448
Comment:

977

Deletions are marked like this.  Additions are marked like this. 
Line 1:  Line 1: 
= How do I produce random variables which follow a negatively skewed distribution? = Most distributions such as the exponential and logNormal distributions are positively skewed with the mode of the distribution occurring for lower values. 

Line 2:  Line 5: 
=  The Weibull distribution is negatively skewed and may be generated [http://www.taygeta.com/random/weibull.xml using random variables which are uniform on the interval (0,1).] 
Line 4:  Line 7: 
Most distributions such as the exponential and logNormal distributions are positive skewed with the model of the distribution for lower values.  The below produces an open ended negatively skewed weibull distribution with parameters, 2 and 20. It has a median of 
Line 7:  Line 10: 
[http://www.uib.no/people/ngbnk/kurs/notes/node31.html The Gamma distribution] which has two parameters, $$\alpha$$ and $$\beta$$ may produce negative skew where the model occurs for higher values (values > 0) when $$/alpha$$ is a lot greater than $$\beta$$. It also has no maximum value.  $$2^text{0.05}ln(2)^text{0.05}$$ = 0.95. which is of form $$A^text{1/B}ln(2)^text{1/B}$$ where A and B are the two parameters of the Weibull distribution. (See [http://www.weibull.com/AccelTestWeb/weibull_distribution.htm here for formulae).] {{{ compute alpha=2. compute beta=20. compute rvw=((1/alpha)*(ln(1rv.uniform(0,1))))**(1/beta). compute med=((alpha)**(1/beta))*(ln(2)**(1/beta)). exe. }}} 
How do I produce random variables which follow a negatively skewed distribution?
Most distributions such as the exponential and logNormal distributions are positively skewed with the mode of the distribution occurring for lower values.
The Weibull distribution is negatively skewed and may be generated [http://www.taygeta.com/random/weibull.xml using random variables which are uniform on the interval (0,1).]
The below produces an open ended negatively skewed weibull distribution with parameters, 2 and 20. It has a median of
$$2^{text{0.05}ln(2)}text{0.05}$$ = 0.95.
which is of form
$$A^{text{1/B}ln(2)}text{1/B}$$
where A and B are the two parameters of the Weibull distribution. (See [http://www.weibull.com/AccelTestWeb/weibull_distribution.htm here for formulae).]
compute alpha=2. compute beta=20. compute rvw=((1/alpha)*(ln(1rv.uniform(0,1))))**(1/beta). compute med=((alpha)**(1/beta))*(ln(2)**(1/beta)). exe.