7.8 随机变数差异

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  随机随机变数

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  Recently, we discussed the process of finding the mean of a discrete random variable . The process resembled that of finding the arithmetic mean of a set of basic numbers, yet had some significant differences as well. Suppose you needed to know the variance or standard deviation of a random variable. Would these values be calculated differently for random variables than for standard numerical data sets, or not?
  ::最近,我们讨论了寻找离散随机变量平均值的过程。该过程类似于寻找一组基本数字的算术平均值,但也有显著差异。假设你需要知道随机变量的差异或标准偏差。这些数值对随机变量的计算是否不同于标准数字数据集的计算?

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  Random Variable Variance 
  ::随机随机变数

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  As we discussed some time ago, sometimes it is not enough to know the average , or mean, value of a data set when trying to get a feel for the trend(s) of the set. It is the same with a random variable, sometimes you need to know about the spread of a variable to get a better idea of the overall behavior.
  ::正如我们早前讨论过的那样,有时,在试图了解数据集的趋势时,光知道数据集的平均值或中值是不够的。 随机变量是一样的,有时你需要知道变量的分布,才能更好地了解整体行为。

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  One of the other additional pieces of information we learned to calculate in order to evaluate sets before was the variance , which is the square of the standard deviation . Both of these measures help to create an understanding of the tendency of values to cluster around the mean. By evaluating the of a random variable, we can get a better idea of the spread of the values than with the mean alone.
  ::我们为了评估之前的组合而计算的其他额外信息之一就是差异,也就是标准偏差的平方。 这两项措施都有助于使人们了解数值围绕平均值的倾向。 通过评估随机变量,我们可以更好地了解数值的分布,而不是单靠平均值。

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  Just as with the mean, or , we have a formula to apply in order to calculate the variance:

  $$\begin{array}{r}{{\sigma }^2}_X=\sum (x_i-{\mu }_x{)}^2{p}_i\end{array}$$

   
  ::与平均值一样, 或者 , 我们有一个公式要应用 来计算差数 : 1212X (xix) 2pi

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  Then, to find the standard deviation , just take the square root of the variance:
  ::然后,为了找到标准偏差, 只需选择差异的平方根:

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  $$\begin{array}{r}{\sigma }_X=\sqrt{{{\sigma }^2}_X}\end{array}$$

  
  ::==================================================================================

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  Calculating the Variance and Standard Deviation 
  ::计算差异和标准偏离

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  1. In another lesson, we calculated the expected value of the number of the number of kids that Sally baby sits on any given day from the data in the table below. Using the table and the mean, calculate the variance and standard deviation of the number of kids she baby sits.
  ::1. 在另一个教训中,我们从下表的数据中计算了Sally婴儿在某一天所坐儿童人数的预期值,用表格和平均数计算她所坐儿童人数的差异和标准差。

  $\begin{array}{r}x\end{array}$	1	2	3	4	5
  $\begin{array}{r}P(X=x)\end{array}$	.35	.40	.15	.05	.05

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  $$\begin{array}{r}{\mu }_X=2\end{array}$$

  
  ::μX=2

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  Use the data given in the question to fill in the formula and find the variance:
  ::使用问题中提供的数据填充公式并找出差异:

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  $$\begin{array}{r}\text{Formula:}{{\sigma }^2}_X=\sum (x_i-{\mu }_x{)}^2{p}_i\end{array}$$

  
  ::公式 : 012X(xIx) 2pi

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  $$\begin{array}{rl}& {{\sigma }^2}_X=(1-2{)}^2\times .35+(2-2{)}^2\times .4+(2-3{)}^2\times .15+(2-4{)}^2\times .05+(2-5{)}^2\times .05\\ & (1\times .35)+(0\times .4)+(1\times .15)+(4\times .05)+(9\times .05)\\ & .35+0+.15+.2+.45\\ & {{\sigma }^2}_X=1.15\end{array}$$

  
  ::282X=(1-2)2×35+(2-2-2)2×4+(2-3)2×15+(2-4)2×15+(2-4)2×2.05+(2-5)2×.05+(1x.35)+(0x.4)+(1x.15)+(4x.05)+(9x.05)+(9x.05)35+0+15+2(2+45-2x=1.15

