At the Science Museum in Boston, there is a really neat exhibit on the normal distribution.
The exhibit has several hundred balls fall through an array of pegs à la plinko into a sequence of bins of equal size. The resulting shape is a normal distribution with mean 0 and some variance.
This is like a random walk. Each ball is a an independent random walk (or at least we can assume so for simplicity) and since the balls wind up in discrete bins, it is a random walk on the integers.
Why should the resulting distribution for final distances of random walks of length $n$ be normal?
Is it because of the CLT? A hint is all I would like.
Well, let's use the conventional definition of a random walk:
Let $X_1,X_2,\dotsc$ be i.i.d. Rademacher random variables, in other words independent and so that $\Pr(X_i = 1) = \Pr(X_i = -1) = \frac{1}{2}$.
This corresponds to the idea that at the $n$-th step of the random walk, we randomly either go left (if $X_n = -1$) or right. In any case, the position of the random walk after the $n$-th step is just:
$$ S_n = \sum_{i=1}^n X_i$$
Of course this type of sum is of the simplest form needed to apply the CLT, i.e.:
$$ \frac{1}{\sqrt{n}}S_n \overset{D}{\to} \mathcal{N}(0,\sigma^2) $$
Here $\sigma^2 = \frac{1}{2} + \frac{1}{2} = 1$ (Variance of the $X_i$, also note the $X_i$ have expectation $0$).