Why is the standard deviation the error on the singular meas

  • #1
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I'm a beginner with the study in data analysis in Physics. I'm trying to understand the meaning, in the field of experimental Physics, of the standard deviation ##\sigma## of a series of data.

There is one fundamental thing about ##\sigma## that I read but I could not understand.

>In a series of ##N## measurements of the same physical quantity (in the same conditions) the standard deviation ##\sigma## of the data represents the error on the singular measurement.

That is I should write the result of one measurement as $$x_i\pm \sigma$$

I'm aware of these facts about ##\sigma## (regarding its meaning):

- ##\sigma=\sqrt{\sum \frac{(x_i-\mu)^2}{N}}## , where ##\mu## is the theoric "true value" of the physical quantity measured
- The flexes of the Gaussian distribution are in ##x_{1,2}=\pm \sigma##
- ##[\bar{x}-\sigma,\bar{x}+\sigma]## contains the ##68\%## of measurements, where ##\bar{x}## is the mean value, which is the best possible approximation of ##\mu##
- There is the ##68\%## of probability to find ##\mu## in ##[x_i-\sigma,x_i+\sigma]## and, which is equivalent, to find ##x_i## in ##[\mu-\sigma,\mu+\sigma]##
- There is the $99.7\%$ of probability to find ##\mu## in ##[x_i-3\sigma,x_i+3\sigma]## and, which is equivalent, to find ##x_i## in ##[\mu-3\sigma,\mu+3\sigma]##

I'm ok with these facts that come from the properties of the Gaussian distribution but still I do not see why ##\sigma## is the error on the singular datum ##x_i##.

In other words I do not understand why the interval of variation of ##x_i## should be ##[x_i-\sigma,x_i+\sigma]##.

Does this interval have particular properties in terms of probability, linked with the error on the singular value?
 

Answers and Replies

  • #2
Dale
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It means that a single measurement can be represented as a normally distributed random variable ##N(\mu,\sigma)##.

This is in contrast to the mean of a sample of n measurements, which could be represented as a normally distributed random variable ##N(\mu,\sigma/ \sqrt{n})##
 

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