### 统计代写|工程统计作业代写Engineering Statistics代考|Continuous Distributions

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• 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|Continuous Distributions

In the previous section, the distributions were related to the number count of events. In contrast, many measurements are continuum-valued. Continuous distributions model the probabilities associated with continuous variables, such as those that describe events such as service life, pressure drop, flow rate, temperature, percent conversion, and degradation in yield strength.

That we measure continuous variables in discrete units or at fixed time intervals does not matter; the variables themselves are continuous even if the measuring devices give data that are recorded as if step changes had occurred. A familiar example is body temperature, a continuous variable measured in discrete increments. Think about it: Even if you have a fever, your temperature does not change from $98.6$ to $101.2^{\circ} \mathrm{F}$ in one step or even in a series of connected $0.2^{\circ} \mathrm{F}$ intervals, just because the thermometer is calibrated that way.
We must acknowledge, however, that the world is not continuous. From an atomic and quantum mechanical view of the universe, no event has a continuum of values. However, on the macroscale of engineering, individual atoms are not distinguishable within measurement discrimination, and so the world appears continuous. For most practical engineering purposes, it is possible to approximate any distribution in which the discrete

variable has more than 100 values with a probability density function of a continuous random variable.
A cumulative continuous distribution function $F(x)$ is defined as
$$C D F(x)=F(x)=\int_{-\infty}^{x} p d f(X) d X$$
where $p d f(X)$ is a continuous probability density function and $X$ is a continuous variable, which could represent time, temperature, weight, composition, etc. $x$ is a particular value of the variable $X$. The units on $x$ and $X$ are identical and are not a count of the number of events as the $x$-variable in the discrete distributions. The $F(x)$ is the area under the $p d f(x)$ curve, is dimensionless, and as $x$ goes from $-\infty$ to $+\infty, F(x)$ goes from 0 to 1 .
$$\int_{-\infty}^{+\infty} p d f(X) d X=1$$
Note, again, the terms $C D F$ and $F$ are used interchangeably.
Additionally, the terms $p d f(x)$ and $f(x)$ are also used interchangeably. In the discrete distributions, $p d f(x)$ would mean point distribution function, and in continuous functions, it means probability density function.

Although both the continuum $p d f(x)$ and discrete $f\left(x_{i}\right)$ represent the histogram shape of data, they are different. The dimensional units of $p d f(x)$ constitute a major difference between a continuous probability distribution function and the $f\left(x_{i}\right)$ of a discrete point probability distribution. The $p d f(x)$ necessarily has dimensional units that are the reciprocal of the continuous variable. For $F(x)$ of Equation (3.31) to be dimensionless, integrating with $d x$, the argument of the integral, $p d f(x)$, must have the units of the reciprocal of $d x$. $p d f(x)$ is often termed a rate, a rate of change of $F(x)$ w.r.t. $x$. By contrast, in a discrete function $F\left(x_{i}\right)$ is the sum of $f\left(x_{i}\right)$, the fraction of the dataset with a value of $x_{j}$, so $f\left(x_{i}\right)$ is dimensionless. You cannot use a discrete point distribution in Equation (3.31) or a continuous function in Equation (3.1) and expect $F(x)$ to remain a dimensionless cumulative probability. Another difference between discrete and continuous probability density functions is that $x$ is used only to represent values of the variable involved throughout the continuous case. For discrete distributions, $x$ was often the number of events in a particular class (category).
So, whether you are using the term $p d f(x)$ or $f(x)$ take care that you are properly using the dimensionless version for distributions of a discrete variable, and the rate version with reciprocal units of $X$ for distributions of continuum variables.
The mean and variance of the theoretical continuum distributions are:
$$\begin{gathered} \mu=\int_{-\infty}^{+\infty} x p d f(x) d x \ \sigma^{2}=\int_{-\infty}^{+\infty}(x-\mu)^{2} p d f(x) d x \end{gathered}$$

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

If a random variable can have any numerical value within the range from $a$ to $b$ and no values outside that range, and if each possible value has an equal probability of occurring, then the probability density function for the uniform continuous distribution is

The random variable $X$ may have any dimensional units which must match that of parameters $a$ and $b$. The mean, $\mu$ will have the same units. Whereas the cumulative distribution function $F(x)$ is dimensionless, the probability density function, $p d f(x)$, has units that are the reciprocal of those of the random variable $X$.

