During bottling 3% of the bottles break. Calculate the range in which the
amount of undamaged bottles lies with a probability of 95% and if 3000
bottles were filled.
I tried to solve that question with the binomial as well as the normal distribution.
Binomial Distribution
$\mathrm{qbinom}\left(\dfrac{1+95\%}{2}, n, (1-3\%)\right)-n\cdot(1-3\%)$ if $\mathrm{qbinom}$ is the inverse cumulative probability-function of my math software and has this signature $\mathrm{qbinom}(p, \text{size}, \text{prob})$
I get a $\Delta x$ of $18$ and therefore my range is: $(2892;2928)$
Normal Distribution
If I calculate it with the normal distribution, however, I get a completely different range:
$$\mu=3000\cdot(1-3\%)$$
$$\sigma=\sqrt{3000\cdot(3\%)\cdot(97\%)}$$
$$\Delta x= F^{-1}(\frac{1+95\%}{2})\cdot\frac{\sigma}{\sqrt{3000}} = 0,334\ldots$$
Any idea what I am doing wrong? I noticed that if I omit the last $\sqrt{3000}$ in the denominator I get $18,312\ldots$ but that might just be a coincidence.
It is not coincidence. The variance in the number of broken bottles (or surviving bottles) is $3000(0.03)(0.97)$. The standard deviation is the square root of that. Since the random variable you are studying is the number of surviving bottles, the appropriate mean is $2910$, and the appropriate standard deviation is $\sqrt{3000(0.03)(0.97)}$.
Remark: When you divided by $\sqrt{3000}$, you found the standard deviation of the proportion of broken (or unbroken) bottles. The proportion is not what you are working with when you want to find information about the number of bottles. You could work with proportion, but then you would need to multiply your proportion confidence interval by $3000$ to get to numbers of bottles.