SOLUTION: Hello. Hoping this email finds you well. I am stuck with this problem and I cannot find a way to solve it. If you be so kind as to indicate steps to solve it and also point at any

Question 1148288: Hello. Hoping this email finds you well. I am stuck with this problem and I cannot find a way to solve it. If you be so kind as to indicate steps to solve it and also point at any lessons you may have in your archive related to these types of problems, I'd appreciate it a lot! Here is my nemesis:
One college requires that scholarship students maintain a GPA of 3.3, when the GPAs at that college are normally distributed with a mean of 3.0 and a standard deviation of 0.6. What percentile must scholarship students remain in?
Thank you

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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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