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
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):
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:
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:
With our known parameters:
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.
Then the omega should be around 1.605681 (negative or positive).
Getting back to mean:
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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