Question

What is standard deviation

Answer 1

For some variable \(x\), where \(⟨x⟩\) denotes the mean (average).

The deviation of some particular \(x\) from the mean is \(\Delta x=x−⟨x⟩\).

For a collection of data (\(x\) values), the average deviation is always going to be zero. (Because data falls evenly about each side of the mean.)

To measure the deviation of values in a distribution, statisticians have invented something called the standard deviation \(\sigma\) (a lowercase greek s);. This is some non-negative number, the square root of a thing called the variance:

The variance is \(\sigma^2\). This is the average of the square of deviations; mathematically equal to difference between the average square value, and the average value squared:
\[\sigma^2=⟨(\Delta x)^2⟩=⟨x^2⟩−⟨x⟩^2\]

For a collection of data, the average standard deviation is going to be some positive number, unless all values in the distribution are the same (in which case is will be zero).

Hence the standard deviation is a useful measure of the spread of a distribution, and is usually calculated with:
\[\sigma=\sqrt{⟨x^2⟩−⟨x⟩^2}\]

Answer 2

Let's try an example:

Consider the data set {11,12,13,16,16,17,20}.

Answer 3

The number of values is\[N=7\]
The mean value is\[\langle x\rangle = \frac{11+12+13+16+16+17+20}{7}=15\]
The mean square is \[\langle x^2\rangle = \frac{11^2+12^2+13^2+16^2+16^2+17^2+20^2}{7}=233.6\]
The variance is\[\sigma^2=⟨x^2⟩−⟨x⟩^2=233.6-15.0^2=8.6\]
And the standard deviation is \[\sigma=\sqrt{8.6}=2.9\]

Answer 4

Most of the data is within one standard deviation of the mean:
\[\langle x\rangle \pm \sigma=15\pm2.9=[12.1,17.9]\]