节: 接近二分制分布 | 9.8 接近二分制分配 | The Way To Learn

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Suppose you were completing a multiple-choice test, and you are worried that you don’t know the information well enough. If there are 75 questions, each with 4 answers, what is the probability that you would get at least 60 correct just by guessing randomly?
::假设你正在完成一个多重选择测试,而你担心你对信息了解不够。如果有75个问题,每个问题都有4个答案,那么仅仅随机猜测就能得到至少60个正确的答案的可能性有多大?

You could probably answer this question if you have completed prior lessons on binomial probability, but it would be quite a calculation, requiring you to individually calculate the probability of getting 60 correct, adding it to the probability of getting 61 correct, and so on, all the way up to 75! At the end of the lesson, we will review this question in light of the normal distribution, and see how much more efficient it can be.
::如果你已经完成了有关二元概率的先前课程, 你也许可以回答这个问题, 但是这是一个相当的计算, 需要你单独计算得到60的正确概率, 加上61的正确概率,等等, 一直到75。在课程结束时, 我们将根据正常分布来审查这个问题, 并且看看它能效率多高。

Approximating the Binomial Distribution
::接近二分制分布

Many real life situations involve binomial probabilities, as we saw in prior lessons on binomial experiments. In fact, even many questions that don’t appear binomial at first can be formatted so that they are, allowing the probability of success or failure of a given study to be calculated as a binomial probability. Unfortunately, if the probability of success spans a wide range of possible values, the calculation can become very burdensome.
::许多现实生活状况都涉及二元概率。正如我们在二元实验的以往课程中所看到的那样。事实上,即使许多起初看起来没有二元性的问题也可以以这样的形式出现,这样就可以将某项研究的成败概率算作二元性概率。不幸的是,如果成功概率包含广泛的可能值,那么计算就会变得非常麻烦。

The good news is that there is another way to approximate the probability of success, and you can see what it is by comparing the following graphs. The first graph displays the probability of getting various numbers of heads over 100 flips of a fair coin, in other words, the distribution of a binomial random variable with $\begin{aligned} P (success) = .50 \end{aligned}$ . The second graph is a normal distribution. Notice any similarities?
::好消息是,有另外一种方法可以估计成功概率,通过比较下图,你可以看到它是什么。第一个图表显示的是将各种头数超过一个公平硬币的100个翻转的概率,换句话说,用P(成功)=50来分配一个二进制随机变量。第二个图表是正常的分布图。注意到任何相似之处吗?

They are extremely similar in shape, in fact, if you follow a “rule of thumb”, you can use a normal distribution to estimate the results of a binomial distribution with quite acceptable accuracy. The rule of thumb for knowing when the normal distribution will provide a good approximation of a binomial distribution with the same mean and standard deviation is:
::事实上,如果遵循“大拇指规则”,可以使用正常的分布法,以相当可接受的精确度估计二进制分布的结果。在知道正常分布法何时能提供相同平均值和标准偏差的二进制分布法下,大拇指规则是:

$\begin{aligned} n \times P > 10 and n (1 - p) > 10 \end{aligned}$

::nxP>10 和 n( 1- p) >10

Where $\begin{aligned} n \end{aligned}$ is the number of trials, and $\begin{aligned} p \end{aligned}$ is the probability of success.
::n是试验次数,p是成功概率。

If you have determined that a given binomial distribution is a candidate for approximation using a normal distribution, you can calculate the $\begin{aligned} μ \end{aligned}$ and $\begin{aligned} σ \end{aligned}$ of the normal distribution using:
::如果您已经确定某个给定的二进制分布是使用正常分布的近似候选, 您可以使用 :

$\begin{aligned} μ & = n p \\ σ & = \sqrt{n p (1 - p)} \end{aligned}$

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不

If you are interested in the comparison between the binomial probability and normal approximation for a particular $\begin{aligned} n \end{aligned}$ or $\begin{aligned} p \end{aligned}$ value , there is an excellent Java applet on Online Stat Book's website that will show the actual values and a graph for any of $\begin{aligned} n \end{aligned}$ and $\begin{aligned} p \end{aligned}$ values.
::如果您对某n或p值的二进制概率和普通近似值之间的比较感兴趣,在线统计手册网站上有一个优秀的 Java Applet ,它将显示任何n和p值的实际值和图表。

Approximating Results
::近似结果

Can the results of a binomial experiment consisting of 40 trials with a 72% probability of success of be acceptably approximated by a normal distribution?
::由40次试验组成的二进制实验的结果,成功概率72%的概率能否被正常分布所近似?

