统计代写|AP统计辅导AP统计答疑|Probability

AP统计学与大学的统计学课程在核心内容上是一致的，只是涉及的深度稍浅，AP统计学主要包含以下四部分内容。 第一部分 如何获取数据，获取数据的方式有哪些呢？ 获取数据的方式主要包括普查、抽样调查、观测研究和实验设计等。

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

统计代写|AP统计辅导AP统计答疑|Probability and Probability Rules

• An understanding of the concept of randomness is essential for tackling the concept of probability. What does it mean for something to be random? AP Statistics students usually have a fairly good concept of what it means for something to be random and have likely done some probability calculations in their previous math courses. I’m always a little surprised, however, when we use the random integer function of the graphing calculator when randomly assigning students to their seats or assigning students to do homework problems on the board. It’s almost as if students expect everyone in the class to be chosen before they are chosen for the second or third time. Occasionally, a student’s number will come up two or even three times before someone else’s, and students will comment that the random integer function on the calculator is not random. Granted, it’s unlikely for this to happen with 28 students in the class, but not impossible. Think about rolling a standard six-sided die. The outcomes associated with this event are random-that is, they are uncertain but follow a predictable distribution over the long run. The proportion associated with rolling any one of the six sides of the die over the long run is the probability of that outcome.
• It’s important to understand what is meant by in the long run. When I assign students to their seats or use the random integer function of the graphing calculator to assign students to put problems on the board, we are experiencing what is happening in the short run. The Law of Large Numbers tells us that the long-run relative frequency of repeated, independent trials gets closer to the expected relative frequency once the number of trials increases. Events that seem unpredictable in the short run will eventually “settle down” after enough trials are accumulated. This may require many, many trials. The number of trials that it takes depends on the variability of the random variable of interest. The more variability, the more trials it takes. Casinos and insurance companies use the Law of Large Numbers on an everyday basis. Averaging our results over many, many individuals produces predictable results. Casinos are guaranteed to make a profit because they are in it for the long run whereas the gambler is in it for the relative short run.

统计代写|AP统计辅导AP统计答疑|Conditional Probability and Bayes’s Rule

• Example 9: Example 5 is a good example of what we mean by conditional probability. That is, finding a given probability if it is known that another event or condition has occurred or not occurred. Knowing whether or not a heart was chosen as the first card determines the probability that the second card is a heart. We can find $P(2 n d$ card heart $\mid$ Ist card heart) by using the formula given in Example 5 and solving for $P(A \mid B)$, read $A$ given $B$.
Thus, $P(A / B)=\frac{P(A \cap B)}{P(B)}$.

When applying the formula, just remember that the numerator is always the intersection (“and”) of the events, and the denominator is always the event that comes after the “given that” line. Applying the formula, we obtain:
$$P(2 n d \text { card heart } / 1 s t \text { card heart })=\frac{P(2 n d \text { card heart } \cap 1 \text { st card heart })}{P(1 s t \text { card heart })}=\frac{\frac{11}{51} \cdot \frac{13}{52}}{\frac{13}{52}}=\frac{12}{51}$$
The formula works, although we could have just looked at the tree diagram and avoided using the formula. Sometimes we can determine a conditional probability simply by using a tree diagram or looking at the data, if it’s given. The next problem is a good example of a problem where the formula for conditional probability really comes in handy.

• Example 10: Suppose that a medical test can be used to determine if a patient has a particular disease. Many medical tests are not $100 \%$ accurate. Suppose the test gives a positive result $90 \%$ of the time if the person really has the disease and also gives a positive result $1 \%$ of the time when a person does not have the disease. Suppose that $2 \%$ of a given population actually have the disease. Find the probability that a randomly chosen person from this population tests positive for the disease.

统计代写|AP统计辅导AP统计答疑|Discrete Random Variables

Now that we’ve discussed the concepts of randomness and probability, we turn our attention to random variables. A random variable is a numeric variable from a random experiment that can take on different values. The random variable can be discrete or continuous. A discrete random variable, $\mathbf{X}$, is a random variable that can take on only a countable number. (In some cases a discrete random variable can take on a finite number of values and in others it can take on an infinite number of values.) For example, if I roll a standard six-sided die, there are only six possible values of $X$, which can take on the values $1,2,3,4,5$, or 6 . I can then create a valid probability distribution for $\mathrm{X}$, which lists the values of $X$ and the corresponding probability that $X$ will occur (Figure 5.8).

