7.10 变换随机变数二

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  变换随机变量二

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  In the previous lesson, Transforming Random Variables I, we considered the following problem, and partially solved it, but did not yet know how to handle the $0.30 per customer materials cost. In this lesson we will learn how to multiply random variables by a constant (like 0.30), so we can finish solving the problem.
  ::在前一课《变换随机变量一》中,我们考虑了以下问题,并部分解决了这一问题,但还不知道如何处理每客户材料成本0.30美元。 在这一课中,我们将学会如何将随机变量乘以一个常数(如0.30),这样我们就能解决问题。

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  Suppose you are running a lemonade stand, and you know that you have a mean of 15 customers per hour, with a standard deviation of 5 customers. If you know that it costs you 30 cents per glass in materials, and that you could hire your little sister to work the stand for you for $4.50 per hour, how could you calculate the mean and standard deviation of the cost per hour of having your sister run the stand?
  ::假设你正在经营一个柠檬水摊位,你知道你每小时平均有15个顾客,标准偏差为5个顾客。 如果你知道每杯每杯材料花费你30美分,而且你可以雇你妹妹为你工作,每小时4.50美元,你怎么能计算你妹妹主持摊位每小时费用的平均和标准偏差?

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  Transforming Random Variables (Part 2) 
  ::变换随机变量(第2部分)

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  We discussed previously the process of adding or subtracting a constant to a random variable, but often, you may need to multiply the random variable by a constant as well, or instead. For instance, suppose you had a random variable that represented the temperature in your classroom in degrees Celsius, and wanted to convert the values to degrees Fahrenheit. You could either start over with your calculations, having first converted each random variable outcome to the Fahrenheit equivalent, or you could just convert the mean and variance you already calculated to Fahrenheit by multiplying by  $\begin{array}{r}\frac{9}{5}\end{array}$ and adding 32.
  ::我们之前讨论过对随机变量添加或减去一个常数的过程,但通常,你可能需要将随机变量乘以一个常数,或者相反。例如,假设你有一个随机变量,以摄氏度表示教室的温度,并想将数值转换为华氏度。你要么从计算开始,先将每个随机变量结果转换为法氏等值,要么通过乘以95和增加32,将你计算的平均值和差异转换为法氏度。

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  Since we discussed the process of adding a constant to a random variable in the previous lesson, let’s look at the multiplication process.
  ::既然我们讨论过在上一个课目中将一个常数添加到随机变量中的过程,

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  The graph below represents the of random variable $\begin{array}{r}Y\end{array}$ , with $\begin{array}{r}{\mu }_Y=3\end{array}$ and $\begin{array}{r}{\sigma }_Y=1\end{array}$ :
  ::下图显示随机变量Y, 即 μY=3 和 Y=1 的随机变量Y:

  

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  Let’s see what happens if we multiply each outcome by 3, and find $\begin{array}{r}{\mu }_{3Y}\end{array}$ and $\begin{array}{r}{\sigma }_{3Y}\end{array}$ :
  ::让我们看看如果我们将每个结果乘以3, 并发现μ3Y和3Y:

  

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  By multiplying every outcome by three, not only was the mean multiplied by three:  $\begin{array}{r}{\mu }_{3Y}=3\times 3=9\end{array}$ , but the standard deviation was also increased by a multiple of three: $\begin{array}{r}{\sigma }_{3Y}=1\times 3=3\end{array}$ . The variance, since it is the square of the standard deviation, would be multiplied by 3 ² or 9.
  ::通过将每个结果乘以3,不仅平均值乘以3: μ3Y=3x3=9,而且标准差也增加了三倍: 3Y=1x3=3。 差异由于是标准差的正方形,将乘以32或9。

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  Transforming Variables 
  ::变变变量

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  1. Given that $\begin{array}{r}{\mu }_X=3\end{array}$ , $\begin{array}{r}{\sigma }_X=.5\end{array}$ , and $\begin{array}{r}{{\sigma }^2}_X=.25\end{array}$ , what is $\begin{array}{r}{\mu }_{3X}\end{array}$ ,  $\begin{array}{r}{\sigma }_{3X}\end{array}$ and $\begin{array}{r}{{\sigma }^2}_{3X}\end{array}$ ?
  ::1. 鉴于μX=3, X=5, 2X=25, μ3X, 3X 和 23X 是什么?

