I have the following problem:
On average, a pearldiver finds a pearl in every 50th shell.
How many shells does he at least need to find to get at least one pearl with a probability of at least 95%?
A hint suggests using the central limit theorem.
Now, I tried using the CLT:
Let $S_n=X_1+X_2+…+X_n$ ,$\quad X_i$ i.i.d. random variables,
$\mathbb{E}(S_n)=n\mu$,$\quad$ Var(X)=$n\sigma^2$,$\quad$ where $\mathbb{E}(X_i)=\mu$ and Var($X_i)=\sigma^2$.
Let $Z_n= \frac{S_n-n\mu}{\sigma\sqrt{n}}$ .
The CLT states that:
$\lim_{n\rightarrow \infty} P(Z_n\leq z)=\phi(z)$$\qquad$$\phi(z):=$CDF of $\mathcal{N}(0,1) \quad \forall z\in \mathbb{R}$
I saw in the standard normal table that $\phi_{0;1}(1.65)=0.95953$ is the smallest number whose probability is $\geq0.95$.
But where do I go frome here?
Since I don't know $\sigma^2$ and only $\mu$, I really can't do anything with the CLT.
I would really appreciate some advice.
Best Answer
Note that you haven't used the first sentence of the question statement yet. In this particular problem, I think you are supposed to assume that $X_1,\ldots, X_n$ are i.i.d. $\text{Bernoulli}(1/50)$. From there you can compute the mean and variance and follow the rest of your work.