Let $X$ be such that $S=e^x$. You are given that $M_X(u)=e^{5u+2u^2}$ Suppose, for the stock market, the price of a certain stock S has density function $f_S(s)=\frac{1}{ts\sqrt{2\pi}} e^{\frac{-1}{2}\left(\frac{\ln(s)-m}{i}\right)^2}$ where $S>0$  and $-\infty<m<\infty $ and $t>0$ are constants.
Let $X$ be such that $S=e^x$. You are given that $M_X(u)=e^{5u+2u^2}$.
Given that $S$ is greater than $50$, what is the probability that is between $70$ and $90$?
I know if $M_X(u)=e^{5u+2u^2}$ then  $X$ is a normal distribution with parameters $\mu= 5 $ and $\sigma^2= 4 $ now I don´t kwon that I should do
Thanks for yours help have a nice day :)

We are told that $X\gt \ln(50)$, and want to find the probability that $\ln(70)\lt X\lt \ln(90)$. 
Let $A$ be the event $\ln(70)\lt X\lt \ln(90)$, and $B$ the event $X\gt \ln(50)$. We want $\Pr(A|B)$.
This is $\dfrac{\Pr(A\cap B)}{\Pr(B)}$.
Note that in our case we have $A\cap B=A$. So we need to find $\Pr(A)$ and $\Pr(B)$.
These are standard normal distribution calculations.
