Apply z-score
Given data
[tex]\begin{gathered} \operatorname{mean}\text{ }\mu\text{ = 100.5} \\ \text{Standard deviation }\sigma\text{ = 1.5} \\ x\text{ = 102.1} \end{gathered}[/tex]Here, we use the normal distribution formula below to calculate the answer.
[tex]\begin{gathered} z\text{ = }\frac{x\text{ - }\mu}{\sigma} \\ \text{Next, substitute the values of }\sigma,\text{ }\mu\text{ and x.} \end{gathered}[/tex][tex]\begin{gathered} z\text{ = }\frac{102.1\text{ - 100.5}}{1.5} \\ z\text{ = }\frac{1.6}{1.5} \\ z\text{ = 1.067} \end{gathered}[/tex]Next, use the normal distribution table to find probability.
From the normal distribution table,
The probability that the mean content is less than 102.1 mL = 0.3554
Final answer = 0.3554