Answer: 0.0668
Explanation:
Given : The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce.
i.e.
and

Let x denotes the volume of soda a dispensing machine pours into a 12-ounce can.
Then, the proportion of the soda cans contain less than the advertised 12 ounces of soda will be :-
![P(x<12)=P((x-\mu)/(\sigma)<(12-12.45)/(0.30))\\\\=P(z<-1.5)\ \ [\because z=(x-\mu)/(\sigma)]\\\\=1-P(z<1.5)\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=1-0.9332\ [\text{By z-table}]\\\\=0.0668](https://img.qammunity.org/2020/formulas/mathematics/college/rq2btdod2r83r7lekx56mp6w2z21ygof6k.png)
Hence, the proportion of the soda cans contain less than the advertised 12 ounces of soda = 0.0668