Here, $\begin{aligned} n = 40 \end{aligned}$ , and $\begin{aligned} p = 0.72 \end{aligned}$
::此处, n=40, p=0.72

First, is $\begin{aligned} n \times p > 10 \end{aligned}$ ?
::首先,是nxp>10吗?

$\begin{aligned} n \times p & = 40 \times .72 = 28.8 \\ 21.6 & > 10 YES \end{aligned}$

::nxp=40x.72=28.821.6>10 是

Second, is $\begin{aligned} n (1 - p) > 10 \end{aligned}$ ?
::其次, n(1-p) > 10?

$\begin{aligned} n (1 - p) = 40 (1 & - 0.72) = 40 (.28) = 11.2 \\ 11.2 & > 10 YES \end{aligned}$

::n( 1 - p) = 40(1 - 0.72) = 40 (. 28) = 11. 21.2 > 10 是

Yes, based on our rule of thumb, you could use a normal distribution to approximate the results of this binomial experiment.
::是的,根据我们的拇指规则, 你可以使用正常的分布来估计这个二进制实验的结果。

Estimating Probability
::估计概率

If Kaile wants to estimate the probability of correctly guessing at least 9 answers out of 50 on a true/false exam, can she estimate using a normal distribution?
::如果Kaile想估计在真实/假考50个答案中至少9个答案正确猜测的概率,

Here, $\begin{aligned} n = 50 \end{aligned}$ and $\begin{aligned} p = 0.50 \end{aligned}$ (true/false):
::在此, n=50 和 p= 0. 50 (真/假):

$\begin{aligned} n \times p & = 50 \times .50 = 25 \\ 25 & > 10 Yes \\ n (1 - p) = 50 (1 & - 0.50) = 50 \times .50 = 25 \\ 25 & > 10 Yes \end{aligned}$

::nxp=50x50=50x50=2525=2525 > 10 Yesn(1-p)=50(1-0.50)=50x50=2525=2525 > 10

Real-World Application: Production Plant
::现实世界应用:生产厂

Ciere works in a production plant. Due to the balance of speed and accuracy in production, each part off the line has a 98.8% probability of defect free production.
::Ciere在生产厂中工作,由于生产速度和准确性之间的平衡,每条线以外的每一条线都有98.8%的无缺陷生产概率。

a. Can a binomial experiment based on 98.9% probability be approximated if Ciere produces 1000 parts?
::a. 如果Ciere生产1000个部件,能否根据98.9%的概率进行二进制实验?

Here, $\begin{aligned} n = 1000 \end{aligned}$ and $\begin{aligned} p = 0.988 \end{aligned}$ , does this satisfy our ‘rule of thumb’?
::N=1000和p=0.988,这符合我们的“拇指规则”吗?

$\begin{aligned} n \times p & = 1000 \times 0.988 = 988 Yes \\ n (1 - p) & = 1000 (1 - 0.988) = 1000 \times 0.012 = 12 Yes \end{aligned}$

::nxp=10000x0.988=988 Yesn(1-p)=1000(1-0.988)=1000x0.988=1000x0.0012=12

If Ciere produces 1000+ parts, this experiment may be approximated using a normal distribution.
::如果Ciere生产1000+部件,该实验可以使用正常分布法进行近似。

b. What would be the mean and standard deviation of the appropriate normal distributino
::b. 适当的正常分配的平均值和标准偏差

The mean, $\begin{aligned} μ \end{aligned}$ , and standard deviation, $\begin{aligned} σ \end{aligned}$ , can be evaluated as follows:
::平均值, 微克和标准偏差, , 可评估如下:

$\begin{aligned} μ & = n p = 1000 \times .988 = 988 \\ σ & = \sqrt{n p (1 - p)} = \sqrt{988 (0.012)} = 3.44 \end{aligned}$

::np=1000×988=988\np(1-p)=988(0.012)=3.44

c. What is the probability that Ciere will produce at least 990 parts without a defect in a 1000 part run?
::c. Ciere在1000个部分没有缺陷的情况下至少生产990个部件的可能性有多大?

