Question 1162828
Using the central limit theorem...
z>(20-19)/9/sqrt(55)
z>0.82 and that is a probability of 0.2061

for 16, z=(16-19)/9/sqrt(55) or -2.47
probability of z between -2.47 and 0.82, the value for $20, is 0.7871

half-interval for 95% CI is 1.96*sigma/sqrt(n)=1.96*9/sqrt(55)=$2.38

the CI is ($16.62, $21.38), the $2.38's being subtracted from and added to the mean.