The question is: The scores on a test have a mean of 100 and a standard deviation of 8. A personnel manager wishes to select the top 60% of applicants. Find the cutoff score. Assume the variable is normally distributed.
What I don't know: I don't know how to set up this problem. I know z = (x - mean) / Standard deviation
so its z = (x - 100) / 8 and I am looking for x, the cutoff score. How should z look like? Its looking for the top 60% so the answer will be less than the mean because its the same as looking for the cutoff score for the bottom 40%. Is 95.2 = x the correct answer (I used -.6 as z to get that), or is 96.8 (-.4 used as z) the correct answer? or are neither of those correct? This is the only problem I missed and I'm just not sure what z needs to be set as. Thanks for any help.
You have the idea of how to do it correctly. You just need to find the correct value $z$ such that $\Pr(Z<z)=.4$. After looking online at some tables, I found the value to be somewhere around $-0.25$. This would give the cutoff score of $x=98$.