I'm given a mean and a standard deviation. The question asks: How much chicken do you have to eat in order to be in the top $5%$ of everyone else.
Mean = $55$
Standard Deviation $=9.2$
The answer is $X=70.13$ but I can't figure out how to get that without a $z$ score.
The question here requires you to find the $z$-score. The question is "what would the $z$-score have to be in order to be in the top $5\%$?" Instead of calculating a $z$-score and looking up the probability in a table, you need to look up the probability in the table and find the corresponding $z$-score.
Once you know the $z$-score, you can use $z=\frac{x-\mu}{\sigma}$ to calculate the corresponding $x$.