### 统计代写|工程统计作业代写Engineering Statistics代考|Poisson Distribution

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|工程统计作业代写Engineering Statistics代考|Poisson Distribution

The Poisson distribution is concerned with the number of events occurring during a given time or space interval. The interval may be of any duration or in any specified region. The Poisson distribution, then, can be used to describe the number of breaks or other flaws in a particular beam of finished cloth, or the arrival rate of people in a queuing line, or the number of defectives in a paint weathering trial, or the number of defective beakers per line per shift. The Poisson distribution describes processes with the following properties:

1. The number of events, $X$, in any time interval or region is independent of those occurring elsewhere in time or space.
2. The probability of an event happening in a very short time interval or in a very small region does not depend on the events outside this interval or region.
3. The interval or region is so short or small that the number of events in the interval is much smaller than the total number of events, $n$.
The point Poisson distribution (point probability) function $f(x)$ can be expressed as
$$f(x)=\frac{\lambda^{x} e^{-\lambda}}{x !}$$
where $x$ is the number of events, $f(x)$ is the probability of $x$ events occurring in an interval, $\lambda$ is the expected average number of events per interval, and $e=2.7182818 \ldots$ is the base of the natural logarithm system.
The cumulative Poisson distribution function $F(x)$ is
$$C D F(x)=F(x)=P(X \leq x)=\sum_{k=0}^{x} \frac{\lambda^{k} e^{-\lambda}}{k !}$$
where $e$ is the base of the natural logarithm system.

## 统计代写|工程统计作业代写Engineering Statistics代考|Negative Binomial Distribution

In cases in which the binomial distribution governs the probability of occurrence of one of two mutually exclusive events, we calculated the probability of success exactly $s$ times out of $n$ trials. The negative binomial distribution is used in a complementary way, that is, for calculating the probability that exactly $n$ trials are required to produce $s$ successes. The probabilities of success and failure remain fixed at $p$ and $q$, respectively. The only way this situation can occur is for exactly $(s-1)$ of the first $(n-1)$ trials to be a success, and for the next, or last, trial also to be a success. The probability of $x=n$, the number of trials needed to produce $s$ successful outcomes, then
$$f(x=n \mid s)=\left(\begin{array}{c} n-1 \ s-1 \end{array}\right) p^{s} q^{n-s}, s \leq n$$
is the negative binomial distribution. The cumulative negative binomial distribution is
$$F(x=n \mid s)=P(s \leq x \leq n)=\sum_{i=s}^{n}\left(\begin{array}{c} i-1 \ s-1 \end{array}\right) p^{s} q^{i-s}$$
Figure $3.4$ illustrates the negative binomial distribution for $p=0.5$ and $s=3$. The mean and variance of the negative binomial distribution are given by
$$\mu=\frac{s}{p}$$

and
$$\sigma^{2}=\frac{s q}{p^{2}}$$
The units on $x, s, n$, and $\mu$ are the numbers of trials. The point and cumulative distribution functions, $f\left(x_{i}\right)$ and $F\left(x_{i}\right)$, are dimensionless. The units on $p$ and $q$ are the probabilities of success or failure.
Example 3.5: Suppose one of your power sources for an analytical instrument in the quality control laboratory has died with a snap and a wisp of smoke. You have finally located the trouble as a faulty integrated circuit (IC). You have been able to find five replacement ICs. You have also found that for this service the chance of failure of an IC is $12 \%$. What is the probability that you will have to use all five ICs before getting one that does not burn out?

Let us define burnout as failure, so $q=0.12$ and $p=0.88$. As $x=5$ and $s=1$, using Equation (3.19),
$$f(x=5 \mid 1)=\left(\begin{array}{l} 4 \ 0 \end{array}\right)(0.88) 0.12^{4}=1.825 \times 10^{-4} \text { or } 0.02 \%$$
the probability is less than $0.02 \%$ that you will have to try all five of the ICs to repair the power supply.

