WISC V Standard Deviation and Percentile
By Inderbir Kaur Sandhu, Ph.D
Q:
My daughter was tested using the WISC V. Her full scale IQ was listed at
142 and her IQ percentage was reported at 99.7%. I am seeking
clarification of the percentage.
For adults, the IQ mean is 100 with a standard deviation of 15. So that
means an adult with an IQ of 145 would be 3 standard deviations above
the mean, or have an intelligence equal to or above that of 95% of the
population. My daughter's IQ was 142, yet her intelligence was reported
as equal to or smarter than 99.7% of the population. Why?
Thank You for your answer. I've been searching online and am just more
confused.
A:
For the WISCV, normative sample included 2,200 children and adolescents
divided into 11 age groups. Each age group consisted of 200
participants. The sample was nationally representative with respect to
Race/Ethnicity, Parent Education Level, and Geographic Region. Standard
deviation is 15. Current IQ tests are developed with the median raw
score of the norming sample as IQ 100 and scores each standard deviation
(SD) up or down at 15 IQ points greater or less (StanfordBinet uses SD
16).
The higher the IQ score, the smaller the percentage difference due to
rarity. Rarity can be calculated using certain statistical function
(e.g., NORMDIST function in Excel). Highest percentile is based on
ceiling on different tests and is 99.99%. For an IQ of 145, the SD is
99.87%. The percentile ranks are theoretical values for a normal
distribution. Errors are taken into account so calculations cannot be
done directly. Percentile ranks are not on an equalinterval scale; that
is, the difference between any two scores is not the same between any
other two scores whose difference in percentile ranks is the same. It
all depends on many factors that the tests would have.
Hope it is a little clearer. For more reading, check this link:
http://www.wrightslaw.com/advoc/articles/tests_measurements.html#19
Good luck!
