Question 1038548
SAT scores are reported on a scale from 400 to 1600. 
The SAT scores for 1.4 million students in the same graduating class were roughly normal with mean = 1026 and standard deviation = 209.

How well must Abigail do on the SAT in order to place in the top 32% of all students?


the mean is 1026 and the standard deviation is 209.


if she wants to be in the top 32% of all students, then she will have to have a score that is greater than or equal to the score of at least 68% of the students.


you would look in the z-score table for a z-score that has 68% or more of the area under the distribution curve to the left of it.


the z-score table i looked at shows:


a z-score of .46 has .6772 of the area under the normal distribution curve to the left of it.


a z-score of .47 has .6808 of the area under the normal distribution curve to the left of it.


the cutoff z-score from this table would be a z-score of .47 since more than 68% of the z-scores would have areas under the normal distribution curve less than it.


this means that less than 32% of the z-scores would have areas under the normal distribution curve greater than it.


you would then need to translate this z-score to a raw score.


the formula for z-score is:


z = (x-m)/s


z is the z-score
x is the raw score you are comparing against the mean.
m is the mean.
s is the standard deviation.


you know the following:
z = .47
m = 1026
s = 209


the formula becomes:


.47 = (x - 1026) / 209


solve for x to get x = .47 * 209 + 1026 = 1124.23


she would have to get a SAT score greater than or equal to 1124.23 in order to be placed in the top 32% of the students taking the test with a little room to spare.


if you used a calculator instead of the table, it would tell you that the z-score needed to be .4676988012.


this would have generated a raw score of 1123.749049.


that's slightly more lenient than 1124.23.


you would probably round your requirement to a score of greater than or equal to 1124.


the table i used can be found here:


<a href = "http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf" target = "_blank">http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf</a>


an online z-score calculator that can be used can be found here:


<a href = "http://davidmlane.com/hyperstat/z_table.html" target = "_blank">http://davidmlane.com/hyperstat/z_table.html</a>