节: 预期值 | 7.7 预期价值

章节大纲

Suppose you were given a six-sided die that was weighted to land on one value more often than normal. You roll the die one hundred times, record the results, and display them on the frequency table below. How could you use this information to determine the probability of a particular value appearing on any given roll, and the value you would expect to be the average, if you were to continue rolling?
::假设有人给了你六面性死亡,而六面性死亡被加权为以比正常高一个值降落。您将死亡翻了一百次,记录结果,并在下面的频率表上显示。您如何使用这些信息来确定特定滚动中特定值的概率,以及如果继续滚动,您期望的平均值?

Roll Value	1	2	3	4	5	6
Frequency	11	11	34	11	11	11

Expected Value
::预期值

A random variable yields outputs that are random by definition, however that does not necessarily mean that all possible values have the same chance of appearing. In the video above, the instructor uses a golf player’s past performance to calculate the expected value of future performance in a similar situation. We can see from the video that the player more often completes similar holes in three strokes than in two or four. To find the expected value, the instructor calculates the mean of the random variable.
::随机的变量产生随机的输出,但这并不必然意味着所有可能的值都有相同的出现机会。在以上视频中,教练用高尔夫球员过去的表现来计算类似情况下未来表现的预期值。我们从视频中可以看出,玩家通常以三下比两下或四下完成相似的孔。要找到预期值,教官计算随机变量的平均值。

Since the expected value of a random variable is the mean output of the observed trials, it is value we would expect the average of lots (perhaps thousands) of trials to approach over time. Interestingly, the expected value is not necessarily the value we would expect to see on any given roll. In fact, as we can see from the video where the expected number of strokes to complete the hole is 2.65, it is quite possible to have an expected value that never has, and never will, be an outcome of the experiment (kind of difficult to swing a golf club 2.65 times to sink the ball!).
::由于随机变量的预期值是所观察到的试验的平均输出值,因此,我们预计随着时间的推移将接近的试验平均量(也许数千次)是价值。有趣的是,预期值并不一定是任何特定卷子上我们预期值。事实上,我们从视频中可以看到,预期完成洞口的中风次数为2.65次,非常有可能有一个从未、也永远不会成为实验结果的预期值(很难打高尔夫球俱乐部的球要打2.65次! ) 。

The formula we use to calculate the mean is not the arithmetic mean formula you have used in the past, since it does not require you to divide by the count of values, but rather to multiply each value by the probability of it appearing as an outcome. The formula looks like this:
::我们用来计算平均值的公式不是您过去使用的算术平均值公式,因为它不要求您除以数值的计算,而是要乘以每个值作为结果出现的概率。公式是这样的:

\begin{aligned} μ_{x} = \sum [x \times P (x)] \end{aligned}

$\begin{align*}\mu _x=\sum[x \times P(x)]\end{align*}$
::μx [xxxxP(x)]

The mean of random variable $\begin{align*}x\end{align*}$ is the sum of possible outcomes of $\begin{align*}x\end{align*}$ , each multiplied by its percent probability of occurrence.
::随机变量x的平均值是 x 的可能结果的总和,每乘以其发生概率的百分率。

The application of the formula is more straight forward than the explanation, so let’s look at a few examples.
::公式的应用比解释更直截了当,所以让我们来看看几个例子。

Finding the Mean
::寻找平均值

What is the mean of discrete random variable $\begin{align*}Y\end{align*}$ , which has a given by the table below?
::离散随机变数Y(下表给出的变量Y)的含义是什么?

$\begin{align}x\end{align}$	1	2	3	4
$\begin{align}P(Y=x)\end{align}$	.26	.24	.35	.15

To calculate the mean, we simply add up each of the values multiplied by its probability of occurrence:
::为了计算平均值,我们简单地将每个值乘以其发生概率:

\begin{aligned} (1 \times .26) + (2 \times .24) + (3 \times .35) + (4 \times .15) & = 2.39 \\ μ_{Y} & = 2.39 \end{aligned}

$\begin{align*}(1 \times .26)+(2 \times .24)+(3 \times .35)+(4 \times .15) &=2.39 \\ \mu_Y &=2.39 \end{align*}$
:

1x.26)+(2x.24)+(3x.35)+(4x.15)=2.39微克Y=2.39)

The mean of random variable $\begin{align*}Y\end{align*}$ is 2.39.
::随机变量Y的平均值为2.39。

Finding Expected Values
::查找预期值

1. What is the expected value of a weighted six-sided die that has a 50% probability of landing on 5, and an equal probability of landing on each other possibility?
::1. 加权六面死亡的预期值是多少? 加权六面死亡5时着陆概率为50%,彼此着陆概率相等?

