Really need help with this stats problem

I am having a really hard time doing this stats assignment. I finished it all and understood it all until the last question which is just driving me crazy. If someone could help me I would greatly appreciate:

The personnel department of a large manufacturing company has given a test to a large number of applicants for positions on the assembly line. The results obtained from 500 men and women indicated that the frequency distribution of scores closely approximates the normal curve. The mean and standard deviation were as below

mean (mu): 84 standard deviation: 5 N = 500

a. what proportion scored 94 or above?
b. How many people scored a 94 or above?
c. what is the percentile rank of the score 93?
d. What is the raw equivalent for the 99th percentile?

This problem is driving me crazy! I tried to first find the Z-score which i calculated by (500-84)/5 and got 83.2 but thats as far as a got. I have reviewed the study material but to no avail. Did I get the Z-score right? and how do I find the answers to the questions a-d? I can't find those steps in my notes.


TS Contributor
I tried to first find the Z-score which i calculated by (500-84)/5
Why?! 500 is the sample size.

What you have to ask first is, how many standard deviations is 94 away from the mean (84).
That's the z score then, and you can look up the corresponding proportion in a z table.


Ok, so I did the z score by z = (94-84)/5 = 2. Is the z score correct? So then for #a would it be the part of the tail, since its asking how many people scored above? which would be .0228? Thank you very much for your help!!
Thank you so much for your help!! But know how do I find how many people Scored a 94 or above? Since the z-score was 2, which gave a result of 0.0228 do I multiply that by the sample size which is N= 500? which gives a result of 11.4, so then 11 people score above 94?


TS Contributor
You already know which percentile ranks are comprised by scores >= 94, so c) should not really be a problem.

For d) you first have to find out the z value of percentile 99...

With kind regards