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Let's replace $Var(X)$ with $\sigma^2$ in the first equation to give $$P(|X - E(X)| \geq r) \leq \frac{\sigma^2}{r^2}.$$ Now suppose $k= \dfrac{r}{\sigma}$, i.e. $r = k \sigma$, and substitute to give $$P(|X - E(X)| \geq k \sigma) \leq \frac{\sigma^2}{k^2\sigma^2} = \frac{1}{k^2}$$ which is your second equation using $k$ instead of $r$. You can think of ...


It is exactly the same. If you use the first inequality you have $$P(|X−E(X)|\geq r⋅σ)\leq \frac{\operatorname{var}(X)}{(r\sigma)^2}=\frac{1}{r^2}.$$


I decided to move from a comment to an answer because some times I ended up using some facts without digging too much in the whys. So, I decided to dig a little more and hopefully answer the questions of the OP. This answer is mainly about why the RMS of the noise is equal to its standard deviation. As a side note, I will also do some comments about the ...


Note that $Var(X) = \sigma^2$. \begin{align*} P(|X - E(X)| \geq r) \leq \frac{\sigma^2}{r^2} &\implies P((X - E(X))^2 \geq r^2) \leq \frac{\sigma^2}{r^2} \\ &\implies P((X - E(X))^2 \geq r^2\sigma^2) \leq \frac{1}{r^2} \\ &\implies P(|X - E(X)| \geq r\sigma) \leq \frac{1}{r^2} \end{align*}


No, in general, if $X$ follows a normal distribution, then in shorthand it is written as $$X\sim N(\mu,\sigma^2).$$ Thus, for example, the variance of $X_1$ is 3 not $3^2$. Or as you have written it $$\sigma_1^2 = 3$$ not $\sigma_1 = 3$.


You say you take the average of all the averages, but I notice that you have a sample count column. Are these averages over different sample sizes? If so, then you would probably want a weighted average for your aggregate average: $$\text{Aggregate Average} =\frac{\sum_i (\text{sample size})_i(\text{average})_i}{\sum_i (\text{sample size})_i}$$ But without ...


The Standard Deviation is a measure of how spread out the numbers in your data are. So if your numbers have the same difference, even with higher values then you will got the same standard daviation. A small example to clarify, using matlab: If your data is : $ 1 ,2,3,4,5,6,7,8,9$ then std is $2.7386$. If your data is : $ 10 ,20,30,40,50,60,70,80,90$ ...

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