- μ : 0 (mean of distribution)
- PDF : $$
\frac{1}{\sqrt{ 2\pi }}e^{\frac{-x^2}{2}}
- POI at $x=\pm{1}$ ($\mu\pm\sigma$)
- $\sigma^2=1$ (variance)
- we use **Z** to represent the standard normal variable
- $P(Z<z) = \int_{-\infty}^{z} f(x)\, dx$
- cannot be integrated, so we use tables
- ![[Pasted image 20240708114555.png|450]]
- The standard Normal Distribution can be transformed to give other normal distributions with different $\mu$ & $\sigma$.
- e.g. IQ scores ($\mu=100$, $\sigma=15$)
- $\frac{1}{\sqrt{ 2\pi }}e^{\frac{-x^2}{2}}$ (standard PDF)
- -> dilate horizontally by 15
- -> shift horizontally by 100
- -> dilate vertically by 1/15 (to keep area under curve = 1)
-
$$=\frac{1}{15}f\left( \frac{1}{15}(x-100) \right)=\frac{1}{15\sqrt{ 2\pi }}e^{\frac{-(x-100)^2}{2*15^2}}$$
- So, the PDF for a normal distribution with mean $\mu$ and standard deviation $\sigma$ is:
-
$$f(x)=\frac{1}{\sigma \sqrt{ 2\pi }}e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}$$
- Again, cannot integrate so we use tables
- you will only be given the table for the standard normal distribution
- the method is to transform x value back into the standard normal distribution, and use the table
- (a.k.a. 'finding the z score')
- the Z value for any $\mu$ and $\sigma$ is $Z=\frac{x-\mu}{\sigma}$
- The Z score is how many standard deviations your value is away from the mean
- 2 key skills to answer questions:
- using symmetry of the bell curve
- using the compliment (i.e. $1-P(Z<z)$)
-