# Standard Deviation

The standard deviation is defined as follows: $s=n−11 ∑_{i=1}(x_{i}−xˉ)_{2} $

- use $n−1$ for a
**sample**where you use notation $s$, and $n$ if you are calculating the standard deviation for a**population**where you use notation $σ$

The smaller the standard deviation, the more closely the measured values cluster around their mean.

Standard Deviation has a close relationship with Variance (always non-negative), where $σ=σ_{2} $, so $Standard deviation=Variance $, a better notation is $σ_{X}=Var(X) $

Why the error squared?

They talk about this in the KF book, but basically go through a few examples, and the square helps accurately depict the variance. You can take an absolute value, but it doesn’t fully paint the picture.

### Standard Deviation of the Mean

This also known as the standard error.

The standard deviation/error of the mean is given by $σ_{x}=n σ $ I saw this sort of idea at Ericsson while working on generating random normal number using the Ericsson while working on generating random normal number using the Central Limit Theorem. Indeed! This come from the Ericsson while working on generating random normal number using the Central Limit Theorem. Indeed! This come from the Central Limit Theorem. Indeed! This come from the Central Limit Theorem, see note for more information.

Intuitively, the variance from measuring the mean is smaller than between samples. As $n→∞$, you find that \sigma_\overline{x} \rightarrow 0. This makes sense! Serendipity with Serendipity with Curse of Dimensionality, this is why!

To motivate the use of this, consider the following example: The University has always collected the STAT 206 grade of each Software Engineering student. It is known that these grades follow a Normal Distribution with $μ=75$ and $σ=6$. In a class of 120 Software Engineering students, what is the probability that the class average will be 76 or more?

We use our Z-Table. However, here, we are not asking for the probability that a student will have a grade of 76.

- That value would be computed by computing $P(X>76)=P(Z>676−75 )=…$

Instead, we are asking for the probability of the mean, in which case we need to use the Standard Deviation of the mean! We have the following computation: $P(X≥76)=P(Z≥n σ X−μ )=P(Z≥120 6 76−75 )=P(Z≥1.83)=1−0.96638$