Coin throw approximation We have coin and we want to determine the probability $p$, that this coin shows "head" after one throw. For this purpose the coin has been thrown $n$-times and $K_i$ is the event that in the $i$-th throw "head" has been thrown.
a) Why is 
$\hat{p}=\frac{1}{n}\sum_{i=1}^{n} 1_{K_i}$ 
($1_{K_i}$ is the indicator function) an approximation value for p? Justify this with the strong Law of large numbers.
b) Assuming the real value of p is 0.3 and the coin is thrown 100 times. Calculate 
$\mathbb{P}(0.25\leq \hat{p} \leq 0.35)$
c) How many times must the coin be thrown, so that
$\mathbb{P}(0.25\leq \hat{p} \leq 0.35)\geq 0.95$ for $p=0.3$
d) How many times must the coin be thrown, so that
$\mathbb{P}(p-0.05\leq \hat{p} \leq p+0.05)\geq 0.95$ for all $p\in[0,1]$
Hint: Use the function values for the Quantile function $\Phi (1.09)=0.8621,\Phi ^{-1}(0.975)=1.96 $
So with a) i have no idea how to justify this, with b) i used the central limit theorem but i do not get any reasonable values to use with $\Phi (x)$. I assumed that this was a binomial distribution with mean $\mu =E(X_i)=n\times p=100\times 0.3=30$ and $\sigma ^2=Var(X_i)=n\times p(1-p)=21$. With these values i get 
$\mathbb{P}(\frac{-2975}{\sqrt{21\times 100}} \leq \frac{X_1+...+X_{100}-30\times 100}{\sqrt{21\times 100}} \leq \frac{-2965}{\sqrt{21\times 100}})$
which does not bring me any further. c) and d) are then similar to b).
 A: We do b) in some detail, in the hope that will serve as a guide for c) and d).
We can work directly with $\hat{p}$, or with the sum of the $X_i$. I prefer to work with $\hat{p}$. This has mean $0.3$ and variance $\frac{p(1-p)}{100}$, so standard deviation $\sigma=\frac{\sqrt{p(1-p)}}{10}$. (This is not the same as the $\sigma$ in your post, which is the standard deviation of $X_1+\cdots+X_{100}$.)
We have
$$\Pr(0.25\le \hat{p}\le 0.35)=\Pr\left(-\frac{0.05}{\sigma}\le  \frac{\hat{p}-0.3}{\sigma}\le \frac{0.05}{\sigma}\right).$$
By the normal approximation, $\frac{\hat{p}-0.3}{\sigma}$ is roughly standard normal, so our required probability is approximately
$$\Pr\left(-\frac{0.05}{\sigma}\le Z\le \frac{0.05}{\sigma}\right),$$
where $Z$ is standard normal. Now the rest can be computed using values of $\Phi(z)$ for appropriate $z$.
Another way: We want $\Pr(25\le X_1+\cdots +X_{100}\le 35)$. The random variable $W=X_1+\cdots +X_{100}$ has mean $30$ and variance $100p(1-p)$, so standard deviation $\sigma'=10\sqrt{p(1-p)}$. Now we use the same style of calculation as in the first solution. 
You used this approach, which is fine. However, the top of your middle term should have been $X_1+\cdots+X_{100}-30$. The tops of the two sides should have been $25-30$ and $35-30$. The denominators $10\sqrt{p(1-p)}$ that you used are correct.