Start by creating a probability distribution for random variable $\begin{align*}X\end{align*}$ :
::以创建随机变量 X 的概率分布开始 :

$\begin{align}x\end{align}$	1	2	3	4	5	6
$\begin{align}P(X=x)\end{align}$	.10	.10	.10	.10	.50	.10

Now we can apply the formula for calculating the mean:
::现在我们可以应用计算平均值的公式:

\begin{aligned} μ_{X} & = (1 \times .1) + (2 \times .1) + (3 \times .1) + (4 \times .1) + (5 \times .5) + (6 \times .10) = 4.1 \\ μ_{X} & = 4.1 \end{aligned}

$\begin{align*}\mu_X&=(1 \times .1)+(2 \times .1)+(3 \times .1)+(4 \times .1)+(5 \times .5)+(6 \times .10) =4.1\\ \mu_X &=4.1\end{align*}$
::μX=(1x.1)+(2x.1)+(3x.1)+(3x.1)+(4x.1)+(5x.5)+(5x.5)+(6x.10)=4.1微克X=4.1

2. Let random variable $\begin{align*}C\end{align*}$ be one-half of the sum of two standard dice. What is the expected value of $\begin{align*}C\end{align*}$ ?
::2. 随机变数C为两个标准骰子之和的一半。 C的预期值是多少?

To calculate $\begin{align*}\mu_c\end{align*}$ , the mean or expected value of $\begin{align*}C\end{align*}$ , start by creating a probability distribution:
::要计算 μc, C 的平均值或预期值, 首先要创建概率分布 :

The distribution of the sum of two standard dice is:
::两个标准骰子之和的分布如下:

$\begin{align}x\end{align}$	1	2	3	4	5	6	7	8	9	10	11	12
$\begin{align}P(X=x)\end{align}$	0	.028	.056	.083	.111	.139	.167	.139	.111	.083	.056	.028

Which makes the distribution of $\begin{align*}C\end{align*}$ :
::C的分布方式是:

$\begin{align}x\end{align}$	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6
$\begin{align}P(X=x)\end{align}$	.028	.056	.083	.111	.139	.167	.139	.111	.083	.056	.028

Now we can apply the formula for calculating the mean of a discrete random variable:
::现在我们可以应用公式来计算离散随机变量的平均值:

\begin{aligned} (1 \times 0.028) + (1.5 \times 0.056) + (2 \times 0.083) + (2.5 \times 0.111) + (3 \times 0.139) + (3.5 \times 0.167) \\ + (4 \times 0.139) + (4.5 \times 0.111) + (5 \times 0.083) + (5.5 \times 0.056) + (6 \times 0.028) \\ (0.03) + (.083) + (0.17) + (.28) + (.42) + (.59) + (.56) + (.5) + (.42) + (.31) + (.17) = 3.5 \\ μ_{X} = 3.5 \end{aligned}

$\begin{align*}&(1 \times 0.028)+(1.5 \times 0.056)+(2 \times 0.083)+(2.5 \times 0.111)+(3 \times 0.139)+(3.5 \times 0.167)\\ & \qquad \ +(4 \times 0.139)+(4.5 \times 0.111)+(5 \times 0.083)+(5.5 \times 0.056)+(6 \times 0.028) \\ &(0.03)+(.083)+(0.17)+(.28)+(.42)+(.59)+(.56)+(.5)+(.42)+(.31)+(.17) =3.5\\ &\mu_X =3.5\end{align*}$
:

1x0.028)+(1.5x0.0056)+(2xx0.0083)+(2.5x0.001.11)+(3x0.139)+(3x0.10.139)+(3.5xx0.167)+(4xx0.139)+(4.5x0.11)+(5x0.083)+(5.5x0.0056)+(6x0.028)+(0.03)+(0.83)+(0.17)+(.28)+(.28)+(.42)+(5.56)+(.5)+(.42)+(.31)+(17)=3.5μX=3.5)

So, if you were to conduct many, many trials, and find the mean, the expected value of that mean would be apx. 3.5.
::所以,如果你要进行很多,很多的试验, 并找到平均值, 这个平均值的预期值是3.5x。

Earlier Problem Revisited
::重审先前的问题

Suppose you were given a six-sided die that was weighted to land on one value more often than normal. You roll the die one hundred times, record the results, and display them on the probability distribution below. How could you use this information to determine the probability of a particular value appearing on any given roll, and the value you would expect to be the average, if you were to continue rolling?
::假设有人给了你六面性死亡,而六面性死亡被加权为以比正常高一个值降落。你将死亡翻了一百次,记录结果,并在下面的概率分布上显示结果。你如何使用这些信息来确定任何特定滚动中出现特定值的概率,以及如果继续滚动,你预期的平均值?

