Question

Assume that adults have IQ scores that are normally distributed with a mean of 105 and a standard deviation of 20. Find P15, which is the IQ score separating the bottom 15%, from the top 85%.

Answer 1

If \(X\) is the random variable for IQ score, then what you're looking for is the cutoff value \(k\) such that
\[P(X<k)=0.15~~\iff~~P(X>k)=0.85\]
I'll pick the first - it doesn't matter which you choose.

First transform \(X\) to \(Z\) so that it has a standard normal distribution:
\[P(X<k)=P\left(\frac{X-105}{20}<\frac{k-105}{20}\right)=P\left(Z<\frac{k-105}{20}\right)=0.15\]
A left-tail probability of 0.15 is given by a critical \(z\)-value of approximately \(z=-1.35\), courtesy of this table:
http://dsearls.org/courses/M120Concepts/ClassNotes/Statistics/520A_LeftTailTable.htm

Set the critical value equal to the quantity containing \(k\) and solve for \(k\):
\[\frac{k-105}{20}=-1.35\]