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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.

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  • $\begingroup$ You can just use the geometric distribution to get an exact answer. I don't really see what CLT has to do with it. CLT would be helpful for estimating the probability distribution of the total number of pearls found after inspecting some fixed number of shells, where that number is much larger than 50. $\endgroup$ Commented Jan 8, 2021 at 16:28

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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.

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  • $\begingroup$ So $\mathbb{E}(X)=p=\frac{1}{50}$ and Var(X)=$p(1-p)=\frac{49}{2500} \Rightarrow \sigma=\frac{7}{50}$ and now I calculated he needs to find at least 134 shells, which seems to make sense. $\endgroup$ Commented Jan 8, 2021 at 17:29

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