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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 WISC-V, 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 (Stanford-Binet 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 equal-interval 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!


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