1. A software engineer at a local company is interested in the quality of 
   recently developed C++ modules. Assuming that the metric used to measure 
   module quality is normally distributed with a mean of 80 and a standard 
   deviation of 25.

  a. What proportion of modules would you expect to have a quality score 
     greater than 100.

        z=(100-80)/25=0.8 hence desired proportion is 50%-57.63/2

  b. What proportion of modules would you expect to have a quality score 
     between 40 and 70.

        z1=(40-80)/25=-1.6 and z2=(70-80)/25=-0.4 hence desired
        proportion is (89.04/2-31.08/2)%

  c. One of the C++ modules has a quality score of 50. What is the percentile 
     rank of this module? 
    
        z=(50-80)/25=-1.2 hence percentile rank is (50 - 76.99/2)% tile
  
  d. What quality score defines the 20th percentile? 

        For 20th percentile look up z for 60%. Hence x=0.85*25+80=101.25 


2. A colleague is interested in the performance of a data compression 
   algorithm. 

   She is particularly interested in the time required to compress files of a 
   certain size (say 1 Megabyte). Let us say that, for such files, the 
   population mean processing time is 275 microseconds with standard deviation 
   60 microseconds. Considering the sampling distribution of, complete the 
   following. Make and state any necessary assumptions

   a. What proportion of samples of size 10 would you expect to have a mean 
      processing time greater than 300 microseconds.

         n is small so assuming time normally distributed, ybar is
         student t with 9 df and mu(ybar)=275, Sybar=Sy/sqrt(n-1)
         =60/3=20. Hence t=(300-275)/20=1.25. Desired proportion
         is between 10% and 25%.

   b. What proportion of samples of size 100 would you expect to have a mean 
      processing time greater than 300 microseconds. 

         n is large so ybar is normally distributed with mu(ybar)
         =275, Sybar=Sy/sqrt(n)=60/10=6. Hence z=(300-275)/6=4.17
         Desired proportion is 50% - 99.9967/2.

      
3. A usability analyst is interested in evaluating the user interface of a new 
   release of a popular Java compiler. She decides to conduct a controlled 
   experiment and selects a simple random sample of twenty six representative 
   users for observation. She provides them with a set of tasks and observes 
   the number of errors made by each user in completing the tasks. She 
   discovers that the mean error count is 56 with standard deviation 24. 

   a. Considering the population of users, provide a point estimate for the 
      population mean error count.

           mu=ybar=56

   b. Determine the margin of error in your estimate of the population mean. 

           since n=26 hence small then margin of error=t(n-1;2.5)*sybar.
           Sybar=Sy/sqrt(n-1)=24/5=4.8 hence margin of error=2.06*4.8=9.888

   c. Construct and interpret a 90% confidence interval for the population 
      mean. 

           t(n-1;5)=1.71 hence 95% CI is 56 +/- 1.71(4.8) [47.792,64.208]
           You are 90% confident that the population mean is between
           47.792 and 64.208.

   d. The usability analyst is told that a level of accuracy of 2 errors with 
      99% confidence is required. Given the statistics obtained for the 
      experiment above, determine the minimum sample size that would be needed 
      to achieve this level of accuracy and confidence.

           n=(2.6*24/2)^2=973.44 hence 974 users required.