Mean, Standard Deviation, and z-Scores...Oh My!

I happened upon your forum today and I am in need of assistance. I am an elementary school teacher with a very limited math background. I am literally a few statistics test questions away from completing my Master's. I have tried listening to the professor's videos and using notes but, with it being an online class, I am having trouble trying to teach myself. I would be ever grateful for any help offered!

Over the past few years 225 students in sections of EDUR 7130 I've taught have completed the final test of three tests administered in that course. The overall mean is 82.3 with a standard deviation of 9.0. These data are nearly normal in distribution; given this answer the following questions:
(a) What percentage of students will score 85.3 or higher on this test?
(b) What is the percentile rank for a student with a score of 91.3?
(c) What is the raw test 3 score for a student with a z score of -1.5?
(d) What is the raw test 3 score for a student with a z score of 0.00?
The theory here is that you can transform your observed data into a standard normal distribution, Z. Remember that the Z-dist has a mean of 0 and a standard deviation of 1.

Z-Scores: These are how far away a point is from the mean. A Z-score of 0 is at the mean. Z of -1 means its 1-standard deviation below. So lets start with question A

A) first transform the observed data into the Z-dist. Z = (x-mu)/sigma. Thus Z = (85.3-82.3)/9 = 1/3. Now look up in a z-table the value 0.333333 which you should get something around .63

B) Again, find the z-score using the following: Z = (x-mu)/sigma. Look up that Z value in the table and thats the percentile.

C) Do this in reverse: Z = (x-mu)/sigma => Z*sigma = x - mu => x = Z*sigma + mu. (EDIT: this is a '+')
D) this is the easiest of them all.
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TS Contributor
Cave! The proportion of students scoring at least 1/3 SD above the mean can
only be a fraction of all those students scoring at least 0 SD above the mean
(which is 50%, according to d). Archidamus gave you a hint, but not the complete

With kind regards