Arbitrary Normal Distributions

Standardized Normal Distribution

If $X \sim N (μ, σ)$ via Gaussian (Normal) Distribution, then the "standardized" rv $Z$ defined by:

Z = \frac{X - μ}{σ}

has a standard normal distribution. Thus:

P (a \leq X \leq b) = P (\frac{a - μ}{σ} \leq Z \leq \frac{b - μ}{σ}) = Φ (\frac{b - μ}{σ}) - Φ (\frac{a - μ}{σ})

where:

P (X \leq a) = Φ (\frac{a - μ}{σ}), P (X \geq b) = 1 - P (X \leq b)

Conversely, if $Z \sim N (0, 1)$ and $μ, σ$ are contants ( $σ > 0$ ) then the "un-standardized" rv $X = μ + σ Z$ has a normal distribution with mean $μ$ and standard deviation $σ$ .

Proof
The CDF of $Z$ is:

\begin{aligned} F_{Z} (z) & = P (Z \leq z) \\ = P (\frac{X - μ}{σ} \leq z) \\ = P (X \leq μ + z σ) \\ = \int_{- \infty}^{μ + z σ} f (x; μ, σ) d x \\ = \int_{- \infty}^{μ + z σ} \frac{1}{σ \sqrt{2 π}} \exp (\frac{- 1}{2} \cdot {(\frac{x - μ}{σ})}^{2}) d x \end{aligned}

Differentiate both sides:

f_{Z} (z) = σ \cdot \frac{1}{σ \sqrt{2 π}} \exp (\frac{- 1}{2} \cdot {(\frac{μ + z σ - μ}{σ})}^{2}) = \frac{1}{\sqrt{2 π}} \exp (\frac{- 1}{2} \cdot z^{2})

This is just $f (x; 0, 1)$ so then:

F_{Z} (z) = \int_{- \infty}^{x} f (x; 0, 1) d x = Φ (x)

as desired.
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Percentiles for Normal Distributions

$η_{p} = μ + Φ^{- 1} (p) \cdot σ$

for gaussian distributions.

Empirical

If the population distribution of a variable is (at least approximately) normal, then:

Roughly 68% of the values are within one SD of the mean.
Roughly 95% of the values are within two SDs of the mean.
Roughly 99.7% of the values are within three SDs of the mean.

Proof

\begin{aligned} P (in n standard devs) & = P (μ - n σ \leq X \leq μ + n σ) \\ = P (\frac{μ - n σ - μ}{σ} \leq Z \leq \frac{μ + n σ - μ}{σ}) \\ = P (- n \leq Z \leq n) \\ = Φ (n) - Φ (- n) \end{aligned}

Using $n = 1, 2, 3$ gives the values above.
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