }\mathrm{\prime }}\\ & \text{School}\mathrm{\#}5{23}^{\mathrm{\prime }}{10.5}^{\mathrm{\prime }\mathrm{\prime }}\end{array}$$

  
  ::学校#124_10.5_学校#224_8.5/8_学校#324_4.25_学校#424_1.75_学校#523_10.5/5_

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  Since Katrina intends to generalize from her sample data back to the population of jumpers in her state; we need to find the sample variance and corresponding sample standard deviation.
  ::由于Katrina打算从她的样本数据中向她所在州的跳跃者人口进行概括;我们需要找到样本差异和相应的样本标准偏差。

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  • Start by finding the mean distance:  $\begin{array}{r}\mu =\frac{{24}^{\mathrm{\prime }}{10.5}^{\mathrm{\prime }\mathrm{\prime }}+{24}^{\mathrm{\prime }}{8.5}^{\mathrm{\prime }\mathrm{\prime }}+{24}^{\mathrm{\prime }}{4.25}^{\mathrm{\prime }\mathrm{\prime }}+{24}^{\mathrm{\prime }}{1.75}^{\mathrm{\prime }\mathrm{\prime }}+{23}^{\mathrm{\prime }}{10.5}^{\mathrm{\prime }\mathrm{\prime }}}{5}={24}^{\mathrm{\prime }}{4.7}^{\mathrm{\prime }\mathrm{\prime }}\end{array}$
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    • As a decimal :  $\begin{array}{r}{24.39}^{\mathrm{\prime }}\end{array}$
      ::小数点小数点后:24.39
    
    ::以找到平均距离开始 : @ 24_ 10.5 @ 24_ 85. @ 24_ 4. 25 @ 24_ 1. 75 @ 23_ 10. 5 @ 5=24_ 4. 7 小数点: 24. 39_
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  • Deviations and squared deviations of each value:
    • $\begin{array}{r}{24}^{\mathrm{\prime }}{10.5}^{\mathrm{\prime }\mathrm{\prime }}={24.875}^{\mathrm{\prime }}:24.875-24.39=.485\to (.485{)}^2=.235\end{array}$
    • $\begin{array}{r}{24}^{\mathrm{\prime }}{8.5}^{\mathrm{\prime }\mathrm{\prime }}={24.71}^{\mathrm{\prime }}:{24.71}^{\mathrm{\prime }}-{24.39}^{\mathrm{\prime }}=.32\to (.32{)}^2=.102\end{array}$
    • $\begin{array}{r}{24}^{\mathrm{\prime }}{4.25}^{\mathrm{\prime }\mathrm{\prime }}={24.35}^{\mathrm{\prime }}:{24.35}^{\mathrm{\prime }}-{24.39}^{\mathrm{\prime }}=-.04\to (-.04{)}^2=.001\end{array}$
    • $\begin{array}{r}{24}^{\mathrm{\prime }}{1.75}^{\mathrm{\prime }\mathrm{\prime }}={24.15}^{\mathrm{\prime }}:{24.15}^{\mathrm{\prime }}-{24.39}^{\mathrm{\prime }}=-.24\to (-.24{)}^2=.058\end{array}$
    • $\begin{array}{r}{23}^{\mathrm{\prime }}{10.5}^{\mathrm{\prime }\mathrm{\prime }}={23.875}^{\mathrm{\prime }}:23.875-{24.39}^{\mathrm{\prime }}=-.52\to (-.52{)}^2=.270\end{array}$
    
    ::每一值的偏差和正方位偏差: 24_10_524.875_24.39=485_(4852=235)24_852=235 24_852=235)24_71}:24.71_24.39\ 32_(32.2=102)24_24_4.25}24/35_24.35=24.35}24_3904_(-042)=001 24_1.75_24_24_(24_24_24_2=058 23_10.5/5}23.875:23.875:23.875-875-
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  • $\begin{array}{r}\text{Sum of squared deviations}=.235+.102+.001+.058+.270=.666\end{array}$
    ::平方偏差总和=235+.102+.001+.058+.270=666
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  • $\begin{array}{r}\text{Sample variance}=\frac{.666}{4}={.167}^{\mathrm{\prime }}\end{array}$ (Remember to divide by $\begin{array}{r}n-1\end{array}$ , since this is a sample)
    ::样本差异=.6664=167(记住除以n-1,因为这是样本)
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  • $\begin{array}{r}\text{Standard deviation}=\sqrt{.167}={.409}^{\mathrm{\prime }}={4.9}^{\mathrm{\prime }\mathrm{\prime }}\end{array}$
    ::标准差=167=4094.9

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  Earlier  Problem Revisited
  ::重审先前的问题

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  What is standard deviation? How is the standard deviation of a set related to variance? Is the standard deviation of a sample different from that of a population, the way it is with variation?
  ::标准偏差是什么?一组标准偏差与差异有何关系?抽样的标准偏差是否与人口的标准偏差不同,变化如何?