To calculate this, we need the z -score of 990, which we can calculate using: $\begin{aligned} Z = \frac{(x - μ)}{σ} \end{aligned}$ . Once we have the z -score, we can reference a z -score probability table to find the probability of a value above it:
::要计算这一点,我们需要990的z-score, 我们可以用 {(x) {} {} {}} {}} {(x}}}}} {}}} {}}} {}} 。一旦我们有了z- score, 我们就可以参考一个 z- score 概率表, 以找到值高于此值的概率 :

$\begin{aligned} z = \frac{(x - μ)}{σ} = \frac{990 - 988}{3.44} = \frac{2}{3.44} = .52 \end{aligned}$

::z z(x) 990-9883.44=23.44=52

Using a reference table (which you can find online or in the previous lesson), we see that the probability of a z -score greater than .52 is 34.60%
::使用参考表格(可以在网上或上一个课目中找到),我们看到,z-score 大于.52的概率是34.60%。

The probability of Ciere producing at least 990 parts in a row without a defect is about 34.60%.
::Ciere连续生产至少990个部件且无缺陷的概率约为34.60%。

Earlier Problem Revisited
::重审先前的问题

Suppose you were completing a multiple-choice test, and you are worried that you don’t know the information well enough. If there are 75 questions, each with 4 answers, what is the probability that you would get at least 60 correct just by guessing randomly?
::假设你正在完成一个多重选择测试,而你担心你对信息了解不够。如果有75个问题,每个问题都有4个答案,那么仅仅随机猜测就能得到至少60个正确的答案的可能性有多大?

Here, since there are 75 questions, $\begin{aligned} n = 75 \end{aligned}$ , and since each has 4 possible answers, $\begin{aligned} p = .25 \end{aligned}$ . First we check to see if we can use the normal approximation:
::由于这里有75个问题, n=75, 并且每个问题有4个可能的答案, p=25。首先我们检查一下我们能否使用正常的近似值 :

$\begin{aligned} n \times p = 75 & \times .25 = 18.75 \\ 18.75 & > 10 Yes \end{aligned}$

::nxp=75×25=18.7518.75 > 10 是

Since we can use the normal distribution, we need to calculate the mean and SD of the distribution:
::由于我们可以使用正常的分布, 我们需要计算分布的平均值和SD:

$\begin{aligned} μ & = n p = 75 \times .25 = 18.75 \\ σ & = \sqrt{n p (1 - p)} = \sqrt{18.75 (.75)} = 3.75 \end{aligned}$

::np=75×25=18.75np(1-p)=18.75(.75)=3.75

Now we need the z -score of our minimum number of correct guesses, 60:
::现在我们需要我们最低猜对数的 z分数,60:

$\begin{aligned} z = \frac{(x - μ)}{σ} = \frac{60 - 18.75}{3.75} = \frac{41.25}{3.75} = 11 \end{aligned}$

::z z(x) 60 - 18.753.75=41.253.75=11

Ha! We don’t even need a z -score reference for this one, your chances of randomly guessing 60 or more correctly are virtually nil, since most tables only go up to $\begin{aligned} z = 3 \end{aligned}$ or $\begin{aligned} z = 3.99 \end{aligned}$ . Better study more next time!
::哈! 我们甚至不需要z-score reference for this one, 你随机猜测60或60以上正确的可能性几乎是零, 因为大多数表格都只达到z=3或z=3.99。下次最好多研究一下!

Examples
::实例

Example 1
::例1

Is a binomial experiment consisting of 35 trials, each with 55% probability of success, a good candidate for normal curve approximation?
::一个由35次试验组成的二进制实验,每个试验的成功概率为55%,是否是正常曲线近似的好人选?

$\begin{aligned} n = 35 \end{aligned}$ and $\begin{aligned} p = 0.55 \end{aligned}$ , use the rule of thumb from the lesson to evaluate:
::n=35和p=0.55,使用教训中的拇指规则评价:

$\begin{aligned} n \times p & = 35 \times .55 = 19.25 \\ 19.25 & > 10 Yes \\ n \times (1 - p) & = 35 \times .45 = 15.75 \\ 15.75 & > 10 Yes \end{aligned}$

::nxp=35x35x.55=19.2519.25>10 Yesnx(1-p)=35x45=15.75.15.75>10 是

Since the experiment meets both qualifications, a normal approximation should be variable.
::由于试验符合两种条件,通常的近似值应可变。

Example 2
::例2

A binomial random variable has a 0.821 probability of success. If data is collected from 48 trials, can the results be viably approximated with a normal dstribution?
::二进制随机变量的成功概率为0.821。如果从48个试验中收集数据,结果能否与正常分配相近?

$\begin{aligned} n = 48 \end{aligned}$ and $\begin{aligned} p = 0.721 \end{aligned}$ , check with our rule of thumb:
::n=48和p=0.721,检查我们的拇指规则:

$\begin{aligned} n \times p & = 48 \times 0.821 = 39.41 \\ 39.41 & > 10 Yes \\ n \times (1 - p) & = 48 \times 0.179 = 7.05 \\ 7.05 & < 10 No! \end{aligned}$

::nxp=48x0.821=39.4139.41 > 10 Yernxxx(1-p)=48x0.179=7.05.05<10 No!