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  When a random variable is multiplied by a constant, the mean and standard deviation are multiplied by the same constant, and the variance is multiplied by the square of the constant:
  ::当随机变量乘以一个常数时,平均值和标准偏差乘以同一个常数,而差异乘以常数的平方时:

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  • $\begin{array}{r}{\mu }_{3X}={\mu }_X\times 3=3\times 3=9\end{array}$
    ::μ3XX×3=3x3=9
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  • $\begin{array}{r}{\sigma }_{3X}={\sigma }_X\times 3=.5\times 3=1.5\end{array}$
    ::303X_X_Xxx3=5x3=1.5
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  • $\begin{array}{r}{{\sigma }^2}_{3X}={{\sigma }^2}_X\times 3^2=.25\times 9=2.25\end{array}$
    ::23X2X×32=25x9=2.25

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  2. Given random variable  $\begin{array}{r}Y\end{array}$ with $\begin{array}{r}{\mu }_Y=.75\end{array}$ ,  $\begin{array}{r}{\sigma }_Y=0.125\end{array}$ , and $\begin{array}{r}{{\sigma }^2}_Y=0.0156\end{array}$ , what is $\begin{array}{r}{\mu }_{-2.2Y}\end{array}$ ,  $\begin{array}{r}{\sigma }_{-2.2Y}\end{array}$ and $\begin{array}{r}{{\sigma }^2}_{-2.2Y}\end{array}$ ?
  ::2. 如果Y=75, QQY=0.125, 和Q12Y=0.0156的随机变数Y为°Y=.75, QQY=0.125, 和°22Y=0.0156, 那么什么是2.2Y, 2.2Y, 2.2Y 和2-2-2Y?

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  As above, when a random variable is multiplied by a constant, the mean and standard deviation are multiplied by the same constant, and the variance is multiplied by the square of the constant:
  ::如上所示,当随机变量乘以一个常数时,平均值和标准偏差乘以同一个常数,而差异乘以常数的正方形:

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  • $\begin{array}{r}{\mu }_{-2.2Y}={\mu }_Y\times -2.2=0.75\times -2.2=-1.65\end{array}$
    ::
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  • $\begin{array}{r}{\sigma }_{-2.2y}={\sigma }_Y\times -2.2=0.125\times -2.2=-0.275\end{array}$
    ::
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  • $\begin{array}{r}{{\sigma }^2}_{-2.2Y}={{\sigma }^2}_Y\times (-2.2{)}^2=0.0156\times 4.84=0.075\end{array}$
    ::2-2-2-2YY2Y×(-2-2)2=0.0156×4.84=0.075

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  Finding the Mean and Standard Deviation 
  ::查找平均值和标准偏差

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  Tony the troll charges a fee of 3 gold pieces per person + 2 gold pieces per sword in order to cross his bridge. If random variable S represents the number of swords carried by a traveler wanting to cross his bridge, then $\begin{array}{r}{\mu }_S=2.8\end{array}$ , and  $\begin{array}{r}{\sigma }_S=1.095\end{array}$ What is the mean and standard deviation of Tony’s income per traveler
  ::Tony 巨巨人每人收费3金块,每把剑加2金块,以通过他的桥。如果随机变数S代表旅行者携带的要通过桥的剑数,那么,mS=2.8, 和 +S=1.095, Tony每个旅行者收入的平均值和标准差是多少?

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  Since we know the mean and variance of the traffic across Tony’s bridge, we are really just looking to multiply that data by 2 and then add 3 in order to convert the number of swords to the number of gold pieces and add the per traveler charge.
  ::由于我们知道托尼桥上交通流量的平均值和差异, 我们其实只是希望将数据乘以2, 然后加上3, 以便把剑数转换成金块数, 加上每个旅行者的费用。

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  We can describe the converted values mathematically, using the random variable S from the question, as: $\begin{array}{r}{\mu }_{2S+3}\end{array}$ , and $\begin{array}{r}{\sigma }_{2S+3}\end{array}$ .
  ::我们可以用问题的随机变量S, 将转换值用数学描述为: μ2S+3 和 0.002S+3。

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  We know that adding a constant affects the mean, but not the variance or standard deviation, and that multiplying by a constant affects both, so the standard deviation should be multiplied by two, and the mean should be multiplied by 2, and then increased by 3:
  ::我们知道,增加一个常数会影响平均值,但不会影响差异或标准偏差,乘以一个常数会影响两者,因此标准偏差应乘以2,平均值应乘以2,然后增加3:

