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There is certainly a very clear "conceptual relationship" between the standard deviation and Euclidean distance: If we treat the whole available sample (the $x_i$'s) as a vector, then the Euclidean distance is a measure of how much this vector deviates from the vector containing the mean value, which is "the center" of the population. But the standard ...


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Standard deviation is very useful in metrology and when testing analog to digital converters (ADCs) input noise. Given that a ADC input has noise, to measure its noise one can plot an histogram of values x ocurrences and then calculate the standard deviation of this data. The standard deviation value representes de rms noise of the ADC input. Using only ...


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Let $X$ and $Y$ be independent random variables each with normal distribution (the means and variances need not be the same). Let $W=XY$. Then $$\text{Var}(W)=E((XY)^2)-(E(XY))^2.$$ We need to compute the two expectations on the right. By independence, $E(XY)=E(X)E(Y)$. Also, $E(X^2Y^2)=E(X^2)E(Y^2)$. Since $E(X^2)=\text{Var}(X) +(E(X))^2$, with a similar ...


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The formula doesn't actually have anything specifically to do with the normal distribution (as opposed to any other distribution). It simply yields the standard deviation, a measure of how spread out the values are from their average value (their mean). If the values are distributed according to the normal distribution, then about $68$ percent of the ...


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Suppose you have a Normal distribution $X \sim N(\mu ,\sigma^2)$ we can transform the random variable into a standard Normal distribution using $Z=\frac{X-\mu}{\sigma}$. This is done so that we have $Z \sim N(0,1)$ which is easier to work with. The cumulative distribution function of the standard Normal distribution is given by: $$\Phi ...


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If the lower 3-sigma limits are really computed as $\bar X \pm 3s$, where s is the sample SD, or as $\bar X \pm 3\sigma$, where $\sigma$ is some 'known' population SD, then you know $\bar X$ is the midpoint of the 3-sigma interval. If you know the sample size $n$ and it is large, then a 95% CI for the population mean is $\bar X \pm 1.96s/\sqrt{n}$. (For $s$ ...


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In this particular case, it means that you draw $d$ times a $N(0,1)$ (real) random variable, and these random variables are independent. It means that $\varepsilon=(\varepsilon_1,\dots,\varepsilon_d)$, the $\varepsilon_i$ are independent, and are normally distributed, with variance $1$ and mean $0$. If you don't have the identity matrix but another ...


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Technically, the definition of the vector Gaussian likelihood density when you have mean 0 and covariance matrix $K$ is proportional to $f(x) = e^{-(1/2)x^T K^{-1} x}$, with constant of proportionality determined by $K$ so that it is truly a probability density. This is just how it's defined, and if you know numerical sampling methods like MCMC then this is ...


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Based on the various suggestions in the comments above, I reduced the number of elements in the array from 600 to 200. Empirical testing showed me that anything below 200 and the calculation spanned too short of a sample and the accurate sensor would be triggered too soon (before the ml/L of oxygen measured σ had settled enough for it to be an accurate ...


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Problems like this all come down to a choice of error function. Conventional choice would be least squares (in each slot measure the distance between actual and predicted and square, then sum over the slots). This tends to give plausible results in a variety of situations, and it has the huge advantage of being computationally fairly tractable. Try it! It ...


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Hint: Let $X$ and $Y$ be random variables. Then: $\mathbb E(X+Y)=\mathbb EX+\mathbb EY$ (and $\mathbb E(aX)=a\mathbb EX$). Linearity of expectation. If moreover $X$ and $Y$ are independent then $\text{Var}(X+Y)=\text{Var}X+\text{Var}Y$. Apply this on $Y=Y_1+\cdots+Y_{12}$. If you know variance then you can deduce standard deviation and vice versa.


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If you are given a sample of size $n$, and you don't know the mean of the underlying distribution, then you should use the version of estimated variance with $n-1$. This slick trick (replacing $n$ with $n-1$ when computing estimated variance) makes the empirical estimator of variance what is called "unbiased", meaning that the expected value of the variance ...


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Let's take a stab at that portion of the problem this not answered in comments under the question. The three numbers are $1,x,9$. The mean is $$ \bar x = \frac {1+x+9} 3 $$ so the sum of squares of deviations from the mean is \begin{align} & (1-\bar x)^2 + (x- \bar x)^2 + (9-\bar x)^2 \\[10pt] = {} & \left( \frac{ -7-x } 3 \right)^2 + \left( \frac ...


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The expected value of $Z^2$ is $E(Z^2)=E((X+Y)^2)=E(X^2+2XY+Y^2)$. $=E(X^2)+2E(XY)+E(Y^2)$ $cov(X,Y)=-0.2994\neq 0$, therefore $E(XY)=E(X)E(Y)+cov(X,Y)=E(X)E(Y)-0.2994$ Some intermediate results: $E(X^2)=3.75, \ E(Y^2)=23.44,\ E(X)=1.61,\ E(Y)=4.54,\ E(Z)=6.15,\ cov(X,Y)=-0.2994$


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I think you need to check your calculations. For example, I get $E(Z) =6.15$, but I haven't checked $E(Z^2)$ yet. Would it be cheating to chuck this into an excel spreadsheet?



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