Roll value	1	2	3	4	5	6
Frequency	11	11	34	11	11	11

Because there were 100 rolls, we can convert the frequency table to a probability distribution just by considering each frequency as a percentage, which gives us the probability of rolling each value. We can then set a random variable, say, $\begin{align*}D\end{align*}$ to equal the outcome of a roll of the die. The expected value is the mean of the random variable $\begin{align*}D\end{align*}$ , found by applying the formula:
::由于有100个滚动,我们可以将频率表转换为概率分布,只要将每个频率作为百分比来考虑,这样可以给我们滚动每个值的概率。然后我们可以设置一个随机变量,比如,D等于死亡滚动的结果。预期值是随机变量D的平均值,通过应用公式发现:

\begin{aligned} 1 \times .11 + 2 \times .11 + 3 \times .34 + 4 \times .11 + 5 \times .11 + 6 \times .11 & = 3 \\ So μ_{D} & = 3 \end{aligned}

$\begin{align*}1 \times .11+2 \times .11+3 \times .34+4 \times .11+5 \times .11+6 \times .11 &=3 \\ \text{So} \ \ \mu_D &=3\end{align*}$
::1x11+2xx11+3x34+4x11+5x11+6x11=3So 微克D=3

Examples
::实例

Example 1
::例1

Suppose you take all of the number cards two through five from a standard deck, and set random variable $\begin{align*}C\end{align*}$ to be the sum of two cards drawn at random, without replacement. What is the expected value of $\begin{align*}C\end{align*}$ ?
::假设您从标准甲板上拿下所有牌号卡2至5,然后将随机变数C设为随机抽取的两张牌的总和,而没有替换。 C的预期值是多少?

Create a probability distribution for $\begin{align*}C\end{align*}$ :
::创建 C 的概率分布 :

$\begin{align}x\end{align}$	4	5	6	7	8	9	10
# possible combinations	6	16	22	32	22	16	6
$\begin{align}P(C=x)\end{align}$	.05	.133	.183	.267	.183	.133	.05

Apply the expected value formula:
::应用预期值公式 :

\begin{aligned} (4 \times .05) + (5 \times .133) & + (6 \times .183) + (7 \times .267) + (8 \times .183) + (9 \times .133) + (10 \times .05) = \\ .2 + .665 + 1.098 + 1.869 + 1.464 + 1.197 + .5 \approx 7 \end{aligned}

$\begin{align*}(4 \times .05)+(5 \times .133)&+(6 \times .183)+(7 \times .267)+(8 \times .183)+(9 \times .133)+(10 \times .05) = \\ & .2+.665+1.098+1.869+1.464+1.197+.5\approx 7 \end{align*}$

The expected value of the random variable $\begin{align*}C\end{align*}$ is 7.
::随机变量C的预期值为 7。

Example 2
::例2

Over the past year, Sally has compiled a probability distribution of the number of kids she baby sits on each day of the week. Based on her data from the table below, what is the expected number of kids she will baby sit on any random day?
::在过去一年里,Sally汇编了她婴儿每星期每天坐在婴儿座位上的概率分布。根据她从下表获得的数据,她每星期每天坐在婴儿座位上的孩子的预期数量是多少?

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

We already have the probability distribution, just apply the formula:
::我们已经有了概率分布, 只要应用公式 :

\begin{aligned} (1 \times .35) & + (2 \times .4) + (3 \times .15) + (4 \times .05) + (5 \times .05) \\ .35 + .8 + .45 + .2 + .25 = 2.05 \end{aligned}

$\begin{align*}(1 \times .35)&+(2 \times .4)+(3 \times .15)+(4 \times .05)+(5 \times .05) \\ & .35+.8+.45+.2+.25 =2.05\end{align*}$

Sally should expect to averge two kids per day for her baby sitting business.
::莎莉应该希望每天生两个孩子来做她的孩子的坐椅生意

Example 3
::例3

Tuscany works for a hot dog vendor at the Colorado Rockies baseball stadium. Over the last couple of years, she has created the probability distribution below of the number of drinks a spectator will consume during a baseball game. If the drinks cost $5 each, how much is a spectator expected to spend on drinks?
::托斯卡尼为科罗拉多洛基斯棒球场的热狗卖家工作。在过去的几年里,她创造了低于饮料数量的概率分布。在一场棒球比赛中,观众会消费一个比饮料数量低的观众。如果每杯要花5美元,观众在饮料上要花多少钱?