• Example 14: Consider the experiment of rolling a standard (fair) six-sided die and the probability distribution in Figure $5.8$. Find the probability of rolling an odd number greater than 1 .
Solution: Remember that this is a discrete random variable. This means that rolling an odd number greater than 1 is really rolling a 3 or a 5 . Also note that we can’t roll a 3 and a 5 with one roll of the die, which makes the events disjoint or mutually exclusive. We can simply add the probabilities of rolling a 3 and a 5 .
$$P(3 \text { or } 5)=1 / 6+1 / 6=1 / 3$$
• We sometimes need to find the mean and variance of a discrete random variable. We can accomplish this by using the following formulas:
$$\text { Mean } \quad \mu_{x}=x_{1} p_{1}+x_{2} p_{2}+\ldots+x_{n} p_{n} \text { or } \sum x \cdot P(x)$$
Variance $\quad \sigma_{x}^{2}=\left(x_{1}-\mu_{x}\right)^{2} p_{1}+\left(x_{2}-\mu_{x}\right)^{2} p_{2}+\ldots+\left(x_{n}-\mu_{x}\right)^{2} p_{n}$ or
$$\sigma_{x}^{2}=\sum\left(x-\mu_{x}\right)^{2} \cdot P(x)$$
Std. Dev $\quad \sigma_{x}=\sqrt{\operatorname{Var}(X)}$
Recall that the standard deviation is the square root of the variance, so once we’ve found the variance it is easy to find the standard deviation. It’s important to understand how the formulas work. Remember that the mean is the center of the distribution. The mean is calculated by summing up the product of all values that the variable can take on and their respective probabilities. The more likely a given value of $\mathrm{X}$, the more that value of $\mathrm{X}$ is “weighted” when we calculate the mean. The variance is calculated by averaging the squared deviations for each value of $\mathrm{X}$ from the mean.

统计代写|AP统计辅导AP统计答疑|Probability and Probability Rules

• 理解随机性的概念对于解决概率的概念至关重要。随机的东西是什么意思？AP统计学学生通常对随机事物的含义有一个相当好的概念，并且可能在他们以前的数学课程中做过一些概率计算。然而，当我们使用图形计算器的随机整数函数将学生随机分配到他们的座位或分配学生在板上做作业时，我总是有点惊讶。就好像学生们希望班上的每个人都在第二次或第三次被选中之前被选中。有时候，一个学生的数字会比别人的数字高出两到三倍，学生会评论说计算器上的随机整数函数不是随机的。诚然，班上 28 名学生不太可能发生这种情况，但并非不可能。考虑滚动一个标准的六面模具。与此事件相关的结果是随机的——也就是说，它们是不确定的，但在长期内遵循可预测的分布。从长远来看，与掷骰子的六个面中的任何一个面相关的比例就是该结果的概率。
• 从长远来看，理解这意味着什么很重要。当我将学生分配到他们的座位或使用图形计算器的随机整数函数分配学生将问题放在板上时，我们正在经历短期内发生的事情。大数定律告诉我们，一旦试验次数增加，重复独立试验的长期相对频率就会更接近预期的相对频率。短期内看似不可预测的事件在积累了足够的试验后最终会“安定下来”。这可能需要很多很多次的试验。它所进行的试验次数取决于感兴趣的随机变量的可变性。变异性越大，需要的试验就越多。赌场和保险公司每天都在使用大数定律。将我们的结果平均在许多人身上会产生可预测的结果。赌场可以保证获利，因为他们长期参与其中，而赌徒则相对短期参与其中。

统计代写|AP统计辅导AP统计答疑|Conditional Probability and Bayes’s Rule

• 例 9：例 5 是条件概率的一个很好的例子。也就是说，如果已知另一个事件或条件已经发生或未发生，则找到给定的概率。知道红心是否被选为第一张牌就决定了第二张牌是红心的概率。我们可以找磷(2nd卡心∣Ist card heart) 通过使用示例 5 中给出的公式并求解磷(一种∣乙)， 读一种给定乙.
因此，磷(一种/乙)=磷(一种∩乙)磷(乙).

• 示例 10：假设可以使用医学测试来确定患者是否患有特定疾病。很多医学检查都没有100%准确的。假设测试给出了肯定的结果90%如果这个人真的患有这种疾病并且也给出了积极的结果1%一个人没有患病的时间。假设2%的特定人群实际上患有这种疾病。找出从该人群中随机选择的人对该疾病检测呈阳性的概率。

统计代写|AP统计辅导AP统计答疑|Discrete Random Variables

• 示例 14：考虑滚动标准（公平）六面模具的实验和图 1 中的概率分布5.8. 求掷出大于 1 的奇数的概率。
解决方案：记住这是一个离散的随机变量。这意味着滚动大于 1 的奇数实际上是滚动 3 或 5 。另请注意，我们不能用一次骰子掷出 3 和 5，这会使事件脱节或相互排斥。我们可以简单地将滚动 3 和 5 的概率相加。
磷(3 或者 5)=1/6+1/6=1/3
• 我们有时需要找到离散随机变量的均值和方差。我们可以通过使用以下公式来完成此操作：
意思是 μX=X1p1+X2p2+…+Xnpn 或者 ∑X⋅磷(X)
方差σX2=(X1−μX)2p1+(X2−μX)2p2+…+(Xn−μX)2pn或者
σX2=∑(X−μX)2⋅磷(X)
标准。开发σX=曾是⁡(X)
回想一下，标准差是方差的平方根，所以一旦我们找到了方差，就很容易找到标准差。了解公式的工作原理很重要。请记住，均值是分布的中心。平均值是通过将变量可以采用的所有值及其各自概率的乘积相加来计算的。给定值的可能性越大X, 的值越大X当我们计算平均值时是“加权的”。方差是通过对每个值的平方偏差进行平均来计算的X从平均数。

有限元方法代写

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MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。