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

	Answer:

	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
	From the standard normal table, 68.27% of the area under the
	curve is between -1 and 1. You are only interested in the area to the
        right of 1 and so by the empirical rule and symmetry it is:
	                   50 - (68.27/2) = 15.86%

b)	Estimate the percentage of women getting over 600.

	The argument is the same. In this case z = {(600 - 455)/100} = 1.45. 
	From the standard normal table, 85.29% of the area under the curve is 
	between -1.45 and 1.45. So by symmetry:
                           50 - (85.29/2) = 7.36%



Question #2:

In a particular course the scores on the final were approximately normally
distributed with mean 50 and standard deviation 10. About what percentage
of students had scores between 35 and 70?

The argument is a little different. You need to determine the percentage of 
students with scores between 35 and 70. First, standardize 35 and 70.
z1 = {(35 - 50)/10 = -1.5 and z2 = {70 - 50)/10 = 2. You now need to
determine the area under the standard normal curve between -1.5 and 2.0.

One way of doing this is to proceed as follows:

1) 	First find the area between -1.5 and 1.5. Divide this area by 2 (due
	to symmetry) to get that portion of the required area to the left of the
	mean. From our table that is 86.64/2 = 43.32%.
2) 	Next find the area between -2.0 and 2.0 and do the same thing to get
	the area to the right of the mean. From the table that is 95.45/2 = 
	47.73%.
3) 	Finally, add 43.32 and 47.73 to get 91.05 of students had scores 
	between 35 and 70.