1 i. a. Since z1=0.67 and z2=2.33 then the desired proportion is:
                (98.12-48.43)/2 = 24.845%
     b. Since the z required is the tabled value for 20% and it is 
        -ve then:
                time=(-0.25)60+400=385
     c. The z required for the 95th %tile is the tabled value for
        90%:
                time=(1.65)60+400=499
        Since 508 is greater than 499 then 508 is slower than the 95th 
        %tile.
                
 ii. a. Since n=17 is small then if y is normally distributed then ybar
        is Student t with 16df.
        Now, s(ybar)=s(y)/sqrt(16)=15 and t=(370-400)/15=-2. Since df=16 
        then the desired proportion is between 2.5% and 5%.
     b. FALSE. The sampling distribution for samples of size n=8 is
        "t" but that for samples of size n=64 is normal. The proportions
        are 98% and 99.73%.

2 i. By FPDA, mu(y)=$2.60 and sigma(y)=0.9. That is, if we have a probability
     sample.
 ii. a. 50%
     b. Yes. Since z=(6.0-2.6)/0.9=3.8 then 50%-(99.986/2)=0.007% of the
        programs in the portfolio will cost more than $6.00. So, if the
        portfolio contains 300 programs then we would expect about 2 of 
        them to cost more than $6.00 per line of code to fix.