Course: 5.6 计算差异 | The Way To Learn

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计算差异

If you were told that the mean income at a certain company was $35,000, you wouldn’t really know much about the actual income of the majority of the employees, since there could be a few upper-level managers or owners whose income might skew the mean badly. However, if you were also given the variance of the incomes, how would that help?
::如果有人告诉您某家公司的平均收入是35,000美元,那么您不会真正了解大多数雇员的实际收入,因为可能会有一些高层管理人员或业主的收入可能严重扭曲其平均收入。但是,如果同时考虑到收入差异,那又有什么帮助呢?

Calculating Variance
::计算差异

Variance (commonly denoted $\begin{align*}\sigma ^2\end{align*}$ ) is a very useful measure of the relative amount of ‘scattering’ of a given set. In other words, knowing the variance can give you an idea of how closely the values in a set cluster around the mean. The greater the variance, the more the data values in the set are spread out away from the mean.
::差异( 通常表示 312 ) 是一个非常有用的指标, 用来衡量某个集的“ 折叠” 相对数量。换句话说, 了解差异可以让您知道一个集的值在平均值周围的值有多近。差异越大, 数据集中的数据值越离平均值越远。

Variance is an important calculation to become familiar with because, like the arithmetic mean , variance is used in many other more complex statistical evaluations. The calculation of variance is slightly different depending on whether you are working with a population (you do not intend to generalize the results back to a larger group) or a sample (you do intend to use the sample results to predict the results of a larger population). The difference is really only at the end of the process, so let’s start with the calculation of the population.
::差异是需要熟悉的重要计算方法,因为与算术平均值一样,差异被用于许多其他更为复杂的统计评估。差异的计算略有不同,取决于您是从事人口工作(您不打算将结果推广到较大群体)还是从事抽样工作(您确实打算使用抽样结果来预测较大人口的结果 ) 。差异实际上只在过程结束时才能出现,所以让我们从人口计算开始。

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

First, identify the arithmetic mean 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除以数值数来标明数据中的算术平均值。

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。

For each value, square the deviation .
::每个值的偏差平方。

Finally, divide the sum of the squared deviations by the number of values in the set. The resulting quotient is the variance $\begin{align*}(\sigma^2)\end{align*}$ of the set.
::最后,将平方偏差总和除以集中的值数。由此得出的商数是集的差异(1212) 。

To calculate the variance of a sample, the only difference is that in step 4, you divide the sum of squared deviations by the number of values in the sample minus 1 . By dividing the sum of squared deviations by one less than the number of values, you help reduce the effect of outliers in the sample and increase the calculated variance of the sample by a small amount to allow more ‘room’ for the unknown values in the population.
::要计算抽样的差异,唯一的差别是,在第四步中,将平方偏差的总和除以抽样中的数值数乘以负1。通过将平方偏差的总和除以少于数值数的一小部分,帮助减少抽样中离值的影响,并将抽样的计算差数增加一小部分,使人口中未知值的 " 范围 " 增加更多。

Calculating the Variance
::计算差异

1. Calculate the variance of set $\begin{align*}x\end{align*}$ :
::1. 计算集x的差异:

\begin{aligned} x = {12, 7, 6, 3, 10, 5, 18, 15} \end{aligned}

$\begin{align*}x=\left \{12, 7, 6, 3, 10, 5, 18, 15\right \}\end{align*}$
::x12,7,6,6,3,10,5,18,15}

Follow the steps from above to calculate the variance:
::遵循以上步骤计算差异:

First, calculate the arithmetic mean:
::首先,计算算术的意思是:

\begin{aligned} μ = \frac{12 + 7 + 6 + 3 + 10 + 5 + 18 + 15}{8} = 9.5 \end{aligned}

$\begin{align*}\mu =\frac{12+7+6+3+10+5+18+15}{8}=9.5\end{align*}$

Subtract each value from the mean to get the deviation of each value, square the deviation of each value:
::将每个值从平均值中减去,以获得每个值的偏差,使每个值的偏差平方:

