Heads/Tails 900 times, find k such that $P(k \leq T \leq 480) = 0.1$ with T the number of Tails.
I will use the CLT because I know that $$P(z_{1} \leq \frac{S_{N} - \mu N}{\sqrt{N} \sigma} \leq z_{2})$$
We have $S_{900}$ a sum of Bernoulli($\frac{1}{2}$) so $\mu = \frac{1}{2}$ and $\sigma = \frac{1}{\sqrt{2}}$.
But here I just don't know what to do.. Do you have any hints ?
Here is a start working a similar problem:
$T \sim \mathsf{Binom}(n = 900, p = 1/2),$ so start by showing that $E(T) = 450$ and $SD(T) = 15.$ Then $$0.1 = P(k \le T \le 450) \approx P\left(\frac{k - 450}{15} \le \frac{T - E(T)}{SD(T)} \le \frac{450 - 450}{15} \right) = P\left(\frac{k-450}{15} \le Z \le 0\right),$$ where $Z$ is standard normal.
Using standard normal tables (backwards), you should be able to deduce the value of $(k-450)/15$ and solve for $k.$ It may help to draw a sketch.
Of course, it is too much to expect the the resulting $k$ will be exactly an integer, so 'round' appropriately.
The normal approximation should work well enough to get an answer for $k$ that gives a probability very near to the desired $0.1.$
You will be using standard normal and $\mathsf{Norm}(450, 15)$. Below are sketches of both. The area for my problem is between the vertical red lines (on both sketches). You should make one type of sketch for your problem.