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  Since the variance is 1.15, the standard deviation is $\begin{array}{r}\sqrt{1.15}=1.07\end{array}$
  ::由于差异为1.15,标准差为1.15=1.07。

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  $$\begin{array}{r}{\sigma }_X=1.07\end{array}$$

  
  ::X=1.07

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  2. Random variable  $\begin{array}{r}X\end{array}$ has mean 18.84, and the probability distribution show below. Calculate the variance and standard deviation.
  ::2. 随机变数X的平均值为18.84,概率分布如下。计算差异和标准偏差。

  $\begin{array}{r}x\end{array}$	3	11	19	27
  $\begin{array}{r}P(X=x)\end{array}$	.07	.08	.65	.20

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  Using the variance formula: $\begin{array}{r}{{\sigma }^2}_X=\sum (x_i-{\mu }_x{)}^2{p}_i\end{array}$
  ::使用差异公式 : 012X (xI*x) 2pi

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  $$\begin{array}{rl}& {{\sigma }^2}_X=(3-18.84{)}^2\times .07+(11-18.84{)}^2\times .08+(19-18.84{)}^2\times .65+(27-18.84{)}^2\times .2\\ & {{\sigma }^2}_X=(-15.84{)}^2\times .07+(-7.84{)}^2\times .08+(.16{)}^2\times .65+(8.16{)}^2\times .2\\ & {{\sigma }^2}_X=251\times .07+61.5\times .08+.03\times .65+66.59\times .2\\ & {{\sigma }^2}_X=17.6+4.9+.02+13.3\\ & {{\sigma }^2}_X=35.8\end{array}$$

  
  ::282X=(3-18.84)2xx.07+(11-18.84)2x.08+(19-18.84)2x.65+(27-18.84)2x2xxx=(15.84)2x.07+(-7.84)2x.08+(16)2x.65+(8.16)2x2x2.21x=251x0.07+6.5xx.08+.03x.65+.66x.65x.59x2.21x=17.6+4.9+.02+13.3-2X=35.8

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  Since the variance is 36.3, the standard deviation is $\begin{array}{r}\sqrt{35.8}=5.98\end{array}$
  ::由于差异为36.3,标准差为35.8=5.98。

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  $$\begin{array}{r}{\sigma }_X=5.98\end{array}$$

  
  ::X=5.98

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  3. The random variable $\begin{array}{r}Z\end{array}$ has a probability distribution shown below, find $\begin{array}{r}{\mu }_Z\end{array}$ , $\begin{array}{r}{{\sigma }^2}_Z\end{array}$ , and $\begin{array}{r}{\sigma }_Z\end{array}$ .
  ::3. 随机变量Z的概率分布如下: ===================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

  $\begin{array}{r}x\end{array}$	.65	.84	1.03	1.22	1.41
  $\begin{array}{r}P(X=x)\end{array}$	.16	.29	.14	.28	.13

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  Start by finding the mean of $\begin{array}{r}Z\end{array}$ :
  ::从发现Z的平均值开始:

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  $$\begin{array}{r}{\mu }_Z=(.65\times .16)+(.84\times .29)+(1.03\times .14)+(1.22\times .28)+(1.41\times .13)=1.02\end{array}$$

  
  :.65x.16)+(.84x.29)+(.03x.14)+(1.22x.28)+(1.41x.13)=1.02

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  Now that we have the mean, we can use it to find the variance:
  ::既然我们有这个本能, 我们可以用它来找出差异:

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  $$\begin{array}{rl}{{\sigma }^2}_Z& =(.65-1.02{)}^2\times .16+(.84-1.02{)}^2\times .29+(1.03-1.02{)}^2\times .14+(1.22-1.02{)}^2\times .28+(1.41-1.02{)}^2\times .13\\ {{\sigma }^2}_Z& =(-.37{)}^2\times .16+(-.18{)}^2\times .29+(.01{)}^2\times .14+(.2{)}^2\times .28+(.39{)}^2\times .13\\ {{\sigma }^2}_Z& =.14\times .16+.03\times .29+0\times .14+.04\times .28+.16\times .13\\ {{\sigma }^2}_Z& =.02+.01+.01+.02\\ {{\sigma }^2}_Z& =.06\end{array}$$

  
  ::#^lg39#.#^lg39#.#^lg39#.#^lg39#.#^lg39#.#^lg39#.

