1. s(ybar)=10/20=0.5 and so z1=(98-100)/0.5=-4, z2=(102-100)/0.5=4.
   Hence 99.9937% of samples of size n=400 will result in a mean
   between 98 and 102.

2. Since n is small assume y is normally distributed with n-1=16df.
   Hence, s(ybar)=10/sqrt(16)=10/4=2.5 and so t1=(98-100)/2.5=-0.8, 
   t2=(102-100)/2.5=0.8. Notice that 0.8 is between 0.69 and 1.34 for 
   16 df. That is, the area greater than 0.8 is between 10% and 25%. 
   We are interested in the area between -0.8 and 0.8. Hence between 
   50% and 80% of samples of size n=17 will result in a mean between
   98 and 102.