Let $(X_n)$ be a simple random walk that starts from $X_0 = 0$ and on each step goes up one with probability $p$ and down one with probability $q = 1 − p$.
Consider the case $p = 0.6$, so $q = 0.4$.
Using a normal approximation, estimate $P(16 ≤ X_{100} ≤ 26)$. You should use an appropriate ‘continuity correction’, and explain why you chose it. (Bear in mind the possible values $X_{100}$ can take.)
So, I have already calculated that $X_{100}$ can be distributed like so:
$X_{100}$ ~ $N(20, 96)$ since $E[X] = 20, Var(x) = 96$.
I also know that since the amount of steps is even, the value the random variable can take must also be even. So now, how can I calculate the probability given this information? Help would be appreciated.
I cannot be certain what your book would suggest for a 'continuity correction', but I would suspect it would be $$P(X_{100} \le 27.5) - P(X_{100} \le 15.5) \approx 0.454986$$
This is close to the actual probability:
$$\sum_{n=8}^{13}\dbinom{100}{50+n}(0.6)^{50+n}(0.4)^{50-n} \approx 0.458129$$
The upper bound is chosen because you want $$P(X_{100} \le 26) = P(X_{100} < 28) \approx P(X_{100} \le 27.5)$$ and the lower bound: $$P(X_{100} \ge 16) = 1-P(X_{100} < 16) \approx 1-P(X_{100} \le 15.5) = P(X_{100} > 15.5)$$