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| − | The normal distribution is the most widely used statistical distribution and is defined by its mean <math> \mu </math> and standard variance <math> \sigma </math>
| + | Normal distribution |
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| − | <math>P(x)</math> = <math>\frac{1}</math>{{\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 }}}<math></math>
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| − | The ''standard normal distribution'' is scaled back to a mean of 0 and a standard deviation of 1:
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| − | <math>Z= \frac {X-\mu}{\sigma}</math>
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| − | <html>
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| − | <img src="/static/files/MBI/Module%207/stdnormaldist.jpg" width=500
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| − | </html>
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| − | Useful properties of the normal distribution are:
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| − | * The total area under the standard normal distribution curve is 1.
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| − | * At Z=0 the total area under the curve is 0.5
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| − | * To measure the area between two Z-values you can subtract the z values (Look up in the [[Standard Normal Distribution Table|Standard Normal Distribution Table]]) To find the probability of a value between 1 and 1.1 standard deviations above the mean you calculate:
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| − | {{{
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| − | Area above Z = 1 is 0.1587
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| − | Area above Z = 1.1 is 0.1357
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| − | }}}
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| − | Therefore the area between Z=1 and Z=1.1 is 0.3413 - 0.3413 = 0.0230
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| − | * It is symmetric. The probability to be -Z away from the mean is equal to being away +Z.
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| − | The normal distribution can be used to approximate the binomial distribution, <math>np</math> is the mean and <math>\sqrt{np(1-p)}</math> is the standard deviation of a binomial distribution. Therefore Z can be written as:
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| − | Z = <math>\Huge \frac{X=np}{\sqrt {np(1-p)}}</math>
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| − | And the poison distribution:
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| − | Z = <math>\Huge \frac{X-\lambda}{\sqrt \lambda}</math>
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| − | Example of the latter. If the average # of broken eggs is 50. What is the probability there will be more than 70 broken eggs?
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| − | Mean = <math>\lambda</math> = 50
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| − | Standard deviation= <math>\sqrt \lambda</math> = 7.07107
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| − | The area requires is the area above X=70.5 so Z will be:
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| − | Z = <math>(70.5 - 50)/7.07107 = 2.8988</math>
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| − | This makes the probability 1 - 0.5 - 0.4980 = 0.002
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| − | Part of: [[M7-Quantitative Decision Methods-S3 - Normal Distribution and confidence intervals|M7-Quantitative Decision Methods-S3 - Normal Distribution and confidence intervals]]
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| | [[Category:mbi]] | | [[Category:mbi]] |
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