Example 3
::例3

What is the approximate probability of correctly guessing at least 20 questions out of 50, on a true/false exam?
::在真实/虚假考试中正确猜出50个问题中至少有20个问题的概率是多少?

Verify that we can use a normal approximation, using $\begin{aligned} n = 50 \end{aligned}$ and $\begin{aligned} p = 0.50 \end{aligned}$ :
::校验我们使用正常近似值, 使用 n=50 和 p= 0. 50 :

$\begin{aligned} n \times p & = 50 \times .5 = 25 \\ 25 & > 10 Yes \\ n \times (1 - p) & = 50 \times .5 = 25 \\ 25 & > 10 Yes \end{aligned}$

::n×p=50=50×5=2525 > 10 耶尼×(1-p)=50×5=2525>10 是

Since we can use the normal approximation, we need to calculate the mean and SD, so we can get the z -score for 20 (the minimum number of correct answers we want to get).
::既然我们可以使用正常的近似值, 我们需要计算平均值和SD, 这样我们就可以得到20个z-score(我们想要得到的最起码的正确答案)。

$\begin{aligned} μ & = n p = 50 \times .5 = 25 \\ σ & = \sqrt{n p (1 - p)} = \sqrt{25 (.50)} = 3.54 \\ z & = \frac{(x - μ)}{σ} = \frac{20 - 25}{3.54} = 1.41 \end{aligned}$

::np=50x5=25}np(1-p)=25(.50)=3.54z=(x) 20-253.54=1.41

Consulting our reference, we learn that the probability that a z -score greater than 1.41 will occur is approximately 94%.
::咨询我们的参考,我们了解到 z-score 大于1.41的概率大约是94%。

You have approximately a 94% probability of correctly guessing at least 20 questions correctly on a 50 question exam.
::你大约94%的概率正确猜对至少20个问题在50个问题考试中正确猜对。

Review
::回顾

Karen is playing a game of chance with a probability of success of 33%. If she plays the game 43 times, what is the probability that she will win more than 19 times?
::凯伦在玩一个赌博游戏,成功概率为33%。如果她玩了43次,那么她赢得超过19次的概率是多少?

When approximating a binomial distribution, how do you calculate the standard deviation?
::当接近二进制分布时, 您如何计算标准偏差 ?

Gregory has created a card game where you either draw a black card or a red card. If you draw a red card, you get a dollar. But if you draw a black card you owe him a dollar. The chance of drawing a red card is 61%. You decide to play against Gregory 26 times. Can you approximate this situation with a normal curve? Why or why not?
::Gregory 创建了一张牌游戏, 绘制黑卡或红卡。如果您绘制红卡, 就会得到一美元。但是, 如果您绘制黑卡, 你欠他一美元。绘制红卡的机会是 61% 。您决定对Gregory 玩26 次。您能否用正常曲线来大致了解这种情况? 为什么或为什么不?

Sue has organized her closet into summer clothing and winter clothing. She closes her eyes and reaches in her closet to pick an outfit. If she selects summer clothing she will be cold (it is winter). The chance of Sue selecting summer clothing is 41%. She decides to select 27 outfits this way. If you were to approximate this with a normal curve, what would the standard deviation be?
::Sue将衣柜组织成夏衣和冬衣。她闭上眼睛,在衣柜里挑衣服。如果她选择夏衣,她会很冷(是冬天)。苏选择夏衣的可能性是41%。她决定这样选择27件衣服。如果你用正常曲线来接近这个曲线,标准偏差会是什么?

Sharon can’t decide between two guys that she likes. She picks a daisy from the garden and decides to play “I like Greg more, I like Stan more” with the petals. The chance of the last petal being “I like Greg more” is 67%. She decides to go through this process with 48 daisies. What is the probability that she will select Greg more than 36 times?
::沙龙无法在她喜欢的两个男人之间做出抉择。她从花园里挑了一只菊花,决定与花瓣一起玩“我更喜欢Greg,我更喜欢Stan ” 。最后一个花瓣“我更喜欢Greg”的机会是67 % 。她决定用48个菊花来经历这个过程。她选择Greg超过36次的可能性有多大?

Vern has to choose between two summer jobs. He painted a wheel red and blue. If the spinner lands in the red area, he works for a landscaping company. If it lands in the blue, he works for a fast food restaurant. The chance of the spinner landing in the red area is 52%. He decides to spin 33 times. What standard deviation would you use to approximate this situation with a normal curve?
::Vern必须在两个夏季工作之间做出选择。他画了一辆红色和蓝色的车轮。如果旋翼车在红色地区降落, 他为一家美化景观公司工作。如果旋翼车在蓝色地区降落, 他为一家快餐餐厅工作。旋翼车在红色地区降落的可能性为52%。他决定旋转33倍。您用什么标准偏差来将这种情况与正常曲线相近?

