8.3 分民族分布-interactive

章节大纲

• 选择节常规

  折叠 展开

  常规

  全部折叠 展开全部

  • 选择活动新闻通告

    

    新闻通告 讨论区
• 选择节分人口分布

  折叠 展开

  分人口分布

  播放段落

  Lesson Objectives
  ::经验教训目标

  播放段落
  • Analyze decisions and strategies using probability concepts
    ::使用概率概念分析决定和战略
  播放段落
  • Calculate the and standard deviation of a binomial distribution .
    ::计算二进制分布的标准偏差。
  播放段落
  • Analyze decisions and strategies using probability concepts.
    ::利用概率概念分析决定和战略。
  播放段落
  • Use probabilities to make fair decisions.
    ::利用概率作出公平决定。

  ---

  播放段落

  Introduction: A Day At The Carnival Continued
  ::导言:狂欢节继续的一天

  播放段落

  In section 1 of this chapter, we saw that Lorenzo and Cheyenne were playing the following game at a carnival.
  ::在本章第1节中,我们看到洛伦佐和夏安在嘉年华会上玩下一场游戏。

  播放段落

  Lorenzo correctly predicted that when the ball is dropped from the middle, you are more likely to get 5 tickets than any of the other outcomes. To understand why the ball is more likely to land in the middle slot, we will need to learn about binomial distributions.
  ::洛伦佐正确地预测,当球从中间投下时,你比任何其他结果更有可能得到5张门票。 要理解为什么球更可能降落在中间位置,我们需要了解二进制分布。

  ---

  播放段落

  Activity 1: Binomial Distribution
  ::活动1:二分分布

  播放段落

  A binomial distribution is a specific type of discrete frequency distribution that models the possible number of successful outcomes based on a given number of trials. The term discrete implies that we are dealing with a discrete random variable , a variable for which we can count the possible outcomes. Additionally, the data in a binomial distribution is binary, which means that there are only two outcomes which we will refer to as a success, the desired outcome, and a failure, the undesired outcome. To understand a binomial distribution, let's consider a coin flip. Consider that you flip a coin n times; how many times should you get heads? What is the probability that you get heads 1 time, 2 times, 3 times, etc.?
  ::二进制分布是一种特定类型的离散频率分布, 它根据给定的试验量来模拟成功结果的可能数量。 离散术语意味着我们正在处理一个离散随机变量, 一个我们可以计算可能结果的变量。 此外, 二进制分布中的数据是二进制的, 这意味着只有两个结果, 我们称之为成功, 理想的结果, 失败, 不理想的结果。 要理解二进制分布, 我们来考虑翻转一个硬币。 想想您翻转一个硬币 n 次; 您需要多少次头? 您得到头的概率是 1 次, 2 次, 3 次等等 ?

  播放段落

  If you flip a fair coin one time, there is only one possible way to get heads, and the probability is 50%. If you flip a fair coin two times, there are four possible outcomes:
  ::如果您一次抛出一个公平的硬币, 只有一种可能的方法才能得到头部, 概率是50%。 如果您两次抛出一个公平的硬币, 则有四种可能的结果 :

  

  播放段落

  This result gives us the following probability distribution:
  ::这一结果使我们得出了以下概率分布:

  Number of Heads	Number of Ways it Can Happen	Probability
  2 Heads	1	25%
  1 Head	2	50%
  0 Heads	1	25%

  播放段落

  If you flip a fair coin three times, there are eight possible outcomes:
  ::如果你翻了一个公平的硬币三次, 有八种可能的结果:

  

  播放段落

  This result gives us the following probability distribution:
  ::这一结果使我们得出了以下概率分布:

  Number of Heads	Number of Ways it Can Happen	Probability
  3 Heads	1	12.5%
  2 Heads	3	37.5%
  1 Head	3	37.5%
  0 Heads	1	12.5%

  播放段落

  Use the interactive below to explore what would happen as the number of tosses increases.
  ::利用下面的交互式互动,探讨随着抛掷数量增加将会发生什么情况。

  播放段落

  Discussion Question: How does the number of outcomes relate to Pascal's Triangle? Why do you think the binomial distribution relates to binomials in algebra?
  ::讨论问题:结果的数量与帕斯卡尔三角有何关系?你为什么认为二进制分布与代数中的二进制有关?

