In the problem above you were given the height and asked to find the percentile. Now, you'll be going in the opposite direction-- you'll be given the percentile (placement on the normal curve) and asked to find the height. In both cases the first step is to find the z-score. If someone is in the 90 th percentile in height (i.e., is taller than 90 % of the other males in the class), let's work on finding his height. First we'll find the z score. (Do NOT look up the z score for Area = 90 % on the table b/c the Areas given on the table are MIDDLE areas. 90 th percentile means 90 % to the LEFT, and 40 % above the 50th percentile)

1. What is theZ-scorecorresponding to themiddle area? (If the middle area falls between two lines on the table, you may use the closest one. PutZ-score in blank below.)
2. Now change the z score to a height. (Remember the z score tells how many SD's the value is from the average. So a z score of 2 means the person is 2 SDs above average which would be 2*(3") + 68".) What is his height?
3. If someone is in the 10 th/st percentile in height (i.e., is taller than only 10 % of the other males in the class), let's find his height. First find theZ score.
4. Now change the z score to a height. Round to the nearest tenth.