Answer: The percentage of these cans will have less than 15.6 ounces = 2.28%
Explanation:
Let x = amount of soda in a 16-ounce can which is normally distributed.

The probability that cans will have less than 15.6 ounces:
![P(X<15.6)=P((X-\mu)/(\sigma)<(15.6-16)/(0.20))\\\\=P(z<-2)=1-P(Z<2)\\\\=1-0.9772\ \ \ [\text{By p-value table}]\\\\=0.0228=2.28\%](https://img.qammunity.org/2021/formulas/mathematics/college/75354rgpvi599y45nj6q3aqsq0zqwll38r.png)
Hence, the percentage of these cans will have less than 15.6 ounces = 2.28%