  ---

  播放段落

  Activity 2: Expected Value
  ::活动2:预期价值

  播放段落

  Since the chance of getting heads when flipping a fair coin is 50%, regardless of how many coin flips take place, we should see a peak in the distribution at around half the number of times the coin is flipped. For example, if you flipped a coin 4 times, you would expect it to come up heads about 2 times. This value is known as the . In this case, the probability of getting heads is 50% or 0.5. Every time you flip a coin, you have the following probability distribution.
  ::抛开一个公平的硬币时, 拿到头的机会是50%, 不论有多少个硬币, 我们都应该看到分配的峰值大约是硬币翻转次数的一半。 例如, 如果您翻转了一个硬币, 就会期望它翻转4次。 这个数值被称为 。 在此情况下, 拿到头的概率是 50% 或 0.5 。 每次翻转硬币时, 你就会看到接下来的概率分布 。

  Number of Heads	Probability
  1 Head	50%
  0 Heads	50%

  播放段落

  Since a binomial distribution is a specific type of probability distribution, you can use the same approach for finding the expected value. Find the sum of each outcome multiplied by its probability. 
  ::由于二进制分布是概率分布的具体类型,您可以使用相同的方法查找预期值。查找每个结果的总和乘以其概率。

  $$\begin{align*}0(0.5) + 1(0.5) = 0.5\end{align*}$$

  播放段落

  Every time we flip a coin, we can expect to get 0.5 heads. Keep in mind that 0.5 is also the probability of getting heads. The expected value in 1 trial equaling the probability of success in a binomial distribution will always be the case, even if the probability isn’t 0.5, because the probability of failure is multiplied by 0 and will cancel.
  ::每当我们抛硬币时,我们都会期望得到0.5个头。 记住0.5也是获得头的概率。 一试中的预期值等于二进制分布成功概率,即使概率不是0.5, 也总是如此,因为失败概率乘以0, 并且会取消。

  播放段落

  Flipping a fair coin 4 times, we expect to get 0.5 during each of the four flips.
  ::翻了4次金币 我们预计每翻4个硬币 都会得到0.5分

  播放段落

  $$\begin{align*}4 \text{ Flips} → 0.5 + 0.5 + 0.5 + 0.5 = 2 \text{ expected heads.}\end{align*}$$
  ::4 5+0.5+0.5+0.5+0.5+0.5=2预期头部。

  播放段落

  Flipping a fair coin n times, we get the following:
  ::抛出一个公平的硬币,不时,我们得到以下信息:

  播放段落

  $$\begin{align*}\underbrace{0.5+0.5+0.5+...+0.5}\\ \text{n times}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\end{align*}$$
  ::0.5+0.5+0.5+.+0.5+.+0.5+

  播放段落

  We can express the expected value of a binomial distribution using the function below:
  ::我们可以使用以下函数表达二元分布的预期值:

  播放段落

  %20%3D%20np">

  $$\begin{align*}E(n) = np\end{align*}$$
  ::E=np

  播放段落
  • "> $\begin{align*}E(n)\end{align*}$   is the expected value
    ::E 是预期值
  播放段落
  • $\begin{align*}n\end{align*}$   is the number of trials
    ::n 是审判次数
  播放段落
  • $\begin{align*}p\end{align*}$   is the probability of success
    ::p 是成功概率

  播放段落

  Additionally, the farther the value from the expected value, the less likely we will be to obtain that value. For example, we would be unlikely to get 7 heads if we flipped a coin 8 times.
  ::此外,值越远于预期值,我们就越不可能获得这一值。 比如,如果我们翻了8次硬币,我们就不可能得到7个头。

  播放段落

  We use the variable $\begin{align*}p\end{align*}$   to represent the probability of success. The variable $\begin{align*}q\end{align*}$   is used to represent the probability of failure. Since there are only two possible outcomes, failure can be calculated by subtracting the probability of success, $\begin{align*}p,\end{align*}$   from 1.
  ::我们使用变量 p 来代表成功概率。变量 q 用来代表失败概率。 由于只有两种可能的结果, 失败可以通过从 1 中减去成功概率来计算, p 从 1 中减去成功概率。

  播放段落

  Discussion Question: How would the distribution of the number of ones rolled using a fair six-sided die compare to the distribution of heads when flipping a coin?
  ::讨论问题:在抛硬币时,使用公平六面法滚动的号码的分布情况与头的分布情况相比会如何?

