I'm a little rusty on my probability and would appreciate any help. I think I have done the bulk of the work already anyway, but my question is:
If $X \sim LN(1,2)$ find $P(X>1)$
$X$ being LogNormal implies: $X = e^Y$, where $Y \sim N(1,2)$
So $P(X>1) \implies P(e^Y>1) \implies P(Y> ln(1)) \implies P(Y > 0)$
From here, we just need to consult our standard normal table?
But the issue is that $Y$ is not standard normal, so we need to convert it.
$Y = \frac{Z-\mu}{\sigma} \implies Y = \frac{Z-1}{\sqrt{2}}$
So we need to solve : $P(\frac{Z-1}{\sqrt{2}} > 0) \implies P(Z>1)$
Is this correct so far?
If so, now I get my standard normal table out, which finds values to the right of the $Z$ score. (I can also to the left, just with some basic algebraic manipualtion).
Consulting this: http://www.stat.ncsu.edu/people/osborne/courses/xiamen/normal-table.pdf
I scroll down to $P(Z>z) \space \text{where z} = 1.00 = .1587$.
However the answer in my text says $P(Y > 0) = 0.76$
Where is my error?
Any help appreciated.
You are normalising wrong. You should instead write
$$Y = \mu + \sigma Z$$
The rest looks fine