# The standard deviation is given by sqrt{(sum(x_i-x)^2)/(n)}, however when we estimate the standart deviation from a sample, the best estimation is sqrt{(sum(x_i-x)^2)/(n-1)}

Estimating standard deviation from wheighted sample
The standard deviation is given by $\sqrt{\frac{\sum \left({x}_{i}-x{\right)}^{2}}{n}}$, however when we estimate the standart deviation from a sample, the best estimation is $\sqrt{\frac{\sum \left({x}_{i}-x{\right)}^{2}}{n-1}}$
How do I have to adjust the standarddeviation if I want to wheight my samples?
I.e. the standard deviation would be $\sqrt{\frac{\sum w\left({x}_{i}\right)\left({x}_{i}-x{\right)}^{2}}{\sum w\left({x}_{i}\right)}}$, if I had the entire data set. What is the correct estimation of the standard deviation, if I'm only given a subsample of the population?
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Teagan Raymond
We know that irrational numbers never repeat by combining the following two facts:
every rational number has a repeating decimal expansion, and
every number which has a repeating decimal expansion is rational.
Together these facts show that a number is rational if and only if it has a repeating decimal expansion.
Decimal expansions which don't repeat are easy to construct; other answers already have examples of such things.
The irrationality of a number by examining its decimal expansion. While it is true that an irrational number has a non-repeating decimal expansion, you don't need to show a given number has a non-repeating decimal expansion in order to show it is irrational. In fact, this would be very difficult as we would have to have a way of determining all the decimal places. Instead, we use the fact that an irrational number is not rational; in fact, this is the definition of an irrational number (note, the definition is not that it has a non-repeating decimal expansion, that is a consequence). In particular, to show a number like $\sqrt{2}$ is irrational, we show that it isn't rational. This is a much better approach because, unlike irrational numbers, rational numbers have a very specific form that they must take, namely $\frac{a}{b}$ where $a$ and $b$ are integers, $b$ non-zero. The standard proofs show that you can't find such $a$ and $b$ so that $\sqrt{2}=\frac{a}{b}$ thereby showing that $\sqrt{2}$ is not rational; that is, $\sqrt{2}$ is irrational.
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