*$\begin{align}\text{Value} - \text{Mean} = \text{Deviation}\end{align}$*	*$\begin{align}\text{Deviation}^2\end{align}$*
$\begin{align}12-9.5=2.5\end{align}$	6.25
$\begin{align}7-9.5=-2.5\end{align}$	6.25
$\begin{align}6-9.5=-3.5\end{align}$	12.25
$\begin{align}3-9.5=-6.5\end{align}$	42.25
$\begin{align}10-9.5=.5\end{align}$	.25
$\begin{align}5-9.5=-4.5\end{align}$	20.25
$\begin{align}18-9.5=8.5\end{align}$	72.25
$\begin{align}15-9.5=5.5\end{align}$	30.25
TOTAL (sum of deviation ² ):	190.00

Finally, divide the sum of the squared deviations by the count of values in the data set :
::最后,将平方偏差之和除以数据集中的数值计数:

\begin{aligned} \frac{190}{8} & = 23.75 \\ ∴ T h e v a r i a n c e & o f s e t x i s 23.75 \end{aligned}

$\begin{align*}\frac{190}{8} & =23.75\\ \therefore \ The \ variance \ & of \ set \ x \ is \ 23.75\end{align*}$
::1908=23.75 * 成套物品x的差异为23.75

2. Find the variance of set $\begin{align*}z\end{align*}$ :
::2. 查找z集的差值:

\begin{aligned} z = {1, 2, 3, 4, 5, 6, 7, 9} \end{aligned}

$\begin{align*}z=\left \{1, 2, 3, 4, 5, 6, 7, 9\right \}\end{align*}$
::日元1,2,3,4,5,6,7,7,9}

Divide the squared deviation of each value from the mean by the total number of values in the set:
::将每个数值的平方偏差与平均值除以集中的数值总数:

\begin{aligned} μ = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 9}{8} = 4.625 \\ (1 - 4.625)^{2} + (2 - 4.625)^{2} + (3 - 4.625)^{2} + (4 - 4.625)^{2} \\ + (5 - 4.625)^{2} + (6 - 4.625)^{2} + (7 - 4.625)^{2} + (9 - 4.625)^{2} = 49.875 \\ \frac{49.875}{8} = 6.234 \\ ∴ V a r i a n c e (σ^{2}) o f s e t z = 6.234 \end{aligned}

$\begin{align*}& \qquad \qquad \mu =\frac{1+2+3+4+5+6+7+9}{8}=4.625 \\ &(1-4.625)^2+(2-4.625)^2+(3-4.625)^2+(4-4.625)^2\\ &\qquad +(5-4.625)^2+(6-4.625)^2+(7-4.625)^2+(9-4.625)^2 =49.875 \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \frac{49.875}{8} =6.234\\ & \qquad \qquad \qquad \ \therefore \ Variance \ (\sigma^2) \ of \ set \ z = 6.234\end{align*}$
::+1+2+3+3+4+5+6+6+7+7+98=4.625(1-4.625)2+(2-4.625)2+(2-4.625)2+(3-4.625)2+(4-4.625)2+(5-4.625)2+(6-4.625)2+(7-4.625)2+(7-4.625)2+(9-4.625)2+(9-4.625)2=49.875 49.8758=6.234 =6.234

3. Find $\begin{align*}\sigma^2 \ of \ y\end{align*}$ :
::3. 查找 y 的 0.12:

\begin{aligned} y = {13, 14, 15, 16, 17, 18, 19, 20, 21} \end{aligned}

$\begin{align*}y=\left \{13, 14, 15, 16, 17, 18, 19, 20, 21\right \}\end{align*}$
::y 13, 14, 15, 16, 17, 18, 18, 19, 20, 21}

Let’s do this one differently, using a nifty trick known as the “mean of the squares minus the square of the mean.” Start, as before, by finding the arithmetic mean:
::让我们用不同的方法来做这件事,使用“平方的平均值减去平均值的平方 ” ( money of the squals down the squal of the squal of the squal ) 。和以前一样,首先发现算术的意思是:

\begin{aligned} μ = \frac{13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21}{9} = 17 \end{aligned}

