I have a Gaussian probability density function
$$\frac{1}{4\sqrt{2 \pi}} exp[-\frac{(x-5)^2}{32}]$$
clearly $\mu = 5$ and $\sigma=4$
I would like to find $p(x\le n) = 0.0158$. I see in the standard z-score table that $p(x\le -2.149) \approx 0.0158$. However this is for a Gaussian with $\mu = 0$ and $\sigma = 1$.
What would I do to this result in order to find the answer to the non-standard Gaussian?
Thank you for your time
Edit: Fixed z-score