The amount of soda in 16 oz bottles has an unknown distribution with a mean of 16.04 oz and a standard deviation of 0.15 oz. of 36 soda bottles are randomly sample, what is the probability that the mean of this sample is more than the advertised 16 oz?

The applicable formula is;

P(X>16)

Z = (x-mean)/(SD/sqrt (N))

Where;
x = 16 oz
mean = 16.04 oz
SD = 0.15 oz
N = 36 bottles

Substituting;
Z = (16-16.04)/(0.15/Sqrt (36)) = -1.6

Therefore;
P(X>16 oz) = P(Z>-1.6)

From Z-table;
P(Z>-1.6) = 1- P(Z=-1.6) = 1 - 0.0548 = 0.9452 = 94.52%

Therefore, the probability that the sample mean is more than advertised mean is 94.525.