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  Finally, the standard deviation is just the square root of the variance:
  ::最后,标准偏差只是差异的平方根:

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  $$\begin{array}{r}{\sigma }_Z=\sqrt{.06}=.25\end{array}$$

  
  ::06.=25

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  Earlier  Problem Revisited
  
  ::重审先前的问题

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  Suppose you needed to know the variance or standard deviation of a random variable. Would these values be calculated differently for random variables than for standard numerical data sets or not?
  ::假设您需要知道随机变量的差异或标准偏差。这些数值是随机变量的计算方法还是标准数字数据集的计算方法?

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  The variance and standard deviation are the same concept when dealing with random variable as with numerical data sets. However, the process of calculating the values is slightly different. Instead of dividing the squared difference of each number and the mean by the count of values: $\begin{array}{r}\frac{(x-\mu {)}^2}{n}\end{array}$ you multiply the square of the difference of each number and the mean by the probability of that value: $\begin{array}{r}(x-\mu {)}^2\times P(x)\end{array}$ .
  ::在处理随机变量时,差异和标准偏差的概念与处理数字数据集时相同。然而,计算数值的过程略有不同。不是将每个数字的平方差和平均值除以数值的计数,而是将每个数字的平方差和平均值除以数值的计数x)2n,你乘以每个数字的平方差和平均值乘以该数值的概率x)2xP(x)。

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  In either case, the standard deviation is the square root of the variance.
  ::在这两种情况下,标准偏差都是差异的平方根。

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  Examples 
  ::实例

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  Example 1
  ::例1

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  Calculate the variance and standard deviation of random variable  $\begin{array}{r}Y\end{array}$ , given: $\begin{array}{r}{\mu }_Y=43.2\end{array}$ and:
  ::计算随机变量Y的差异和标准差,给定: μY=43.2 和:

  $\begin{array}{r}x\end{array}$	15	30	45	60	75
  $\begin{array}{r}P(Y=x)\end{array}$	.20	.25	.15	.27	.13

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  All of the values we need for this one are given, it is really just a “plug-n-chug” using the variance formula: $\begin{array}{r}{{\sigma }^2}_X=\sum (x_i-{\mu }_x{)}^2{p}_i\end{array}$
  ::给出了我们给此数值所需的全部值, 它实际上只是一个使用差异公式的“ plut- n- cug” : 312X(xix) 2pi

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  $$\begin{array}{rl}{{\sigma }^2}_Y& =(15-43.2{)}^2\times .2+(30-43.2{)}^2\times .25+(45-43.2{)}^2\times .15+(60-43.2{)}^2\times .27+(75-43.2{)}^2\times .13\\ {{\sigma }^2}_Y& =(-28.2{)}^2\times .2+(-13.2{)}^2\times .25+(-1.8{)}^2\times .15+(16.8{)}^2\times .27+(31.8{)}^2\times .13\\ {{\sigma }^2}_Y& =159+43.6+.5+76.2+131.5\\ {{\sigma }^2}_Y& =410.8\\ {\sigma }_Y& =\sqrt{410.8}=20.27\end{array}$$

  
  ::-2Y=(15-43.2)2×2+(30-43.2)2×2×25+(45-43.2)2×15+(60-43.2)2×(60-43.2)2×.27+(75-43.2)2×13-23-2Y=(-28.2)2×(13-2)2×25+(1.8)2×15+(16.8)2×(16.8)2×2.27+(31.8)2×13-12-2Y=159+43.6+5.76.2+131.512-2Y=4.10.8-12-Y=4.10.8Y=4.10.8=20.27

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  Example 2
  ::例2

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  Find $\begin{array}{r}{\mu }_X\end{array}$ ,  $\begin{array}{r}{\sigma }_X\end{array}$ , and  $\begin{array}{r}{{\sigma }^2}_X\end{array}$ given:
  ::查找 μX, X, 和 2X 给定 :

