How many times must you roll a non-fair die to be at least 84% sure that the sample probability will be within 3% from the actual probability.
Since the die is not-fair, we do not know p. My question is, can I assume it to be around 1/6, or not?
And so far, I found the Z-scores for 84% of the probability from the mean to be + & - 1.4. But, I do not know how to move forward with an unknown probability.
As others say in the comments, the question makes no sense because it does not state the probability of what. The probability of the die falling to a surface is presumably one, for example, regardless of how loaded it is. I will assume it means the probability of rolling a six (I know this is by no means the only possibile interpretation of this ambiguous question).
The wording of the question asks for you to be "at least 84% sure that the sample probability will be within 3% from the actual probability." The "at least" is surely significant. You can answer the question by assuming the worst case scenario - the die is loaded so six shows up 50 percent of the time so the variance of the number of sixes is $n(\frac{1}{2})^2$. Solve from there using usual methods.
So my answer is no, you cannot assume p is 1/6. As you have been asked to be conservative ("at least") you must assume the worst case scenario, which is p=1/2.