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  • $\begin{array}{r}{\mu }_{2S+3}=2({\mu }_S)+3=2(2.8)+3=5.6+3=8.6\end{array}$
    ::μ2S+3=2(μS)+3=2(2.8)+3=5.6+3=8.6
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  • $\begin{array}{r}{\sigma }_{2S+3}=2({\sigma }_S)=2(1.095)=2.19\end{array}$  (remember that adding a constant to a random variable doesn’t affect the standard deviation.)
    ::282S+3=2(2-12S)=2(1.095)=2.19(记住在随机变量中添加一个常数并不影响标准偏差 。 )

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  Earlier  Problem Revisited
  ::重审先前的问题

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  Suppose you are running a lemonade stand, and you know that you have a mean of 15 customers per hour, with a standard deviation of 5 customers. If you know that it costs you 30 cents per glass in materials, and that you could hire your little sister to work the stand for you for $4.50 per hour, how could you calculate the mean and standard deviation of the cost per hour of having your sister run the stand?
  ::假设你正在经营一个柠檬水摊位,你知道你每小时平均有15个顾客,标准偏差为5个顾客。 如果你知道每杯每杯材料花费你30美分,而且你可以雇你妹妹为你工作,每小时4.50美元,你怎么能计算你妹妹主持摊位每小时费用的平均和标准偏差?

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  Now we should have all of the pieces to solve this problem. If we let random variable  $\begin{array}{r}D\end{array}$ equal the number of customers per hour, then $\begin{array}{r}{\mu }_D=15\end{array}$ and $\begin{array}{r}{\sigma }_D=5\end{array}$ . We need to multiply the number of customers, $\begin{array}{r}D\end{array}$ , by $0.30, and add that to the base cost of $4.50 that we have to pay our sister regardless of the number of customers, to get the hourly cost in dollars. Mathematically, we need to find the mean and standard deviation of  $\begin{array}{r}.3D+4.5\end{array}$ . In other words, we are looking for  $\begin{array}{r}{\mu }_{.3D+4.5}\end{array}$ and $\begin{array}{r}{\sigma }_{.3D+4.5}\end{array}$ .
  ::现在我们应该有全部的碎片来解决这个问题。 如果我们让随机变量D 等于每小时的客户数量, 那么 μD=15 和 ŠQD=5 。 我们需要将客户数量乘以 D 0. 30 美元, 并加上, 在4.50美元的基本费用中, 我们不论客户数量多少, 都要付给妹妹, 才能以美元支付每小时的费用。 从数学角度讲, 我们需要找到 3D+4.5 的平均值和标准差 。 换句话说, 我们要找 3D+4.5 和 + + + + +4.5 。

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  • $\begin{array}{r}{\mu }_{.3D+4.5}=.3({\mu }_D)+4.5=.3(15)+4.5=4.5+4.5=\mathrm{\$}9.00perhour\end{array}$
    ::μ3D+4.5=3(μD)+4.5=3(15)+4.5=4.5+4.5=9.00美元/小时
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  • $\begin{array}{r}{\sigma }_{.3D+4.5}=.3({\sigma }_D)=.3(5)=\mathrm{\$}1.50perhour\end{array}$
    ::30.3D+4.5=3(2-12D)=3(5)=每小时1.5美元

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  So, based on our mean number of customers, and paying our sister $4.50 per hour, we can expect to have a total of $9.00 per hour in materials and labor, with a standard deviation of $1.50.
  ::因此,根据我们平均的客户数量, 以及支付我们妹妹每小时4.50美元, 我们预计每小时材料和劳力总价值为9.00美元, 标准差为1.50美元。

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  Examples 
  ::实例

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  Example 1
  ::例1

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  If  $\begin{array}{r}{\mu }_X=5.5\end{array}$ , and  $\begin{array}{r}{\sigma }_X=.8\end{array}$ , what is  $\begin{array}{r}{\mu }_{2.5X}\end{array}$  and  $\begin{array}{r}{\sigma }_{2.5X}\end{array}$ ?
  ::如果 μX=5.5 和 X=.8,那么 微克2.5X 和 2.5X 是什么?

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  Multiplying a random variable by a constant affects both mean and standard deviation:
  ::以常数乘以随机变量影响平均差和标准差:

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  • $\begin{array}{r}{\mu }_{2.5X}=(2.5)({\mu }_X)=(2.5)(5.5)=13.75\end{array}$
    ::2.5X=(2.5)(μX)=(2.5)(5.5)=13.75
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  • $\begin{array}{r}{\sigma }_{2.5X}=(2.5)({\sigma }_X)=(2.5)(0.8)=.2\end{array}$
    ::2.5X=(2.5)(2-x)=(2.5)(0.8)=2

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  Example 2
  ::例2

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  Given  $\begin{array}{r}{\mu }_Y=3.1\end{array}$ ,  $\begin{array}{r}{\sigma }_X=.35\end{array}$ , and  $\begin{array}{r}{{\sigma }^2}_X=.1225\end{array}$ , what would  $\begin{array}{r}{\mu }_{-1.3X}\end{array}$ ,  $\begin{array}{r}{\sigma }_{-1.3X}\end{array}$  and  $\begin{array}{r}{{\sigma }^2}_{-1.3X}\end{array}$  be?
  ::鉴于微克Y=3.1, X=35, 2X=1225, 1.3X, 1.3X 和2-1.3X 是什么?

