1. You are told that, for some population, the population parameters are mu=50 
   and sigma=5. Considering the sampling distribution of ybar, answer the 
   following:

   a) What percentage of samples of size n=100 would you expect to have mean 
      between 49 and 51. 

	sigma(ybar)=5/10=0.5 and so z1=(49-50)/0.5=-2, z2=(51-50)/0.5=2.
	Hence 95.45% of samples will have a mean between 49 and 51.
  
   b) What percentage of samples of size n=100 would you expect to have mean 
      greater than 53. 

	z=(53-50)/0.5=6 and so you would not expect any samples to have
	a mean this large.
 

2. You are told that, for some population, the population parameters are mu=80 
   and sigma=12. Considering the sampling distribution of ybar, answer the 
   following:
 
   a) What percentage of samples of size n=36 would you expect to have mean 
      less than 83. 

        sigma(ybar)=12/6=2 and so z=(83-80)/2=1.5. From the z table the 
        desired percentage is 50+86.64/2=93.32%.