When approximating a binomial distribution, what is the mean?
::当接近二进制分布时,这意味着什么?

George has devised a scheme where he flips a coin to earn money. If it lands on heads you get a quarter. If it lands on tails you give him a quarter. The chance of the coin landing on heads, is 68%. You play against George 22 times. Can you approximate this situation with a normal curve? Why or why not?
::George设计了一个计划,他用硬币来赚钱。如果硬币落在头顶上,你得到四分之一。如果它落在尾巴上,你给他四分之一。硬币落在头顶上的机会是68%。你对George玩了22次。你能否用正常曲线来估计这种情况?为什么或为什么不能?

Jade has been practicing shooting a bow and arrow. Based on her target practice she has a 30% chance of hitting the bull’s-eye with the bow and arrow. She shoots the bow and arrow 27 times. Can you approximate this situation with a normal curve? Why or why not?
::杰德一直在练习射箭和弓。根据她的目标练习,她有30%的机会用弓和箭击中牛眼。她射了弓和箭27次。你能用正常曲线来形容这种情况吗?为什么不行?

Steve has created a “grab the marble” game. If you grab a green marble you get a dollar, if you grab a yellow marble you get nothing. There are 31 green marbles, and 69 yellow ones. You decide to reach into the bag and grab a marble 39 times, replacing the marble you grab each time. What is the probability that you will win more than 9 dollars?
::史蒂夫创造了一个“ 大理石大理石”游戏。如果你抢了绿色大理石,你得到一美元,如果你抢了黄色大理石,你一分钱也没有。有31个绿色大理石,69个黄色大理石。你决定伸进袋子里,拿大理石39次,取代你每次抓的大理石。你赢得9美元以上的概率是多少?

Steve asks another friend to play “grab the marble”. If you grab a green marble you get a dollar, if you grab a yellow marble you get nothing. Now, however, there are 49 green and 51 yellow marbles. You decide to reach into the bag and grab a marble 32 times, replacing the marble you grab each time. By approximating this situation with a normal curve, try to predict the expected outcome.
::史蒂夫要求另一个朋友演奏“大理石 ” 。如果你抓住绿色大理石,你能得到一美元,如果你抓住黄色大理石,你一分钱也没有。然而,现在有49个绿色大理石和51个黄色大理石。你决定伸进袋子里,拿大理石32次,用每次抓的大理石替换大理石。通过用正常曲线来接近这种情况,试图预测预期的结果。

Kyle is reading a “Choose Your Own Adventure Book”. He has decided to leave his decisions to chance, so before making a choice between decision ‘a’ and decision ‘b’, he flips a coin. If it lands on heads, he selects choice ‘a’, if it lands on tails, he selects choice ‘b’. The trick is that he uses an unfair coin with a probability of heads of 74%. He has to flip the coin 42 times to get to the end of the story. Can you approximate this situation with a normal curve? Why or why not?
::Kyle正在读一本“选择你自己的冒险书 ” 。他决定让自己的决定成为偶然,所以在对决定`a'和`b'作出选择之前,他翻了一个硬币。如果它落在头上,他选择了选择`a',如果它落在尾巴上,他选择了选择`b'。他的伎俩是,他使用不公平的硬币,头的概率为74%。他必须翻硬币42倍才能达到故事的结尾。你能用正常曲线来形容这个情况吗?为什么或为什么没有?

19 Students at a local high school are all applying to two different colleges. The chance of each of them getting into their first college of choice is 50%. Can you approximate this situation with a normal curve? Why or why not?
::19名本地高中学生都申请两所不同的学院。他们每人进入第一所大学的机会是50%。你能用正常曲线来估计这种情况吗? 为什么不行?

Tammy is at the circus, playing a game where she has to throw darts at pink and green balloons on a spinning dart board. If she hits a pink balloon, she earns a ticket. If she hits a green balloon, she receives nothing. If she misses entirely, the throw is not counted. There are 4 pink and 6 green balloons. She decided to play the game 29 times. Can you approximate this situation with a normal curve? Why or why not?
::塔米在马戏团玩游戏,她不得不在旋转的飞镖板上向粉色和绿色气球扔飞镖。如果她撞到粉色气球,她会赚到一张票。如果她撞到绿气球,她会一分钱也没有。如果她完全失手,球就不算数。有4个粉色和6个绿气球。她决定玩29次。你能用正常曲线来大致了解这种情况吗?为什么或为什么没有?

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