Cutoff score for acceptance

  • Thread starter Thread starter pkaraffa
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pkaraffa

Good afternoon, this is not a homework question, I am trying to brush
up on statistics for my job. I am having a problem with a question. A
4 year college will accept any student ranked in the top 60% on a
National Examination. If the test score is normally distributed with a
mean of 500 and a standard deviation of 100, what is the cutoff score
for acceptance?

The book says 475. If I take =NORMSINV(1-A1) where A1 houses 0.6 it
gives me -0.25 the z-score but how do I come up with 475? If I check
the equasion with the answer =(475-500)/100 it will give me -0.25. How
do I solve the unknown?

Thanks in advance
PJ
 
Hi

x is the unknown.

(x-500)/100 = -0.25

(x-500) = -25

(x-500) +500 = -25 + 500

x = -25 + 500

x = 475
 
Try this:

=100*NORMSINV(1-A1) + 500

(Algebra rather than statistics!).

Hope this helps.

Pete
 
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