  $\begin{array}{r}x\end{array}$	4	8	12	16	20
  $\begin{array}{r}P(Y=x)\end{array}$	.50	.25	.15	.05	.05

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  First, calculate the mean
  ::首先计算平均值

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  $$\begin{array}{rl}& {\mu }_X=(4\times .5)+(8\times .25)+(12\times .15)+(16\times .05)+(20\times .05)=7.6\end{array}$$

  
  ::μX=(4x.5)+(8x.25)+(12x.15)+(16x.05)+(20x.05)=7.6

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  Then, use the mean to calculate the variance:
  ::然后,使用平均值来计算差异:

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  $$\begin{array}{rl}& {{\sigma }^2}_X=(4-7.6{)}^2\times .5+(8-7.6{)}^2\times .25+(12-7.6{)}^2\times .15+(16-7.6{)}^2\times .05+(20-7.6{)}^2\times .05\\ & {{\sigma }^2}_X=20.64\\ & {\sigma }_X=\sqrt{20.64}=4.5\end{array}$$

  
  ::282X=(4-7.6)2x5+(8-7.6)2x.25+(12-7.6)2x.25+(12-7.6)2x.15+(16-7.6)2x.05+(20-7.6)2x.05-2x=20.64-X=20.64=4.5

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  Example 3
  ::例3

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  Marie has a part-time job walking dogs to earn money on weekends. The following probability distribution represents the probability of having a particular number of clients on any given day. If she earns $2.75 per client, how much could she expect to earn each day, on average, and what is the standard deviation of her expected earnings?
  ::Marie有一个兼职的散养狗在周末挣钱。接下来的概率分布是某一天有一定数量客户的概率。 如果她每个客户挣2.75美元,平均每天能挣多少钱,以及预期收入的标准偏差是什么?

  播放段落 # clients ::* 客户	20	25	30	35	40
  播放段落 probability ::概率概率概率	.15	.35	.30	.15	.05

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  Start by finding the mean:
  ::首先,首先发现以下意思:

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  $$\begin{array}{r}{\mu }_X=20\times .15+25\times .35+30\times .3+35\times .15+40\times .05=28\end{array}$$

  
  ::μX=20×15+25×35+30×3+35×15+40×0.05=28

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  Use the mean to find the variance:
  ::使用平均值查找差异:

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  $$\begin{array}{r}{{\sigma }^2}_X=(20-28{)}^2\times .15+(25-28{)}^2\times .35+(30-28{)}^2\times .30+(35-28{)}^2\times .15+(40-28{)}^2\times .05=28.5\end{array}$$

  
  ::-2X=(20-28)2×15+(25-28)2×35+(30-28)2×30+(35-28)2×30+(35-28)2×15+(40-28)2×0.05=28.5

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  Use the variance to find the standard deviation: $\begin{array}{r}{\sigma }_X=\sqrt{28}=5.3\end{array}$
  ::使用差异来查找标准差: {X=28=5.3

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  Now we can find her average income by multiplying the mean, 28 by Marie’s rate, $2.75, to get her average daily income of $77 .
  ::现在,我们可以通过将平均数乘以28乘以Marie的2.75美元率来找到她的平均收入,使平均日收入达到77美元。

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  Finally, we can multiply the calculated standard deviation, 5.3, by the rate, $2.75, to get the standard deviation of her income: $\begin{array}{r}5.3\times \$2.75=\$14.58\end{array}$
  ::最后,我们可以将计算的标准差5.3乘以2.75,以获得其收入的标准差:5.3×2.75=14.58美元。

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  What all this means is that Marie can expect to average $77 per day, on average, give or take about $14.50.
  ::所有这一切意味着,玛丽平均每天平均预期要77美元,平均要14.50美元。

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  Review 
  ::回顾

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  For questions 1 – 9, find the variance and standard deviation of the random variable, given the mean and probability distribution.
  ::对于问题1 - 9, 找到随机变量的差异和标准偏差, 取决于平均值和概率分布 。

