z-test for the lower specification limit

#1
I have a problem to understand the sample question and answer about z-test (from A2 below), and would like to ask your advices. The original text was in german, and I tried to translate it using Google translator :

Q. When examining the menthol content of candies you will get the values in the following - 18.2, 18.4, 18.4, 17.6, 17.7, 17.9, 18.2, 17.4, 18.6, 17.4, 17.3, 17.8. You have a specification in which you specify a minimum content of 16.5 mg / g and an error rate (i.e. non-conformities) of max. 5%. Check whether the specification has been met.

A. It is not correct to compare the mean value of the data with the minimum content of 16.5 mg / g, since it is not the mean value that is specified, but the minimum content. The 5% relates to the proportion that may be below the minimum content and not to the probability of error (significance level).
Finally, using the normal distribution curve on the distribution of random variables
  1. arithmetic mean = 17.91, standard deviation = 0.43
  2. Calculation of the z-value : z = -1.98 (where come this from?)
  3. Correct application of the z-value table : 2.24% and recognition that the specification has been met. (i don't understand the sentence.)
Thank you so much.