Again, the population coefficients, $a$ and $b$, represent the true values. You might not know what they are exactly, but certainly you can do enough experiments to get good estimates for them.

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

A proportion is the probability of an event, a fraction of outcomes, and is a continuumvalued variable. In flipping a coin, the probability of a particular outcome is $p=0.5$. In rolling a die the probability of getting a 5 is $p=0.166 \overline{66}$. In rolling 10 dice and winning means getting at least one five in the 10 outcomes, the probability is $p=0.83949441$…. Although the events are discrete, the probability could have a continuum of values between 0 and 1. $0 \leq p \leq 1$.

If the proportion is developed theoretically, then it is known with as much certainty as the basis and idealizations allow. Then the variance on the proportion is 0 .
$$\sigma_{p}^{2}=0$$
Alternately, the proportion could be determined from experimental data. For example, a trick die could be weighted to have $p=0.21$ as the probability of rolling a 5 . Here, proportion, $p$, is the ratio of number of successes, s, per total number of trials, $n, p=s / n$, as $n \rightarrow \infty$. Alternately, the proportion would be estimated as the average after many trials.
$$\hat{\mu}{p}=\hat{p}=\frac{s}{n}=\frac{\sum s{i}}{\sum n_{i}}$$
If experimentally determined, the variance on the proportion would be estimated by
$$\hat{\sigma}_{p}{ }^{2}=\frac{\hat{p}(1-\hat{p})}{n}=\frac{\hat{p} \hat{q}}{n}=\frac{s(n-s)}{n^{3}}$$
Note this is similar to the mean and variance of the binomial distribution, but here the statistics are on the continuum-valued proportion. In the binomial distribution, the statistics

are on the number count of a particular type of event. The variance of the count of successes of $n$ samples from a population would be given by Equation (3.14).
Example 3.9: What are the mean and sigma when the probability of an event (outcome $=1$ ) is and unknown $p$, and the probability of a not-an-event (outcome $=0$ ) is $q=(1-p)$ ? A sequence of $n$ dichotomous events might be
$$0,0,1,1,1,0,1,0,0,1,0,0,1, \ldots$$
Whether we call the event a $\mathrm{H}$ or a $\mathrm{T}$, a success or a fail, the ${1,0}$ notation is equivalent. Experimentally, there are $s=21$ successes out of $n=143$ trials. From Equation (3.40) the estimate of $p$ is
$$\hat{p}=\frac{s}{n}=\frac{21}{143}=0.14685314 \ldots$$
From Equation (3.41) the standard deviation on $\hat{p}$ is
$$\hat{\sigma}{p}=\sqrt{\hat{\sigma}{p}^{2}}=\sqrt{\frac{s(n-s)}{n^{3}}}=\sqrt{\frac{21(143-21)}{143^{3}}}=0.02959957 \ldots$$
acknowledging the uncertainty on $\hat{p}$ and $\hat{\sigma}{p}$ one might report $$\hat{p}=0.148$$ and $$\hat{\sigma}{p}=0.03$$

## 统计代写|工程统计作业代写Engineering Statistics代考|Continuous Distributions

variable 有超过 100 个值，具有连续随机变量的概率密度函数。

CDF(X)=F(X)=∫−∞XpdF(X)dX

∫−∞+∞pdF(X)dX=1

μ=∫−∞+∞XpdF(X)dX σ2=∫−∞+∞(X−μ)2pdF(X)dX

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

σp2=0

μ^p=p^=sn=∑s一世∑n一世

σ^p2=p^(1−p^)n=p^q^n=s(n−s)n3

0,0,1,1,1,0,1,0,0,1,0,0,1,…

p^=sn=21143=0.14685314…

σ^p=σ^p2=s(n−s)n3=21(143−21)1433=0.02959957…

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## MATLAB代写

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