Question #1:

On the math SAT men have a distinct edge. In 1988 for instance the men
averaged about 500 on the math SAT while the women averaged about 455.
If both standard deviations are 100 and assuming a normal distribution
answer the following:

a) 	Determine the percentage of men getting over 600 on this
	test in 1988.

	Answer (using Ott table):

	The mean for men is 500 and the standard deviation is 100. Since we 
	need to determine the percentage of men getting over 600, we need to 
	determine the area under the normal curve to the right of the vertical
	line defined by 600. 

	You first need to obtain the z value corresponding to a SAT score of 
	600. You therefore need to standardize the value 600 using the formula:
			z = {(600 - 500)/100} = 1
	You are only interested in the percentage greater than 1. From Ott
        the entry for 1 is 0.8413. This is a decimal and so the desired 
        percentage is:
	                   100 - 84.13 = 15.87%

	Answer (using table posted on course homepage):

	You are only interested in the percentage greater than 1. For the
        the posted table the proportion between -1 and 1 is 68.27%. By symmetry
        the desired percentage is:
	                   50 - 68.27/2 = 15.865%

b)	Estimate the percentage of women getting over 600.

	Answer (using Ott table):

	The argument is the same. In this case z = {(600 - 455)/100} = 1.45. 
                           100 - 92.65 = 7.35%

	Answer (using table posted on course homepage):

        In this case:
	                   50 - 85.29/2 = 7.355% 


Question #2:

For CTI-2002 graduates starting salaries are approximately normally
distributed with mean 50000 and standard deviation 10000. About what 
percentage of graduates obtained salaries greater than 35000?

Answer (using Ott table):

Since z = (35000 - 50000)/10000 = -1.5 then the desired proportion is
100 - 6.68 = 93.32%

Answer (using table posted on course homepage):

Again, by symmetery the desired proportion is 50% + 86.64/2 = 93.32%