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  As above, in $\begin{array}{r}Q1\end{array}$ :
  ::如上文所述,问题1:

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  • $\begin{array}{r}{\mu }_{-1.3X}=(-1.3)({\mu }_X)=(-1.3)(3.1)=-4.03\end{array}$
    ::1.3X=(-1.3( μX) =(- 1.3( 3.1) ) 4.03
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  • $\begin{array}{r}{\sigma }_{-1.3X}=(-1.3)({\sigma }_X)=(-1.3)(.35)=-0.455\end{array}$
    ::1.3X=(-1.3(-)-(-)-(-)-(-)-(-)-(-)-(.)(.)(.)-(.)-(.)-(.)-(.)-(.)-(.)-(.)-(.)-(.)-(.)-(.)-(.)-(.)-(.)
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  • $\begin{array}{r}{{\sigma }^2}_{-1.3X}=(-1.3{)}^2({{\sigma }^2}_X)=(1.69)(0.1225)=0.207\end{array}$
    ::~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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  Example 3
  ::例3

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  If random variable  $\begin{array}{r}Z\end{array}$  has  $\begin{array}{r}\mu =14.9\end{array}$ , and  $\begin{array}{r}{\sigma }^2=16\end{array}$ , what is the mean and the standard deviation of  $\begin{array}{r}4Z+2\end{array}$ ?
  ::如果随机变数 Z 有 +14.9, 和 °12=16, 则平均值和标准偏差是 42 ?

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  Multiplying by a constant affects both mean and variance, adding a constant affects only the mean, so here we get:
  ::一个常数乘以一个常数影响中值和差异,加上一个常数只影响中值,所以我们得到:

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  • $\begin{array}{r}{\mu }_{4Z+2}=4({\mu }_Z)+2=4(14.9)+2=59.6+2=61.6\end{array}$
    ::μ42=4()+2=4(14.9)+2=59.6+2=61.6
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  • $\begin{array}{r}{{\sigma }^2}_{4Z+2}=4^2({\sigma }^{2}Z)=(16)(16)=256\end{array}$
    ::=(16)(16)=256 =(16)=(16)=(256)=(262)=(16)=(262)=(16)=(16)=(16)=(266)=(16)=(16)=(16)=(266)

