How to find the mean variable of a normal distribution with a given probability and standard deviation?

Question

We have a machine that produces µ g of pasta to be stored in their package, with a standard deviation of 20g. It follows a normal distribution. And we don't want it to produce more than the package's storage capacity more than 2% of the time. What should be the µ value?

Answer 1

If $c$ is the capacity of the package, and $X$ is a random variable for the amount of pasta produced, then you want to be sure that $P(X > c) \leq .02$.

So, then you want to normalize by writing $P(\frac{X-\mu}{\sigma} > \frac{c-\mu}{\sigma}) \leq .02 $.

$P(Z > \frac{c-\mu}{20}) \leq .02$.

$1-P(Z \leq \frac{c-\mu}{20}) \leq .02$

$P(Z\leq \frac{c-\mu}{20}) \geq .98$

Which would happen if $\frac{c-\mu}{20} \geq z_{.98}$

And so now solve for $\mu$ to get $\mu \leq -z_{.98}20+c$