$\begin{align*}\mu =\frac{13+14+15+16+17+18+19+20+21}{9}=17\end{align*}$

Then, to find the variation, divide the sum of the squares of each value by the number of values (this is the “mean of the squares”), then square the mean we calculated above, 17 (the “square of the mean”), and subtract it from the mean of the squares:
::然后,为了找到变异,将每个值的平方平方之和除以数值数(这是“平方的平均值”),然后将我们计算出的平均值平方17(“平方的平均值”),从平方的平均值中减去:

\begin{aligned} σ^{2} = \frac{13^{2} + 14^{2} + 15^{2} + 16^{2} + 17^{2} + 18^{2} + 19^{2} + 20^{2} + 21^{2}}{9} - 17^{2} = 6.6 \bar{6} \\ ∴ σ^{2} o f y = 6.6 \bar{6} \end{aligned}

$\begin{align*}&\sigma ^2 = \frac{13^2+14^2+15^2+16^2+17^2+18^2+19^2+20^2+21^2}{9}-17^2=6.6\overline{6} \\ &\qquad \qquad \qquad \qquad \qquad \quad \therefore \ \sigma ^2 \ of \ y=6.6\overline{6}\end{align*}$
::312=132+142+142+152+162+162+172+182+182+192+202+202+2129-172=6.66=3.66 y=6.66

Earlier Problem Revisited
::重审先前的问题

If you were told that the mean income at a certain company was $35,000, you wouldn’t really know much about the actual income of the majority of the employees, since there could be a few upper-level managers or owners whose income might skew the mean badly. However, if you were also given the variance of the incomes, how would that help?
::如果有人告诉您某家公司的平均收入是35,000美元,那么您不会真正了解大多数雇员的实际收入,因为可能会有一些高层管理人员或业主的收入可能严重扭曲其平均收入。但是,如果同时考虑到收入差异,那又有什么帮助呢?

By learning the variance of the set of incomes, you could get a feel for how representative the $35,000 figure was of the likely salary of a common employee.
::通过了解一套收入的差异,你可以体会到这个35,000美元的数字与普通雇员可能的工资的多少代表性。

Examples
::实例

Example 1
::例1

Find $\begin{align*}\mu\end{align*}$ and $\begin{align*}\sigma ^2\end{align*}$ of set $\begin{align*}z\end{align*}$ .
::查找 μ 和立方公尺 z 的 0. 12 和 0. 12 。

Let’s use the “mean of the squares minus the square of the mean” method:
::让我们使用“平方除以平方”的方法:

First find the mean of the set: $\begin{align*}\frac{3.25+3.5+2.85+3.4+2.95+3.02+3.17}{7}=3.16286\end{align*}$
::第一个找到集的平均值 : 3.25+3.5+2.85+3.4+2.95+3.02+3.177=3.16286

Now divide the sum of each of the values squared by the number of values:
::现在将每个平方值的和除以数值数:

$\begin{align*}\frac{3.25^2+3.5^2+2.85^2+3.4^2+2.95^2+3.02^2+3.17^2}{7}-10.0036=10.0524-10.0036=0.049\end{align*}$ is the variance.
::3.252+3.52+2.852+3.42+2.952+3.022+3.1727+10.0036=100524-10.0036=0.049是差异。

Example 2
::例2

If all values of set $\begin{align*}z\end{align*}$ , above, were increased by 5, what would the new mean and variance be?
::如果以上z项的所有数值增加5,那么新的平均值和差异是什么?

Find the mean of the new set: $\begin{align*}\frac{8.25+8.5+7.85+8.4+7.95+8.02+8.17}{7}=8.16286\end{align*}$
::查找新套件的平均值:8.25+8.5+7.85+8.4+7.95+8.02+8.077=8.16286

Divide the sum of the values squared by the number of values: $\begin{align*}\frac{466.7668}{7}=66.681\end{align*}$
::将平方值之和除以数值数: 466.76687=66.681

Subtract the squared mean from the mean of the squares: $\begin{align*}66.681-66.632=0.049\end{align*}$ is the variance.
::从正方平均值中减去平方平均值:66 681-66.632=0.049是差异。

