1.a. By FPDA, mu(y)=ybar=350; sigma(y)=s(y)=100
1.b. z=(365-350/100=0.15. Required area is 50-(11.92/2)=44.04%.
1.c. z=(250-350)/100=-1. Percentile rank is 50-(68.27/2)=15.87 percentile.
1.d. n=26 and since y is normally distributed then ybar is Student t with
     25df. Also, s(ybar)=100/sqrt(25)=20 hence the 95% CI is:
		ybar plus/minus t(25,2.5)*s(ybar)
		350 plus/minus 2.06*20
		[308.8, 391.2]
     I am 95% confident that mu(y) is between 308.8 and 391.2 days.

2.a. H0: mu(y)=$62,500; Ha: mu(y)<$62,500 
2.b. First, ybar=$60,000 is consistent. Assume H0 true and since n is
     large determine the z value. s(ybar)=10000/sqrt(100)=1000 hence
     z=(60000-62500)/1000=-2.5. The p-value is 0.62% (or 0.0062 
     expressed as a probability) which is highly significant and so we
     reject the null hypothesis.
2.c.i. b=0.8(10000/5)=1600; a=60000-1600*20=28000. Hence the regression line
       equation is:
		income=28000 + 1600(yrs_of_tenure)
2.c.ii. For unit increase in yrs_of_tenure we can expect income to 
        increase by $1600. The intercept indicates that a tenured faculty
        member with zero yrs_of_tenure can expect, on average, to earn
        $28,000 which may be reasonable.

3.a. Since r=0.9891 is positive then as document size increases, processing
     time also increases. Furthermore r^2=0.9783 hence 97.83% of the
     variability in processing time is accounted for by considering
     document size. 
3.b  Since r=0.9891, the slope is 0.9891(s(y)/s(x))=0.9891(24.95/66.33)
     =0.372 (rounded) and the intercept is ybar-0.372(xbar)=103-0.372(148)
     =47.937. Hence regression equation is:
                     time=47.937+0.372(size)
3.c  For unit increase in size we can expect time to increase by 0.372
     time units. For a document of size zero units we can expect a
     processing time of 47.937 time units which may make sense since
     we would expect some overhead in processing an empty document.

4.a. H0: mu(y)=$2.10; Ha: mu(y)<$2.10
4.b. First, ybar=$2.04 is consistent. Assume H0 true and since n is
     large determine the z value. s(ybar)=0.30/sqrt(36)=0.05 hence
     z=(2.04-2.10)/0.05=-1.2. The p-value is 11.505% (or 0.11505 
     expressed as a probability) which is non-significant and so we
     have insufficient evidence to reject the null hypothesis. The
     directors claim is therefore reasonable.
4.c. Since the null hypothesis was not rejected then the null hypothesis 
     mu=$2.10 is reasonable. Some other point estimate for the population
     mean would therefore be inappropriate.