Answer: The percentage of these cans will have less than 15.6 ounces = 2.28%
Step-by-step explanation:
Let x = amount of soda in a 16-ounce can which is normally distributed.
[tex]\mu=16\text{ ounces},\ \ \ \ \sigma=0.20\text{ ounces}[/tex]
The probability that cans will have less than 15.6 ounces:
[tex]P(X<15.6)=P(\dfrac{X-\mu}{\sigma}<\dfrac{15.6-16}{0.20})\\\\=P(z<-2)=1-P(Z<2)\\\\=1-0.9772\ \ \ [\text{By p-value table}]\\\\=0.0228=2.28\%[/tex]
Hence, the percentage of these cans will have less than 15.6 ounces = 2.28%
