Question 1148288
the mean is 3.0 and the standard deviation is .6.
you want to know what the percentile needs to be in order to meet that requirement.
the z-score is found using the following formula.
z = (x-m) / s
z is the z-score
x is the raw score requirement
m is the mean
s is the standard deviation
you get z = (3.3 - 3.0) / .6 = .3 / .6 = 3 / 6 = .5
the z-score is .5
the area under the normal distribution to the left of that z-score is equal to .6914524568.
round to 4 decimals to get .6915
multiply by 100 to get 69.15%.


if i did this correctly, the scholarship student must be in the 69.15 percentile.
this means that 69.15% of the student body got scores less than that.


graphically, it looks like this.
first graph is with z-score.
second graph is with raw score.
with z-score, mean is 0 and standard deviation is 1
with raw score, mean is 3.0 and standard deviation is .6


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