This test is standard, so the mean of the scores on this test is 500 and the standard deviation is 100. Consider Ali, a student who has earned a score of 400 in this test. Hence the score 400, is his raw score. Using raw score the colleges cannot decide where he stands academically in comparison to his peer. So they need the z-score and they also want to know his standing in comparison to other students taking the same test. So first find the z-score for Ali and then find what percentage of students in his cohort who performed better than him?
(Do not forget the characteristics of normal curve especially that normal curve is symmetrical=whatever you have for one side you have for the other side or half. Do not forget that the numbers in the Table of Normal Curve are probabilities. You need to multiply them by 100 to get percentage).
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 400
u = mean = 500
s = standard deviation = 100
Thus,
z = (x - u) / s = -1 [ANSWER, Z SCORE]
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Thus, using a table/technology, the right tailed area of this is
P(z > -1 ) = 0.8413
Hence, 84.13% performed better than him.
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