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It's not the average difference, exactly. The squaring creates a weighting effect: large fluctuations contribute disproportionately more to the standard deviation than small ones. For your die roll, you get rather large deviations from the mean when you hit 1 or 6, which happens 1/3 of the time. The standard deviation being larger than the "median deviation" ...


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The short answer is "Not a whole lot"...standard deviations are useful when they have meaningful probability statements (e.g., like with the Normal distribution). In your cases, all SD really telling you is that typical fluctuations seem to be on the same order of magnitude as your mean, so you cannot treat it as "essentially constant" but the mean has ...


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Answer in development Let a $d$-sided coin have sides labeled $X\in\{1,2,3,\ldots,d\}$, and let the state space be $$\{p_1,p_2,p_3,\ldots,p_d\}$$ this is, $p_i$ is the probability of throwing the side labeled $i$. Note that $$\sum_{i=1}^dp_i=1$$ Then, the standard deviation is the square root of the variance, and: $$\begin{array}{rcl} ...


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Wikipedia has a discussion of this. For a large sample, your sample standard deviation is very close to the population standard deviation. I would be more concerned about bias from taking the first samples. It would be better to take a random sample in case the first data is somehow different from the rest.


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We interpret your question as asking under what conditions is it reasonable to use a model in which the population standard deviation is known, Let us suppose that we are using a high precision scientific instrument to determine say the mass of an object. We will do this by making a series of $n$ measurements of the mass of the object. The behaviour of the ...


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Variance measures how spread out they are, so with five numbers, put two of them at the one endpoint and three at the other. Thus: $0,0,0,100,100$. With an even number of numbers, put have of them at one endpoint and half at the other. For $0,0,0,100,100$, the variance is $\left(\frac 3 5\cdot\frac 2 5\right)\cdot100^2$. (Here I am using $\frac 1 n ...



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