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  1. $\begin{array}{r}{\mu }_x=4.435\end{array}$
  ::1. 微x=4.435

  $\begin{array}{r}x\end{array}$	4.1	4.4	4.7	4.9	5.1
  $\begin{array}{r}P(X=x)\end{array}$	.30	.45	.10	.05	.10

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  2. $\begin{array}{r}{\mu }_x=7.6\end{array}$
  ::2. 微克x=7.6

  播放段落 $\begin{array}{r}x\end{array}$ ::x x	4	8	12	16	20
  播放段落 $\begin{array}{r}P(X=x)\end{array}$ ::P(X=x)	.50	.25	.15	.05	.05

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  3. $\begin{array}{r}{\mu }_x=43.2\end{array}$
  ::3. 微克x=43.2

  播放段落 $\begin{array}{r}x\end{array}$ ::x x	15	30	45	60	75
  播放段落 $\begin{array}{r}P(X=x)\end{array}$ ::P(X=x)	.20	.25	.15	.27	.13

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  4. $\begin{array}{r}{\mu }_X=93\end{array}$
  ::4. μX=93

  播放段落 $\begin{array}{r}x\end{array}$ ::x x	30	60	90	120	150	170
  播放段落 $\begin{array}{r}P(X=x)\end{array}$ ::P(X=x)	.18	.16	.24	.22	.20	.00

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  5. $\begin{array}{r}{\mu }_X=12.92\end{array}$
  ::5. μX=12.92

  播放段落 $\begin{array}{r}x\end{array}$ ::x x	5	9	13	17
  播放段落 $\begin{array}{r}P(X=x)\end{array}$ ::P(X=x)	.07	.08	.65	.20

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  6. $\begin{array}{r}{\mu }_X=21.80\end{array}$
  ::6. μX=21.80

  播放段落 $\begin{array}{r}x\end{array}$ ::x x	13	17	21	25	29	33	37
  播放段落 $\begin{array}{r}P(X=x)\end{array}$ ::P(X=x)	.15	.17	.23	.30	.10	.03	.02

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  7. $\begin{array}{r}{\mu }_X=57.98\end{array}$
  ::7. μX=57.98

  播放段落 $\begin{array}{r}x\end{array}$ ::x x	26	39	52	65	78
  播放段落 $\begin{array}{r}P(X=x)\end{array}$ ::P(X=x)	6%	14%	30%	28%	22%

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  8. $\begin{array}{r}{\mu }_X=64.99\end{array}$
  ::8. μX=64.99

  播放段落 $\begin{array}{r}x\end{array}$ ::x x	22	43	64	85	106
  播放段落 $\begin{array}{r}P(X=x)\end{array}$ ::P(X=x)	10.5%	22.5%	31.5%	22.8%	12.7%

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  9. $\begin{array}{r}{\mu }_X=7.46\end{array}$
  ::9. 微X=7.46

  播放段落 $\begin{array}{r}x\end{array}$ ::x x	3.65	5.84	7.03	9.22	11.41
  播放段落 $\begin{array}{r}P(X=x)\end{array}$ ::P(X=x)	.16	.25	.18	.24	.17

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  10. Dorian works for a construction company, where he earns $11.50 per hour. The number of hours he works each week varies between 25 and 40. Based on prior experience, Dorian has compiled the probability distribution below describing the probability that he will work a given number of hours. Can Dorian afford to buy a new truck that has a payment of $525/month, if he wants to be sure not to put more than 25% of his average monthly income into car payments? What is the standard deviation of his monthly income?
  ::10. Dorian为一家建筑公司工作,每小时收入11.50美元,每周工作时数从25小时到40小时不等,根据以往的经验,Dorian将概率分布汇编在下面,说明他将工作一个小时的概率;Dorian能否购买一辆新卡车,支付525个月的工资,如果他想确定不会将月平均收入的25%以上用于汽车付款,那么他能否买得起一辆新卡车,支付525个月的工资?他月收入的标准偏差是什么?

  播放段落 # hours ::# 小时	25	28	31	34	37	40
  播放段落 probability ::概率概率概率	.15	.14	.26	.18	.14	.13

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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 。