Just look after how the mean and variance of a skew normal distribution can be computed and you got the answer! Knowing that the mean looks like:
and ![alt text](https://i.stack.imgur.com/9pyKn.png)
You can see, that with a xi=0 (location), omega=1 (scale) and alpha=0 (shape) you really get a standard normal distribution (with mean=0, standard deviation=1):
![alt text](https://i.stack.imgur.com/1vra9.png)
If you only change the alpha (shape) to 5, you can except the mean will differ a lot, and will be positive. If you want to hold the mean around zero with a higher alpha (shape), you will have to decrease other parameters, e.g.: the omega (scale). The most obvious solution could be to set it to zero instead of 1. See:
![alt text](https://i.stack.imgur.com/hXPut.png)
Mean is set, we have to get a variance equal to zero with a omega set to zero and shape set to 5. The formula is known:
![alt text](https://i.stack.imgur.com/8UUsS.png)
With our known parameters:
![alt text](https://i.stack.imgur.com/n2DxD.png)
Which is insane :) That cannot be done this way. You may also go back and alter the value of xi instead of omega to get a mean equal to zero. But that way you might first compute the only possible value of omega with the formula of variance given.
![alt text](https://i.stack.imgur.com/9D9B6.png)
Then the omega should be around 1.605681 (negative or positive).
Getting back to mean:
![alt text](https://i.stack.imgur.com/1NKGX.png)
So, with the following parameters you should get a distribution you was intended to:
location = 1.256269 (negative or positive), scale = 1.605681 (negative or positive) and shape = 5.
Please, someone test it, as I might miscalculated somewhere with the given example.
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