The find a percentile rank we first find the z-score associated with a score of 80 = (80-100)/15 = -1.333. Now we lookup the area that is 1.33 standard deviations BELOW the mean and we get .0912 making a score in the 9.1 percentile.

For between area we find the areas for both scores (just as in part a, and subtract the smaller area from the larger area).For a score of 120, we get a z-score of 1.333, which is in the 90.9^{th} percentile The percent of scores in between 80 and 120 is then 90.9-9.1 = .8175 or 81.7%

The standard score is just a z-score for 90 = (90-100)/15 = -.6667.

We find the z-score of 120 (from part b) = 1.3333 and the area above this in the normal curve = .0912 or 9.12%

We find the z-score associated with the 87^{th} percentile using excel's formula =NORMSINV(.87) and we get a z-score of 1.1264. So we just need to know what values is 1.1264 standard deviations above the mean. Multiplying this times the sd we get 1.1264*15= 16.8959. Now we just add that to the mean of 100 and we get a cutoff score of 116.8959.