Normal Random Variable - uniform distribution So here's the question I'm trying to solve:

A stock price movement model supposes that if the current stock price
  is s, then, after one period, the stock price will be $us$ with
  probability $p$ and $ds$ with probability $1 - p$. Assuming that successive
  movements of the stock are independent, approximate the probability
  that the stock’s price will be up at least $50\%$ after the next $1000$
  periods if $u = 1.1$, $d = 0.95$, and $p = 0.49$. What is the minimal value
  of $p$ to guarantee that the stock price will be over $60\%$ up after $1000$
  periods?

And I don't even know where to start. Any suggestions as to how I should approach this problem?
thank you.
 A: You're multiplying 1000 times by something that's equal to either $u$ or $d$ each time.  That means you're adding something equal to $\log u$ or $\log d$ 1000 times.  The central limit theorem applies to independent random variables that are added; that's why we're taking logarithms.
We have
$$
\mathbb{E}\left.\begin{cases}
\log 1.1 & \text{with probability } 0.49 \\
\log 0.95 & \text{with probability } 0.51
\end{cases}\right\} = 0.49\log 1.1 + 0.51\log 0.95,
$$
and
$$
\operatorname{var}\left.\begin{cases}
\log 1.1 & \text{with probability } 0.49 \\
\log 0.95 & \text{with probability } 0.51
\end{cases}\right\} = (0.49)(0.51)(\log 1.1 - \log 0.95)^2.
$$
Hence
$$
\frac{\left.\begin{cases}
\log 1.1 & \text{with probability } 0.49 \\
\log 0.95 & \text{with probability } 0.51
\end{cases}\right\} - (0.49 \log1.1+0.51\log 0.95)}{\sqrt{(0.49)(.051)(\log1.1-\log0.95)}}
$$
has expected value $0$ and standard deviation $1$.
Adding up $1000$ independent copies of this, we get
$$
\frac{\text{sum}-1000(0.49\log1.1+0.51\log0.95)}{\sqrt{1000(0.49(0.51)(\log1.1-\log0.95)}}
$$
has an approximately standard normal distribution.  The problem is: what is the probability that this is more than
$$
\frac{\log 1.5 - 1000(0.49\log1.1+0.51\log0.95)}{\sqrt{1000(0.49(0.51)(\log1.1-\log0.95)}}.
$$
(I.e. going up at least $50\%$ is the same as multiplying by $1.5$ or more.)
So that last item is what you put into the table of values of the c.d.f. of the standard normal distribution.