The variance is the same as before! Does that surprise you? It should, because they actually aren’t the same, it just appears that way due to rounding. The new set actually has a variance closer to 0.048688, and the original is more accurately 0.04873469. Obviously they are very close, but not exactly the same.
::差异和以前一样!你觉得奇怪吗?它应该,因为它们实际上不同,只是因为四舍五入而看起来是这样。新的组合实际上有接近0.048688的差异,原版更准确地说是0.04873469。显然,它们非常接近,但并不完全相同。

Example 3
::例3

If all values of set $\begin{align*}z\end{align*}$ from question #1 were doubled, how would that affect $\begin{align*}\mu\end{align*}$ and $\begin{align*}\sigma ^2\end{align*}$ ?
::如果问题1的所有z值都翻了一番,那会如何影响 μ 和 □ 12 ?

The question is what would happen if all of the values were doubled. Do the mean and variance also double? Let’s see:
::问题是,如果所有数值都翻一番,会发生什么。平均值和差异是否也翻一番? 让我们看看:

The mean of the new set is $\begin{align*}\frac{6.5+7+5.7+6.8+5.9+6.04+6.34}{7}=\frac{44.28}{7}=6.326\end{align*}$ , which is twice the mean of the original set. So far so good.
::新套套件的平均值为6.5+7+5.7+6.8+6.8+5.9+6.04+6.347=44.287=6.326,是原套件的平均值的两倍。

The “mean of the squares” is $\begin{align*}\frac{6.5^2+7^2+5.7^2+6.8^2+5.9^2+6.04^2+6.34^2}{7}=\frac{281.47}{7}=40.21\end{align*}$ , which is four times the original mean of the squares, not double after all (which makes sense, given that each doubled value was squared).
::“方块平均值”为6.52+72+5.72+5.72+6.82+5.92+6.042+6.3427=281.477=40.21,是方块原始平均值的四倍,但毕竟不是两倍(考虑到每个双倍值是平方值,这是有道理的)。

Finally, subtract the two values: $\begin{align*}40.21-6.326^2 = .192\end{align*}$ is the variance. If we compare this to the original: $\begin{align*}\frac{.192}{.049}\approx 4\end{align*}$ , we can see that doubling the original values quadruples the variance.
::最后,减去两个值:40.21-6.3262=.192是差异。如果我们将这个值与原来的值(.192.049)4相比较,我们可以看到,原来的值翻了一番,使差异翻了四倍。

Review
::回顾

Questions 1-12: find $\begin{align*}\sigma ^2\end{align*}$
::问题1-12: 找到212

$\begin{align*}y=\left \{4, 50, 63, 2, 82, 99\right \}\end{align*}$
::y4,50,63,2,82,99}

Set $\begin{align*}x\end{align*}$ is a random sample from a population with 38 members: $\begin{align*}x=\left \{8, 13, 5, 10\right \}\end{align*}$
::Set x 是来自38个成员组群的随机抽样: x8,13,5,10}

Set $\begin{align*}z\end{align*}$ is a random sample from a larger population: $\begin{align*}z=\left \{4,3,5,15,5\right \}\end{align*}$
::Set z是来自较大人口的随机抽样: z 4,3,5,15,5}

$\begin{align*}y=\left \{3,26,5,1,1\right \}\end{align*}$
::y3,26,5,5,1,1}
22, 21, 13, 19, 16, 18

Sample: 1, 2, 5, 1
::样本: 1, 2, 5, 1

Sample: 10, 6, 3, 4
::样本:10、6、3、4
8, 11, 17, 7, 19
15, 17, 19, 21, 23, 25, 27, 29

Sample: 15, 17, 19, 21, 23, 25, 27, 29
::样本:15、17、19、21、23、25、27、29
.25, .35, .45, .55, .26, .75

Find the variance of the data in the table:
::查找表格中数据的差异:

HEIGHTS (rounded to the nearest inch)	FREQUENCY OF STUDENTS
60	35
61	33
62	45
63	4
64	3
65	4
66	7
67	4

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。

Section outline

General

计算差异

Calculating Variance ::计算差异

Calculating the Variance ::计算差异

Earlier Problem Revisited ::重审先前的问题

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)