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

$\Large P(x)$ = $\LARGE \frac{1}{{\sigma \sqrt {2\pi } }}e^{{{ - \left( {x - \mu } \right)^2 } \mathord{\left/  {\vphantom {{ - \left( {x - \mu } \right)^2 } {2\sigma ^2 }}} \right. } {2\sigma ^2 }}}$

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

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

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<img src="/static/files/MBI/Module%207/stdnormaldist.jpg" width=500
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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, $\large np$ is the mean and $\large \sqrt{np(1-p)}$ is the standard deviation of a binomial distribution. Therefore Z can be written as:
Z = $\Huge \frac{X=np}{\sqrt {np(1-p)}}$

And the poison distribution:
Z = $\Huge \frac{X-\lambda}{\sqrt \lambda}$

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]]
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mbi_public
created
Tue, 08 Feb 2011 20:04:10 GMT
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dirkjan
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Tue, 08 Feb 2011 20:04:10 GMT
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dirkjan
tags
Decision
M7
Methods
Quantitative
Term
creator
dirkjan