# How do you solve for the mean in a Normal Distribution?

I am working on a problem and am a little bit stuck on how to solve it.

The problem: Find a Normal Distribution with SD 2.5 and 5% Quantile at -15.2.

What I have done so far: $$X=\mu+2.5Z$$ $$.05=P(\mu+2.5Z\le-15.2)$$ How do we reduce this further? It looks like we have two unknowns.

I know that Q 95% = 1.645 based on the Z Table and that this would correspond to 5% being $$-1.645$$

Also, $$\sigma^2=6.25$$, so I know it needs to be something along the lines of $$X\sim\mathcal N(x, 6.25)$$. I am a little bit confused on the next step though.

Manipulate the expression inside $$P()$$: $$\mu+2.5Z\le-15.2$$ $$Z\le\frac{-15.2-\mu}{2.5}$$ But we know that the critical $$Z$$-value that will give the 5th percentile is $$-1.645$$. Thus $$-1.645=\frac{-15.2-\mu}{2.5}$$ $$\mu=-11.0875$$