Question 918974
 mean of 32.0 oz and a standard deviation of .19 oz
n = 48
{{{z =blue (x - mu)/blue(sigma/sqrt(n))}}} 
x-bar come first
z(31.8) = (31.8 - 32)/(.19/sqrt(48) = -.2/ .0274 ~ -7.3
P(z < -7.3) basically zero
1 at the very most... will have LESS THAN 31.8oz
None would be more likely