InView has a mean of 100 and SD of 16. If you look up a bell curve/chart that has z scores corresponding to percentiles, you can easily find the range of z scores that could round to the relevant percentiles. Multiply the z score by the SD (16, in this case), and add (or subtract, if a negative z score) to the mean (100), and that generates your standard score. Once you top the 99th, of course, this won't work to spread the scores. In this case, I used the z scores for 96.5 to 97.4 %ile and 97.5 to 98.4 %ile:
97th %ile should be 129-131
98th %ile should be 131-134
However, schools are provided with multiple sets of norms. If your child attends a private school or high-performing public, your local norms may be stronger than the national norms, in which case it is quite possible that, though her national percentiles are as you've reported, her local or specialized (prep, private, Christian/ACSI) percentiles are lower (due to being compared to a generally higher-performing population).
ETA: and yes, I also get 126 for the 95th %ile, and 121 for the 91st %ile.
Last edited by aeh; 06/08/15 10:59 AM.
