To answer this question, first calculate the z-score:
test score - mean score 65 - 70
z = ----------------------------------- = ------------------- = -5/10 = -0.5
std. dev. 10
We need to determine the area under the normal curve to the left of z = -0.5.
Refer to a table of z-scores, and look for z=-0.5. What area appears in connection with this z-score? It's less than 0.5000.
Using my TI-83 Plus calculator (specifically, its "normalcdf( " function, I found that the area under the std. normal curve to the left of z = -0.5 is 0.309.
Thus, the percentile is 30.9, or roughly 31 (31st percentile).
Roughly 31% of students earned a grade lower than 65, and roughly 69% earned a higher grade.
