Problem: On average, every 50th shell has a pearl. What is the minimum amount of shells you have to open to get at least one pearl with probability greater or equal to 0.95. Calculate using the exact approach, poisson approximation and normal approximation of the binomial distribution.
My approach:
Let $X_n\sim \mathrm{Ber}(n,p)$ where $p=1/50$. We are looking for the smallest $n$ such that $0.95\le P(X_n\ge1)=1-P(X_n=0)$. Exact calculation gives me $n\approx 148.28$ thus $n\ge149$, the poisson approximation gives me $n\approx 150$. I am stuck at the normal approximation. I know that the centralised binomial distribution approaches a standard normal distribution for large $n$. Thus we are looking at:
$0.95\le 1-P(X_n=0)=1-P(-0.5\le X_n\le0.5)=1-P(\frac{-0.5-np}{\sqrt{np(1-p)}}\le S_n \le\frac{0.5-np}{\sqrt{np(1-p)}})$ which can be rewritten as $0.05\ge P(\frac{-0.5-np}{\sqrt{np(1-p)}}\le S_n \le\frac{0.5-np}{\sqrt{np(1-p)}})$ where $S_n=\dfrac{X_n-np}{\sqrt{np(1-p)}}$. How do I solve this?
Edit: I know that $0.05\ge P(\frac{-0.5-np}{\sqrt{np(1-p)}}\le S_n \le\frac{0.5-np}{\sqrt{np(1-p)}})\approx\Phi(\frac{0.5-np}{\sqrt{np(1-p)}})-\Phi(\frac{-0.5-np}{\sqrt{np(1-p)}})$ where $\Phi$ is the CDF of the standard normal distribution. I would be able to solve it if I only had "one of the $\Phi$'s", but don't know how to approach it in this case.
Hint: I think you're making this more complicated than necessary. What you really want is $$ 0.95 \leq P(X \geq 1) $$ Now, granted, $P(X \geq 1) = 1 - P(X = 0)$, since "exactly zero" and "at least one" are complementary events in the discrete distribution.
However, the following equation is also true: $$ P(X \geq 1) = 1 - P(X < 1) $$
Using this formulation, you can approximate using just a single instance of the cdf for the normal distribution, which I think is what you asked for.
If we take a step back and examine what you were trying to do: at a certain point you were trying to use $P(-0.5 \leq X \leq 0.5)$ to measure the probability of "failure". If you think about it, basically this would mean you were treating anything higher than $0.5$ pearls gathered as a success, but you would also be treating anything below $-0.5$ as a success, which obviously doesn't make sense: even if gathering $-0.5$ pearls made sense as a physical measurement (which we have to accept in order to use the normal approximation), you still wouldn't want a negative number of pearls to count as a success.