## 统计代写|工程统计作业代写Engineering Statistics代考|Hypergeometric Distribution

The hypergeometric distribution is often used to obtain probabilities when sampling is done without replacement. As a result, the probability of success changes with each trial or experiment. The point hypergeometric probability function is
$$P(X=s)=f(s)=\frac{\left(\begin{array}{l} S \ s \end{array}\right)\left(\begin{array}{c} N-S \ n-s \end{array}\right)}{\left(\begin{array}{l} N \ n \end{array}\right)}, s \leq \min (n, S)$$
where $N$ is the population size, $n$ is the sample size, $S$ is the actual number of successes in the population, $s$ is the number of successes in the sample, and $n \leq N$ and $(n-s) \leq(N-S)$. The cumulative hypergeometric distribution is
$$F(x)=P(X \leq x)=\sum_{k=0}^{x} \frac{\left(\begin{array}{l} S \ k \end{array}\right)\left(\begin{array}{c} N-S \ n-k \end{array}\right)}{\left(\begin{array}{c} N \ n \end{array}\right)}$$
Examples of the point and cumulative hypergeometric distributions are shown in Figure $3.5$ for $N=20, S=15$, and $n=5$.
The mean of the point hypergeometric distribution is
$$\mu=\frac{n S}{N}$$

and the variance is
$$\sigma^{2}=\frac{N-n}{N-1} n \frac{S}{N} \frac{N-S}{N}$$
The units of $\mu, s, N, n$, and $S$ are the number of items, populations, or successes. The point and cumulative probability functions $f(x)$ and $F(x)$ are dimensionless.
Example 3.6: In the production of avionics equipment for civilian and military use, one manufacturer randomly inspects $10 \%$ of all incoming parts for defects. If any of the parts is defective, all the rest are inspected. If 2 of the next box of 50 diodes are actually defective, what is the probability that all of the diodes will be checked before use? This question is really whether the quality control sample of 5 will contain at least one of the defective parts.
For this problem, $N=50, n=5$, and $\mathrm{s}=2$, as we choose to define success as finding a defective diode. The probability is found from
\begin{aligned} F(0) &=P(X \geq 1)=\sum_{k=1}^{2} \frac{\left(\begin{array}{c} 2 \ k \end{array}\right)\left(\begin{array}{c} 50-2 \ 5-k \end{array}\right)}{\left(\begin{array}{c} 50 \ 5 \end{array}\right)} \ &=0.1918367 \text { or } 19 \% \end{aligned}
With the current sampling procedure, there is approximately a $20 \%$ chance of finding a defective part.

## 统计代写|工程统计作业代写Engineering Statistics代考|Poisson Distribution

1. 事件的数量，X，在任何时间间隔或区域中，独立于在时间或空间其他地方发生的那些。
2. 在很短的时间间隔或很小的区域内发生事件的概率不取决于该间隔或区域之外的事件。
3. 区间或区域太短或太小，以至于区间内的事件数远小于事件总数，n.
点泊松分布（点概率）函数F(X)可以表示为
F(X)=λX和−λX!
在哪里X是事件的数量，F(X)是概率X间隔内发生的事件，λ是每个间隔的预期平均事件数，并且和=2.7182818…是自然对数系统的底。
累积泊松分布函数F(X)是
CDF(X)=F(X)=磷(X≤X)=∑ķ=0Xλķ和−λķ!
在哪里和是自然对数系统的底。

## 统计代写|工程统计作业代写Engineering Statistics代考|Negative Binomial Distribution

F(X=n∣s)=(n−1 s−1)psqn−s,s≤n

F(X=n∣s)=磷(s≤X≤n)=∑一世=sn(一世−1 s−1)psq一世−s

μ=sp

σ2=sqp2

F(X=5∣1)=(4 0)(0.88)0.124=1.825×10−4 或者 0.02%

## 统计代写|工程统计作业代写Engineering Statistics代考|Hypergeometric Distribution

F(X)=磷(X≤X)=∑ķ=0X(小号 ķ)(ñ−小号 n−ķ)(ñ n)

μ=n小号ñ

σ2=ñ−nñ−1n小号ññ−小号ñ

F(0)=磷(X≥1)=∑ķ=12(2 ķ)(50−2 5−ķ)(50 5) =0.1918367 或者 19%

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## MATLAB代写

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