I have this exercise to do:
Consider a stock with an initial price of $40, an expected return of 16% per annum and a volatility of 20% per annum. Find the probability distribution of the future stock price in 6 months' time. Determine the 95% confidence of interval.
$S_0=40$
$\mu=0.16$
$\sigma=0.20$
$T=0.5$
Using the lognormal property of stock prices $(\ln{S_T}\approx\varphi\left[\ln{S_0}+\left(\mu-\frac{\sigma^2}{2}\right)T,\ \sigma^2T\right]$, I have $\ln{S_T}\approx\varphi\left[3.759,\ 0.02\right]$.
The results of the textbook says "there is a 95% probability that a normally distributed variable has a value within 1.96 standard deviaton of its mean."
Viewing another topic from this forum, I found this possible solution: $P[lnS_T>ln95]=1−P[lnS_T<ln95]=1−P[Z<z=\frac{ln95-3.759}{\sqrt{0.02}}]$ If I want to compute $z$, as result I will have $z=5.62$. Thus, $N(5.62)=1$ (is not possible for this exercise).
Supposing a 95% c.i. I can't solve the problem. Is anyone can explain me how can I approach it?
I would be beyond thankful for any help and advice.