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  By now you should know that standard deviation is a measure of the spread of data, and is calculated as the square root of the variance. Since variance is calculated slightly differently for a sample than for a population, the deviation will differ similarly.
  ::现在,你应该知道,标准偏差是衡量数据分布的尺度,是作为差异的平方根计算的。由于对抽样的差值计算方式与对抽样的差值计算方式略有不同,因此偏差也有所不同。

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  Examples   
  ::实例

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  Example 1
  ::例1

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  Find the mean, variance, and standard deviance of set z.
  ::查找中值、差异和标准偏差 集合 z 。

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  $$\begin{array}{r}z=\left\{12.3,12.5,12.2,11.9,12.6,12.35\right\}\end{array}$$

  
  ::12.3、12.5、12.2、11.9、12.6、12.35}

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  Let’s start by finding the mean, since we will need it to calculate the others:
  ::让我们首先找到一种方法, 因为我们需要它来计算其他方法:

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  $$\begin{array}{rl}\frac{12.3+12.5+12.2+11.9+12.6+12.35}{6}& =12.30833\\ Themean(\mu )& =12.30833\end{array}$$

  
  ::12.3+12.5+12.2+11.9+12.6+12.356=12.30833 平均值(微克)=12.30833

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  Now we calculate the deviation of each value from the mean and square it:
  ::现在我们计算每个值与平均值的偏差和平方值的偏差:

  $\begin{array}{r}12.3-12.30833=-0.00833\to -{0.00833}^2=0.00007\end{array}$

  $\begin{array}{r}12.5-12.30833=0.19167\to {0.19167}^2=0.03674\end{array}$

  $\begin{array}{r}12.2-12.30833=-0.10833\to -{0.10833}^2=0.01174\end{array}$

  $\begin{array}{r}11.9-12.30833=-0.40833\to -{0.40833}^2=0.16673\end{array}$

  $\begin{array}{r}12.6-12.30833=0.29167\to {0.29167}^2=0.08507\end{array}$

  $\begin{array}{r}12.35-12.30833=0.04167\to {0.04167}^2=0.00173\end{array}$

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  Now we sum the squared deviations: $\begin{array}{r}0.00007+0.03674+0.01174+0.16673+0.08507+0.00173=0.30208\end{array}$ , and divide the total by the number of values:  $\begin{array}{r}\frac{0.30208}{6}=0.050347\end{array}$ to get the variance.
  ::现在,我们将平方差数相加:0.0007+0.03674+0.01174+0.16673+0.08507+0.00173+0.00173=0.30208,将总数除以数值数:0.302086=0.050347,以得出差异。

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  $$\begin{array}{r}Thevariance({\sigma }^2)=0.05347\end{array}$$

  
  ::差异(212)=0.05347

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  Finally, to get the standard deviation $\begin{array}{r}(\sigma )\end{array}$ , just take the square root of $\begin{array}{r}{\sigma }^2\end{array}$ .
  ::最后,为了获得标准偏差( 12), 只需选择 12 的平方根 。

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  $$\begin{array}{r}Thestandarddeviation(\sigma )is\sqrt{0.05347}=0.23124\end{array}$$

  
  ::标准偏差( 12) = 0.05347=0. 23124

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  Example 2
  ::例2

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  Find the mean, variance, and standard variance of the following set.
  ::查找下一组的平均值、差异和标准差异。

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  $$\begin{array}{r}y=\left\{9.1,10.1,8.27,7.9,8.6,10.0\right\}\end{array}$$

  
  ::y*9,10.1,8.27,7.9,8.8.6,10.00}

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  Start by finding  $\begin{array}{r}\mu :\frac{9.1+10.1+8.27+7.9+8.6+10.0}{6}=8.995\end{array}$
  ::开始查找 μ:9.1+10.1+8.27+7.9+8.6+10.06=8.995

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  Next, find the squared variation from the mean for each value:
  ::接下来,找到每个值平均值的正方形变异:

