Normal distribution

The normal distribution is the most widely used statistical distribution and is defined by its mean $\mu$ and standard variance $\sigma$

$P(x)$ = Failed to parse (syntax error): {\displaystyle \frac{1}} {{\sigma \sqrt {2\pi } }}e^{{{ - \left( {x - \mu } \right)^2 } \mathord{\left/ {\vphantom {{ - \left( {x - \mu } \right)^2 } {2\sigma ^2 }}} \right. } {2\sigma ^2 }}}

The standard normal distribution is scaled back to a mean of 0 and a standard deviation of 1:

$Z={\frac {X-\mu }{\sigma }}$

Useful properties of the normal distribution are:

• The total area under the standard normal distribution curve is 1.
• At Z=0 the total area under the curve is 0.5
• To measure the area between two Z-values you can subtract the z values (Look up in the Standard Normal Distribution Table) To find the probability of a value between 1 and 1.1 standard deviations above the mean you calculate:

{{{ Area above Z = 1 is 0.1587 Area above Z = 1.1 is 0.1357 }}} Therefore the area between Z=1 and Z=1.1 is 0.3413 - 0.3413 = 0.0230

• It is symmetric. The probability to be -Z away from the mean is equal to being away +Z.

The normal distribution can be used to approximate the binomial distribution, $np$ is the mean and ${\sqrt {np(1-p)}}$ is the standard deviation of a binomial distribution. Therefore Z can be written as: Z = Failed to parse (unknown function "\Huge"): {\displaystyle \Huge \frac{X=np}{\sqrt {np(1-p)}}}

And the poison distribution: Z = Failed to parse (unknown function "\Huge"): {\displaystyle \Huge \frac{X-\lambda}{\sqrt \lambda}}

Example of the latter. If the average # of broken eggs is 50. What is the probability there will be more than 70 broken eggs?

Mean = $\lambda$ = 50 Standard deviation= ${\sqrt {\lambda }}$ = 7.07107

The area requires is the area above X=70.5 so Z will be: Z = $(70.5-50)/7.07107=2.8988$ This makes the probability 1 - 0.5 - 0.4980 = 0.002

Part of: M7-Quantitative Decision Methods-S3 - Normal Distribution and confidence intervals