Workaround for the expinv function in statistics toolbox

2022 6 21
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更新时间：2022 6 21
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Hi,
I have been working with the norminv function for a while using my own code as follow:
function z = norminv(p,sigma,mu)
z = sigma.*(-sqrt(2)*erfcinv(2*p))+mu;
end
However, I would like now to use the expinv function but I don't have access to the statistics toolbox. I was wondering if there is a way to work around using that particular function ?
Thanks !

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Torsten
Torsten 2022-6-21
编辑：Torsten 2022-6-21
Ran in:

0 个投票

Ran in:
p = 0.3;
mu = 1/4;
x = -mu.*log(1-p)
x = 0.0892
x2 = expinv(p,mu)
x2 = 0.0892

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