Answer:
115 mg
Step-by-step explanation:
The 50th percentile has a corresponding z-score of z = 0. The z-score, for any given value of caffeine, X, is given by:
[tex]z = \frac{X-mean}{SD}[/tex]
Therefore, if z=0, then X must be equal to the mean value of the normal distribution, which is 115 mg. Thus, the 50th percentile value of caffeine content is 115 mg.
