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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|抽样理论作业代写sampling theory 代考|The Gaussian Model: The Normal Probability Distribution

The Gaussian model is the most common model in use in conventional statistics. This model is characterized by a unique property. If the set of the possible contents of a certain component is from large enough samples, the distribution of the average values obtained from those samples will tend toward a bell-shaped Gaussian distribution. This property is called the central limit theorem. A random variable obeys a normal law of distribution when its probability density $f(x)$ can be written as follows:
$$f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\left[(x-\bar{x})^2 / 2 \sigma^2\right]}$$
where $\bar{x}$ is the arithmetic average and $\sigma$ is the true unknown standard deviation, in which the distances $\pm 1 \sigma$ from the average $\bar{x}$ are defined as the $\mathrm{x}$-coordinates of both inflexion points of the above function $f(x)$. Most of the time, however, we only have access to the estimated standard deviation s. Then investigating the normality of a distribution consists of comparing the characteristics of the estimated standard deviation $s$ with those of an ideal model; for example, if we have:
$P{-s<x<s}$ approaches $68 \%$
$P{-2 s<x<2$ s $}$ approaches $95 \%$
$P{-3 s<x<3 s}$ approaches $99.7 \%$ thus, the distribution is approximately normal. An illustration of a typical standard normal probability distribution is shown in Figure 3.6.

统计代写|抽样理论作业代写sampling theory 代考|The Double Poisson Process

When primary samples taken from the deposit contain the constituent of interest in a limited average number $\mu$ of discrete grains or clusters of such grains (i.e., $P[y=n])$, and they are subsampled in such a way that the subsamples also contain discrete grains of reduced size in a limited average number $\theta$ (i.e., $P[x=r]$ ), a double Poisson distribution of the assay values is likely.

The probability $P$ of $r$ grains of mineral appearing in any sub-sample is determined by the sum of the probabilities of $r$ grains being generated from samples with $n$ grains (not to be confused with $n$ trials as defined earlier).
Let’s define the ratio $f$ :
$$f=\frac{\theta}{\mu}$$
With $\theta=\mu \cdot f$ or $\theta=n \cdot f$ for each possibility, the equation for the resulting, compounded probability of the double Poisson distribution is:
$$P(x=r)=\sum P(y=n) \cdot P(x=r)=\sum \frac{\mu^n e^{-\mu}}{n !} \cdot \frac{(n f)^r e^{-n f}}{r !}=\frac{f^r e^{-\mu}}{r !} \sum_{n=0}^{n=\infty} \frac{\mu^n e^{-n f} n^r}{n !}$$
for $\mathrm{r}=0,1,2,3, \ldots$
This is the probability of obtaining a sample with $r$ grains of the constituent of interest. The equation could be modified using improved Stirling approximations given later on in Part $\mathrm{V}$ of this book.

In practice, one does not usually count grains; concentrations are measured. The conversion factor from number of grains to, percent $X$ for example, is $C$, the contribution of a single average grain. Because the variance of a Poisson distribution is equal to the mean:
$$s=C \sqrt{\mu}$$
Therefore:
$$s^2=C^2 \mu$$
But variances of random variables are additive, then:
$$s^2=C^2 \mu+c^2 \theta$$
The data available are usually assays in \% metal, gram/ton, ppm, or ppb. They are related by the equation:
$$x_i=a_H+c r_i$$

抽样理论代考

统计代写|抽样理论作业代写sampling theory代考|高斯模型:正态概率分布

$$f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\left[(x-\bar{x})^2 / 2 \sigma^2\right]}$$
，其中$\bar{x}$是算术平均值，$\sigma$是真正的未知标准差，其中$\pm 1 \sigma$到平均值$\bar{x}$的距离定义为上述函数$f(x)$的两个拐点$\mathrm{x}$ -坐标。然而，大多数时候，我们只能得到估计的标准差s。然后，研究一个分布的正态性包括比较估计的标准差$s$与理想模型的特征;例如，如果我们有:
$P{-s<x<s}$接近$68 \%$
$P{-2 s<x<2$ s $}$接近$95 \%$
$P{-3 s<x<3 s}$接近$99.7 \%$因此，分布近似正态分布。典型的标准正态概率分布如图3.6所示

统计代写|抽样理论作业代写sampling theory代考|双泊松过程

$$f=\frac{\theta}{\mu}$$

$$P(x=r)=\sum P(y=n) \cdot P(x=r)=\sum \frac{\mu^n e^{-\mu}}{n !} \cdot \frac{(n f)^r e^{-n f}}{r !}=\frac{f^r e^{-\mu}}{r !} \sum_{n=0}^{n=\infty} \frac{\mu^n e^{-n f} n^r}{n !}$$

$$s=C \sqrt{\mu}$$

$$s^2=C^2 \mu$$

$$s^2=C^2 \mu+c^2 \theta$$

$$x_i=a_H+c r_i$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师