  ---

  播放段落

  Activity 3: Winning the Game
  ::活动3:赢得比赛

  播放段落

  Lorenzo and Cheyenne want to explore if where they place the ball influences where the ball will land and which starting location will produce the most tickets on average if they play the game repeatedly. Let's begin by mapping out the possible outcomes for starting in the middle.
  ::Lorenzo和Cheyenne想探究他们是否把球的影响放在哪里,球会在哪里落地,如果球反复玩游戏,哪个起始位置会平均出票最多。我们先从绘制从中间开始的可能结果开始。

  播放段落

  The pattern 1, 4, 6, 4, 1 should look familiar. It is no coincidence that the result is a row from Pascal's triangle since this scenario represents a binomial distribution. As the ball makes its way through the pegs, a series of simple events occur. The pegs are structured in such a way that as the ball falls from one level to the next, the ball can either fall to the right or left of a peg. The possible outcomes give us the following binomial distribution:
  ::模式 1 、 4 、 6 、 4 、 1 应该是熟悉的。 并非巧合的是, 此结果是从 Pascal 三角形的一行, 因为此假想代表了二进制分布 。 随着球穿过钉子, 一系列简单的事件会发生 。 钉子的结构可以让球从一个层次跌到下一个层次, 球可以落在一个钉子的右边或左边。 可能的结果会给我们以下的二进制分布 :

  Outcome	Probability
  4 Tickets	1/16 = 0.0625
  1 Ticket	4/16 = 0.25
  5 Tickets	6/16 = 0.375
  2 Tickets	4/16 = 0.25
  3 Tickets	1/16 = 0.0625

  播放段落

  We can use these probabilities to determine the expected number of tickets. If we played the game 100 times, we could multiply each probability by 100 to find the expected number of times each outcome should occur. We can then multiply that probability by the number of tickets earned for each respective outcome to find the expected number of tickets from each outcome.
  ::我们可以用这些概率来决定预期的票票数。 如果我们玩了100次, 我们可以把每场比赛的概率乘以100, 以找到每个结果的预期次数。 然后我们可以把这一概率乘以每个结果的预期票数, 以找到每个结果的预期票数。

  Outcome	Expected Outcomes for 100 Trials	Expected Number of Tickets
  4 Tickets	0.0625 • 100 = 6.25	6.25 • 4 = 25
  1 Ticket	0.25 • 100 = 25	25 • 1 = 25
  5 Tickets	0.375 • 100 = 37.5	37.5 • 5 = 187.5
  2 Tickets	0.25 • 100 = 25	25 • 2 = 50
  3 Tickets	0.0625 • 100 = 6.25	6.25 • 3 = 18.75

  播放段落

  Adding the expected number of tickets for each outcome will give us the total expected number of tickets for 100 trials if we drop the ball from the middle:
  ::加上每个结果的预期票数,如果我们把球从中间掉下去,那么100场审判的预期票数总数将达到:

  播放段落

  $$\begin{align*}25+25+187.5+50+18.75=306.25 \text{ tickets}\end{align*}$$
  ::25+25+187.5+50+18.75=306.25

  播放段落

  We can take the same approach to find the expected number of tickets from playing the game 1 time. Expressing the outcome in terms of 1 trial may seem counterintuitive, but the idea is similar to finding a unit rate.
  ::我们可以采取同样的方法来寻找第一次比赛的预期票数。 以1场审判来表达结果可能看起来是反直觉的,但这个想法与寻找单价类似。

  Outcome	Expected Outcome for 1 Trial	Expected Number of Tickets
  4 Tickets	0.0625 • 1 = 0.0625	0.625 • 4 = 0.25
  1 Tickets	0.25 • 1 = 0.25	0.25 • 1 = 0.25
  5 Tickets	0.375 • 1 = 0.375	0.375 • 5 = 1.875
  2 Tickets	0.25 • 1 = 0.25	0.25 • 2 = 0.5
  3 Tickets	0.0625 • 1 = 0.0625	0.0625 • 3 = 0.1875

  播放段落

  If we play the game repeatedly, dropping the ball from the middle, we should expect to win an average of 3.0625 tickets per attempt. Use this strategy to determine where to drop the ball to maximize the number of tickets won.
  ::如果我们反复玩游戏,从中间投下球,我们就能期望每次尝试平均赢得3.0625张门票。 使用这一策略来确定投球地点,以最大限度地增加赢票数。

  播放段落

  Discussion Question: Checking the remaining starting positions, which starting position will have the highest average tickets won per trial?
  ::讨论问题:检查其余的起步职位,哪个起步职位的每场审判平均票数最高?

