Normal Distribution
The MOST powerful distribution. Normal distribution is a probability distribution that is symmetric about the mean $μ$, where data near the mean occur more frequently.
Normal Distribution (Definition)
We say that a r.v. $X$ is normally distributed, $X∼N(μ,σ_{2})$, if its p.d.f., $f$, is
$f(x)=σ2π 1 e_{−21(σx−μ)_{2}}$ where
 $μ$ is the mean or expectation of the distribution (and also its median and mode),
 $σ$ is its Standard Deviation
Normal Distribution is symmetrical around $μ$, i.e. $f(x+μ)=f(x−μ)$ where $μ$ is the mean, mode, and median.
Expectation and Variance
 $E(x)=∫_{−∞}xf(x)dx=μ$ (the calculation for this is pretty messy)
 $Var(X)=σ_{2}$
Standard Normal Distribution (Definition)
The standard Normal Distribution has $μ=0$ and $Var(X)=1$, i.e. $Z∼N(0,1)$. We have the p.d.f. $ϕ$ $ϕ(z)=σ2π 1 e_{−2z}$
I learned the matrix form from the SLAM book.
Multidimensional Normal Distribution (MatrixForm Definition)
The probability density function $f$ is
$f(x)=(2π)_{N}det(Σ) 1 exp(−21 (x−μ)_{T}Σ_{−1}(x−μ))$
Also see Gaussian Variable.
Where do we see gaussians in the real world?
 Height Distribution: Human heights follow a Gaussian distribution.
 IQ Scores: IQ scores typically follow a Gaussian distribution.
 Measurement Errors: Instrument measurement errors often have a Gaussian distribution.
 Stock Returns: Daily stock returns frequently exhibit a Gaussian distribution.
Ztable (Standard Normal ztable)
The ztable table gives values of the CDF $F(x)=P(X≤x)$ for $X∼N(0,1)$ and $x≥0$.
What is really cool is that we can go from the standard normal distribution to another normal distribution with a particular mean and variance, using the theorem below:
Theorem
If $X∼N(μ,σ_{2})$ and $Z=σX−μ $, then $Z∼N(0,1)$
Look at the page on ZScore
Example Exercise
Example: Suppose that final grades in this course follow a normal
distribution with $μ=75$ and $σ_{2}=64$.

Find the probability that a student has a final grade of 83 or more.

Calculate the 95th percentile.

We have that $X∼N(75,64)$, so find $P(X≥83)$ $Z=σx−μ =883−75 ≤1.6449$ $P(Z≤1.6449)=0.94950$

We solve this by using the $N(0,1)$ Quantile table. $P(Z≤z)=0.95⟹P(Z≤1.6449)=0.95$ $Z=σx−μ =8x−75 ≤1.6449$ Therefore, we conclude that $X≤75+8⋅1.6449$
Part 2: A sample of 9 students is taken. Find
 $P(S≥700)$
 $P(X<73)$
I’m not sure if I understand correctly, but there is this Variance of the mean value that I need to use, instead of the direct variance.
$S∼N(75⋅9,9⋅64)$ $P(S≥700)=P(Z_{s}≤1.04)=1−0.85083$
 $X∼N(75,64/9)$ $P(Z<4−3 )=1−0.77337$ (I don’t know if $Z<4−3 $ is correct, might be a typo)
The 689599 rule
The 689599 rule is based on the mean and standard deviation. It says:
 68% of the population is within 1 standard deviation of the mean.
 95% of the population is within 2 standard deviation of the mean.
 99.7% of the population is within 3 standard deviation of the mean.
Generating Normal Distributions in Code
To randomly sample from the Normal Distribution, we have a few options (Learned from Ericsson):