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  Review 
  ::回顾

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  1. If $\begin{array}{r}{\mu }_X=.23\end{array}$ , and $\begin{array}{r}{\sigma }_X=.03\end{array}$ , what is  $\begin{array}{r}{\mu }_{3.1X}\end{array}$ and $\begin{array}{r}{\sigma }_{3.1X}\end{array}$ ?
     ::如果 μX=23, 和 X=. 03, 那么 μ3.1X 和 3.1X 是什么 ?
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  2. If $\begin{array}{r}{\mu }_C=124\end{array}$ , and $\begin{array}{r}{\sigma }_C=12\end{array}$ , what is $\begin{array}{r}{\mu }_{4C}\end{array}$  and $\begin{array}{r}{\sigma }_{4C}\end{array}$ ?
     ::如果 μC=124, 和 □C=12, m4C 和 □4C 是什么?
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  3. If $\begin{array}{r}{\mu }_Y=19.2\end{array}$ , and $\begin{array}{r}{\sigma }_Y=2.3\end{array}$ , what is $\begin{array}{r}{\mu }_{-1.9Y}\end{array}$ and $\begin{array}{r}{\sigma }_{-1.9Y}\end{array}$ ?
     ::如果μY=19.2和Y=2.3,什么是1.9Y和1.9Y?
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  4. If $\begin{array}{r}{\mu }_Z=48.38\end{array}$ , and $\begin{array}{r}{\sigma }_Z=2.27\end{array}$ , what is the  $\begin{array}{r}\mu \end{array}$ and  $\begin{array}{r}\sigma \end{array}$ of  $\begin{array}{r}2.3Z\end{array}$ ?
     ::如果48.38和2.27,2.3Z的微米和是多少?
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  5. If $\begin{array}{r}{\mu }_X=7.84\end{array}$ , and $\begin{array}{r}{\sigma }_X=.72\end{array}$ , what is the $\begin{array}{r}\mu \end{array}$ and $\begin{array}{r}\sigma \end{array}$ of $\begin{array}{r}-1.98X\end{array}$ ?
     ::如果 μX=7.84, 和 X=.72, 那么- 1. 98X 的 微克 和 是什么 ?
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  6. If $\begin{array}{r}{\mu }_A=\frac{5}{7}\end{array}$ , and $\begin{array}{r}{\sigma }_A=\frac{1}{7}\end{array}$ , what is the  $\begin{array}{r}\mu \end{array}$ and $\begin{array}{r}\sigma \end{array}$ of $\begin{array}{r}\frac{3}{14}A\end{array}$ ?
     ::如果 μA=57, 和 □A=17, 314A 的 微克和 □ 和 □ 是什么?
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  7. If $\begin{array}{r}{\mu }_B=14.11\end{array}$ , and $\begin{array}{r}{\sigma }_B=1.15\end{array}$ , what is the $\begin{array}{r}\mu \end{array}$ and $\begin{array}{r}\sigma \end{array}$ of $\begin{array}{r}-0.35B+2\end{array}$ ?
     ::如果 μB=14.11, 和 B=1.15, 则- 0. 35B+2 的 微克 和 是什么 ?
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  8. If $\begin{array}{r}{\mu }_A=\frac{3}{7}\end{array}$ , and $\begin{array}{r}{\sigma }_A=\frac{1}{7}\end{array}$ , what is the  $\begin{array}{r}\mu \end{array}$ and $\begin{array}{r}\sigma \end{array}$ of $\begin{array}{r}2A-\frac{3}{14}\end{array}$ ?
     ::如果 μA=37, 和 A=17, 2A-314 的 微克 和 是什么 ?
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  9. If $\begin{array}{r}{\mu }_Y=21.3\end{array}$ , and $\begin{array}{r}{\sigma }_Y=2.94\end{array}$ , what is the  $\begin{array}{r}\mu \end{array}$ and  $\begin{array}{r}\sigma \end{array}$ of $\begin{array}{r}-3.8Y+2.13\end{array}$ ?
     ::如果μY=21.3, 和 .12/Y=2.94, -3.8Y+2.13的微克和是多少?
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  10. If $\begin{array}{r}+1SD\end{array}$ of $\begin{array}{r}{\mu }_X=4.75\end{array}$ and $\begin{array}{r}-1SD\end{array}$ of $\begin{array}{r}{\mu }_X=3.25\end{array}$ , what is $\begin{array}{r}{\mu }_{3X+2.25}\end{array}$ and $\begin{array}{r}{\sigma }_{3X+2.25}\end{array}$ ?
      ::如果+1SDofμX=4.75 和-1-SDofμX=3.25, μ3X+2.25 和 □3X+2.25 是什么?
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  11. If $\begin{array}{r}+1SD\end{array}$ of $\begin{array}{r}{\mu }_X=11.25\end{array}$ and $\begin{array}{r}-1SD\end{array}$ of $\begin{array}{r}{\mu }_X=9.75\end{array}$ , what is $\begin{array}{r}{\mu }_{2X-2.25}\end{array}$ and $\begin{array}{r}{\sigma }_{2X-2.25}\end{array}$ ?
      ::如果+1SDofμX=11.25和-1-SDofμX=9.75,什么是μ2X-2.25和□2X-2.25?
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  12. If $\begin{array}{r}+2SD\end{array}$ of $\begin{array}{r}{\mu }_X=25\end{array}$ and $\begin{array}{r}-2SD\end{array}$ of $\begin{array}{r}{\mu }_X=5\end{array}$ , what is $\begin{array}{r}{\mu }_{3X-4}\end{array}$ and $\begin{array}{r}{\sigma }_{3X-4}\end{array}$ ?
      ::如果+2SDofμX=25和-2-SDofμX=5,那么μ3X-4和3X-4是什么?
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  13. If $\begin{array}{r}-2SD\end{array}$ of $\begin{array}{r}{\mu }_Z=9\end{array}$ and $\begin{array}{r}+2SD\end{array}$ of $\begin{array}{r}{\mu }_Z=15\end{array}$ , what is $\begin{array}{r}{\mu }_{2Z-.75}\end{array}$ and $\begin{array}{r}{\sigma }_{2Z-.75}\end{array}$ ?
      ::如果 -2SDof%9 和+2SDof%15, 是什么?

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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 。