  ---

  播放段落

  Activity 4: Standard Deviation
  ::活动4:标准偏离

  播放段落

  We can use what we know about calculating the standard deviation and variance of a probability distribution to derive the formula for calculating the standard deviation of a binomial distribution.
  ::我们可以利用我们所了解的关于计算概率分布的标准偏差和差值的信息,得出计算二进制分布标准偏差的公式。

  播放段落

  Example
  ::示例示例示例示例

  播放段落

  The binomial distribution below represents the probabilities for flipping a coin 1 time. Find the standard deviation of flipping a fair coin n times.
  ::下面的二进制分布表示翻硬币一次的概率。 查找翻硬币一次的标准偏差。

  Number of Heads	Probability
  1 Heads	50%
  0 Heads	50%

  播放段落

  We will use the variance of a probability distribution formula to find the variance for 1 coin flip. Once we have the variance, we can use the square root to get the standard deviation.
  ::我们将使用概率分布公式的差异来查找一个硬币翻转的差异。 一旦我们发现差异, 我们可以使用平方根来获取标准偏差 。

  播放段落

  $$\begin{align*}𝝈^2 &= Σ \left(P(x)・(x - 𝜇)^2 \right) \\\\ 𝜇 &= 0.5\end{align*}$$
  ::282(P(x)-(x)-(x)2)________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

  x	x - μ	(x - μ) 2	P(x)	P(x) • (x - μ) 2
  0	0.5	0.25	0.5	0.125
  1	0.5	0.25	0.5	0.125

  播放段落

  The $\begin{align*}Σ\end{align*}$   in the formula tells us to add each $\begin{align*}P(x)•(x-μ)^2: \end{align*}$ 0.125 + 0.125 = 0.25.
  ::公式中的 表示我们添加每个 P(x) • (x) 2: 0.125 + 0.125 = 0.25 = 0.25。

  播放段落

  The variance, $\begin{align*}𝝈^2\end{align*}$  , of one flip is 0.25. Thus the variance of n flips is 0.25n. From here, we can take the square root to see that the standard deviation of n flips.
  ::0.25。在这里,我们可以选择平方根来查看 n 平方根的标准偏差。

  播放段落

  Answer:  $\begin{align*} \sqrt{0.25n}\end{align*}$  
  ::答复:0.25n

  播放段落

  We used variance above because the variance of each simple event can be added to get the variance of n simple events, but the standard deviation of each simple event cannot be added to get the standard deviation of n simple events. This is because $\begin{align*}n\sqrt{0.25}\end{align*}$   is not the same as  $\begin{align*}\sqrt{0.25n}\end{align*}$   for all values of n.  Use the interactive to derive the general formula for the standard deviation of a binomial distribution.
  ::我们使用上述差异是因为每个简单事件的差异可以添加来获取 n 简单事件的差异,但无法添加每个简单事件的标准偏差来获取 n 简单事件的标准偏差。 这是因为 n/ 0. 25 与 n 的所有值不相同。 使用交互式公式来计算二进制分布的标准偏差的一般公式 。

  Summary

  播放段落 A binomial distribution is a specific type of discrete frequency distribution that models the possible number of successful outcomes based on a given number of trials. ::二进制分布是一种特定类型的离散频率分布,它根据一定数量的试验,对成功结果的可能数量进行模型计算。 播放段落 Binomial distributions only have two outcomes: success or failure. ::分民族分布只产生两个结果:成败。 播放段落 The expected value of a binomial distribution can be expressed as the product of the number of trials n and the probability of success p. %20%3D%20np"> $$\begin{align*}E(n) = np\end{align*}$$ ::二进制分布的预期值可以表示为试验n数量和成功概率的产物,第E=np页

  ---

  播放段落

  Wrap-Up: Review Questions
  ::总结:审查问题