I have a problem that says
The random variable $Y$ has lognormal distribution with $u = 2$ and $o = 0.4$. $z = \frac xy$. (recall log properties)
Find $P(Z\leq 6)$.
The solution begins saying $\ln Z = \ln X-\ln Y$ so $Z$ is lognormal with $u = 3-2 = 1$ and $o = \sqrt{(0.5)^2 + (0.4)^2} = \sqrt{0.41}$.
$P(Z\leq 6) = P\left(Z \leq \frac{\ln6 - 1}{\sqrt{0.41}}\right) = F(1.24) = 0.8925$
This is confusing to me, I don't know how they got $u = 3-2 = 1$ and $o = \sqrt{(0.5)^2 + (0.4)^2}$. I have been analyzing the book and my notes in this section, I tried applying the main formula for $Y\sim \log(u,o^2)$, and I thought I'd use the given $2$ and $0.4$ for $u$ and $o$ in final $P(Z\leq 6)$ equation.
I also thought it might be the $X-Y\sim N(ux-uy,o^2x + o^y)$ , but then I have no idea why hes using $ux = 3$ and $o^x = 0.5$.
Can someone please explain the process of how they found the $u = 1$ and $o = \sqrt{0.41}$ steps?