# of drinks	1	2	3	4	5
Probability	28%	42%	20%	7%	3%

Start by applying the probability distribution values to the expected value formula:
::开始将概率分布值应用到预期值公式中 :

\begin{aligned} 1 \times .28 & + 2 \times .42 + 3 \times .2 + 4 \times .07 + 5 \times .03 \\ .28 + .84 + .6 + .28 + .15 = 2.15 \end{aligned}

$\begin{align*}1 \times .28&+2 \times .42+3 \times .2+4 \times .07+5 \times .03\\ & .28+.84+.6+.28+.15=2.15\end{align*}$

So the average customer is expected to buy 2.15 drinks per game. Since the drinks cost $5 each, that means that Tuscany's hot dog cart can expect to average $10.75 per customer in drink sales.
::因此,平均顾客每场比赛要买2.15杯饮料。由于每杯饮料费用为5美元,这意味着托斯卡尼的热狗车预计每个顾客平均售酒10.75美元。

Review
::回顾

For questions 1 – 10, calculate the expected value of the random variable with the given probability distribution:
::对于问题1 - 10, 使用给定概率分布计算随机变量的预期值 :

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

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

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

$\begin{align}x\end{align}$	30	60	90	120	150	170
$\begin{align}P(X=x)\end{align}$	.18	.16	.24	.22	.20	.00

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

$\begin{align}x\end{align}$	13	17	21	25	29	33	37
$\begin{align}P(X=x)\end{align}$	.15	.17	.23	.30	.10	.03	.02

$\begin{align}x\end{align}$	26	39	52	65	78
$\begin{align}P(X=x)\end{align}$	6%	14%	30%	28%	22%

$\begin{align}x\end{align}$	22	43	64	85	106
$\begin{align}P(X=x)\end{align}$	10.5%	22.5%	31.5%	22.8%	12.7%

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

10.

$\begin{align}x\end{align}$	$\begin{align}\frac{3}{8}\end{align}$	$\begin{align}\frac{1}{2}\end{align}$	$\begin{align}\frac{5}{8}\end{align}$	$\begin{align}\frac{3}{4}\end{align}$	$\begin{align}\frac{7}{8}\end{align}$
$\begin{align}P(X=x)\end{align}$	$\begin{align}\frac{2}{5}\end{align}$	$\begin{align}\frac{1}{4}\end{align}$	$\begin{align}\frac{1}{5}\end{align}$	$\begin{align}\frac{1}{10}\end{align}$	$\begin{align}\frac{1}{20}\end{align}$

11. Carrie shines shoes for money on weekday mornings, and she has compiled the following probability distribution of the number of clients she is likely to get each day. If she earns $3.50 per shine, how much should she expect to earn each day, on average?
::11. Carrie在周日早上为赚钱而擦鞋,她汇编了她每天可能得到的客户人数的下列概率分布,如果她每日赚取3.50美元,平均每天要挣多少?

# clients	20	25	30	35	40
probability	.15	.35	.30	.15	.05

12. Vincente works for a fast food restaurant, where he earns $8.50 per hour. The number of hours he works each week varies between 20 and 40, based on how busy the restaurant is during the week. Over the past year, he has compiled the probability distribution below describing the percent probability of his getting 20, 25, 30, 35, or 40 hours in any random week. If Vincente wants to move out, and knows that he shouldn’t spend more than $\begin{align*}\frac{1}{3}\end{align*}$ of his average income on housing, how much can he afford for rent?
::12. Vincente为一家快餐餐厅工作,每小时收入8.50美元,每周工作时数在20至40小时之间不等,这取决于餐厅每周的繁忙程度,过去一年中,他汇编的概率分布低于20、25、30、35或40小时的概率百分比。如果Vincente想搬出去,并且知道他不应将平均收入的13小时以上花在住房上,那么他能付多少房租?

# hours	20	25	30	35	40
probability	.15	.28	.32	.15	.10

13. How much more could Vincente afford for rent if he were given a raise of $2 per hour?
::13. 文森特如果得到每小时2美元的加薪,他能承担多少租金?

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。

$\begin{align}x\end{align}$	$\begin{align}\frac{3}{8}\end{align}$	$\begin{align}\frac{1}{2}\end{align}$	$\begin{align}\frac{5}{8}\end{align}$	$\begin{align}\frac{3}{4}\end{align}$	$\begin{align}\frac{7}{8}\end{align}$
$\begin{align}P(X=x)\end{align}$	$\begin{align}\frac{2}{5}\end{align}$	$\begin{align}\frac{1}{4}\end{align}$	$\begin{align}\frac{1}{5}\end{align}$	$\begin{align}\frac{1}{10}\end{align}$	$\begin{align}\frac{1}{20}\end{align}$

章节大纲

Expected Value ::预期值

Finding Expected Values ::查找预期值

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)