# What kind of errors does normal distribution account for?

No Experiment can give 100% accurate results and are bound to have some errors and normal distribution gives us an idea of distribution of outcomes, But what errors doees it account for, Are they random errors, Or some other kind?

• Would Cross Validated be a better home for this question? – Qmechanic Jun 5 '18 at 16:52
• Keep in mind, also, that not all experiments yield normally (or even Poisson) distributed outcomes. Gaussians are common but universal. – dmckee --- ex-moderator kitten Jun 5 '18 at 17:02
• a gaussian does not account well for errors that are human-caused, like truncation errors, aliasing errors due to improper sampling, and the effects of having more than one distinct and disjoint population improperly lumped together in a sample. – niels nielsen Jun 5 '18 at 18:18

The Gaussian/normal distribution is a continuous probability distribution which accounts for the sampling of random variables.

Imagine you are measuring the length of a spring. You take a series of measurements $x_1, x_2, ..., x_{n-1}, x_n$. Each $x_i$ represents a single measurement, and $n$ is the total number of measurements you made. Then, your mean/average - a measure of central tendency in your data set -- is given by

$$\left<x\right> = \frac{1}{n} \sum_{i=1}^n x_i.$$

The Gaussian distribution provides the probability for what a random measurement will be, as given by the standard deviation. Approximately 68% of the measurements will fall within one standard deviation

$$s_x = \sum_{i=1}^n \sqrt{ \frac{|x_i - \left<x\right>|^2}{n} }.$$

Likewise, 95 percent of random values will fall within $2\sigma$, and something like 99 percent for $3\sigma$. This is why $5\sigma$ is sometimes considered a standard for statistical significance: there is only a $\sim 1$ in $10^6$ chance a measurement could fall outside of $5\sigma$.

But let's get back to our spring. We want to know our uncertainty the length we have measured, $\left<x\right>$. The standard error of the mean is given by

$$\sigma_x = \frac{s_x}{\sqrt{n}}.$$

This value provides an error in our spring length from its true length. Thus you would report the measurement of its length as

$${\rm spring\,\, length} = \left<x\right> \pm \sigma_x.$$

In this case the error bars are due to random errors. As such, the measurement could actually be anywhere, so we can only say that there is a 68% chance that the true length falls within $\pm 1\sigma_x$ of $\left< x \right>$.