  • $\begin{array}{r}9.1-8.995=0.105\to {0.105}^2=0.011025\end{array}$
  • $\begin{array}{r}10.1-8.995=1.105\to {1.105}^2=1.221025\end{array}$
  • $\begin{array}{r}8.27-8.995=-0.725\to {0.525625}^2=0.276281\end{array}$
  • $\begin{array}{r}7.9-8.995=-1.095\to -{1.095}^2=1.199025\end{array}$
  • $\begin{array}{r}8.6-8.995=-0.395\to {0.105}^2=0.156025\end{array}$
  • $\begin{array}{r}10.0-8.995=1.005\to {0.105}^2=1.010025\end{array}$

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  Sum the squared deviations and divide by the number of values to get the variance:
  ::平方偏差和除以数值数以得出差数的平方差和除法 :

  $$\begin{array}{r}{\sigma }^2=\frac{3.873406}{6}=0.6455677\end{array}$$

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  Finally, take the square root of the variance to get the standard deviation:
  ::最后,选择差异的平方根来获得标准偏差:

  $$\begin{array}{r}\sigma =\sqrt{0.6455677}=.8034723\end{array}$$

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  Example 3
  ::例3

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  Which of the following sets has the greater standard deviation?
  ::以下哪几组标准差较大?

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  $$\begin{array}{r}x=\left\{2,4,6,8,10\right\}y=\left\{3,5,7,9,11,13\right\}\end{array}$$

  
  ::x2,4,6,6,8,10}y$3,5,7,9,11,13}

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  Follow the same series of steps to find the standard deviation of each set.
  ::遵循同样的一系列步骤来查找每套标准偏差。

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  • $\begin{array}{r}x=\left\{2,4,6,8,10\right\}:\mu =6,{\sigma }^2=10,\sigma =3.16228\end{array}$
    ::=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================
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  • $\begin{array}{r}y=\left\{3,5,7,9,11,13\right\}:\mu =8,{\sigma }^2=14,\sigma =3.74166\end{array}$
    ::y3,5,7,9,11,13}8,12=14,3.74166

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  Set $\begin{array}{r}y\end{array}$ has the greater standard deviation
  ::Sety 具有更大的标准偏差

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  Example 4
  ::例4

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  Kevin takes a random sample of students in his class, and gets the following values, what is the sample variance and standard deviation of this set?
  ::Kevin抽取了他的班级学生的随机抽样, 并获得以下值, 抽样差异和标准差是什么?

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  $$\begin{array}{r}a=\left\{15,16,16,15,17,17,18,16,17,16,18,18,15\right\}\end{array}$$

  
  ::a15、16、16、15、17、17、18、16、17、16、18、18、15}

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  There are 13 values, with  $\begin{array}{r}\mu =16.46154\end{array}$
  ::共有13个值,其中16.46154

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  • The sum of the squared deviations is: 15.2308, divide by 12 (since this is a sample!), to get the sample variance : $\begin{array}{r}\frac{15.2308}{12}=1.26923\end{array}$
    ::平方偏差的总和是:15 2308,除以12(因为这是抽样!)),得出抽样差异:15 230812=1.26923。
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  • The square root of the sample variance is the sample standard deviation: $\begin{array}{r}\sqrt{1.26923}=1.0824\end{array}$
    ::抽样差异的平方根是抽样标准差:1.26923=1.0824

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  Review 
  ::回顾

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  Find  $\begin{array}{r}\mu ,{\sigma }^2\end{array}$ and $\begin{array}{r}\sigma \end{array}$ :
  ::查找 μ、 1212 和 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

  1. 265, 280.7, 293, 279, 314.2, 300, 289

  2. 7200, 7020, 7165.9, 7100, 7196, 7112, 7116.1

  3. 27, 20.3, 30.7, 40, 46, 36, 40, 33

  4. 3607, 3600, 3600, 3631, 3600.6

  5. 700, 700, 712, 736, 741, 716, 782

  6. 3370, 3300.5, 3366, 3306.6, 3310, 3336, 3301.3

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  Calculate the sample standard deviation:
  ::计算抽样标准差:

  7. 34.4, 34, 34.7, 34.6, 34, 34.1, 31, 31.3

  8. 989.22, 990.6, 992, 996.9, 981.1, 986, 975

  9. 10, 16, 10.33, 10.63, 18, 17, 16.36, 10.46

  10. 3240, 3260, 3250, 3280, 3280, 3300, 3310, 3270

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  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。