Given $\mu=19$ and $P(Y<22.5)=0.63683$, find $\sigma$.

I did the following:

$$Z=\frac{Y-\mu}{\sigma}$$

$$\sigma=\frac{Y-\mu}{Z}$$

Then I went to the normal table to find that $P(Z<0.35)=0.63683$

$$\sigma=\frac{Y-\mu}{Z}=\frac{22.5-19}{0.35}=10$$

What do you think?