What is the normal approximation of binomial distribution? I can't find a specific formula for this problem where I have to use the normal approximation of the binomial distribution.
Suppose we have a Bernoulli experiment with $p=0.8$ and we want to know how many trials we need for $\mathbb{P}(X\ge 30)$ to get the probability $0.975$
So what we have is
$n$ = unknown
$p$ = 0.8
$q$ = 0.2
$\mathbb{P}(X\ge 30)$
$0.975$
I have to use the normal approximation of the binomial distribution to solve this problem but I can't find any formula for this.
The only thing I found is $$Z=\frac{Y-np}{\sqrt{np(1-p)}}\rightarrow N(0,1)$$
Will be this the approximation formula?
 A: For large $n$ and/or $p$ close to $1/2$, the $\text{Bin}(n,p)$ distribution looks roughly like a normal distribution with the same mean and standard deviation. Since the mean of a $\text{Bin}(n,p)$ random variable is $np$ and the standard deviation is $\sqrt{np(1-p)}$, this means we can approximate a $\text{Bin}(n,p)$ random variable by a $N(np,np(1-p))$ random variable.
To actually do that approximation, we have to be a little careful because binomial random variables take on whole number quantities, but normal random variables take on real values. If $X \sim \text{Bin}(n,p)$ and $Y\sim N(np, np(1-p))$, then for any whole number $k$,
$$P(X=k) \approx P\left(k-\frac{1}{2} \leq Y \leq k+\frac{1}{2}\right).$$
That is, to account for the fact that $Y$ does not always take on whole number values, we consider every value of $Y$ in the interval $\left[k-\frac{1}{2},k+\frac{1}{2}\right]$ to be roughly $k$. Similarly, if we have integers $k<\ell$, then 
\begin{align*}
P(k \leq X \leq \ell) &\approx P\left(k-\frac{1}{2} \leq Y \leq \ell+\frac{1}{2}\right)\\
P(X \leq \ell) &\approx P\left(Y \leq \ell+\frac{1}{2}\right)\\
P(k \leq X) &\approx P\left(k-\frac{1}{2} \leq Y\right).
\end{align*}
All of the probabilities on the right-hand side can be found by computing the $z$-scores (with $\mu=np$ and $\sigma=\sqrt{np(1-p)}$) and using a $z$-chart. 
For your problem, $n$ is unknown, so we have to work backward. We want
\begin{align*}
0.975 &= P(X \geq 30) \\
&\approx P(Y \geq 29.5) \\
&= P\left( Z \geq \frac{29.5-np}{\sqrt{np(1-p)}}\right) \\
&= 1-P\left( Z \leq \frac{29.5-np}{\sqrt{np(1-p)}}\right) 
\end{align*}
where $Y \sim N(np,np(1-p))$ and $Z\sim N(0,1)$. Now you can use a $z$-chart to find the value of $\frac{29.5-np}{\sqrt{np(1-p)}}$, plug in the value of $p$, and solve for $n$. If $n$ turns out not to be in integer, then you need to figure out whether it should be rounded up or down. (Hint: you want $P(X \geq 30)$ to be at least $0.975$. Does this probability get bigger or smaller as $n$ increases?)
