statistics
posted by Abby .
Scores on a test have a mean of 66 and Q3 of 81. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.

Q3 = 75 percentile
In a table in the back of your stat book labeled something like "areas under normal distribution," find the Z score equivalent. Put value in formula below.
Z = (x  μ)/ SD = (8166)/SD
Solve for SD. 
(8166)/SD = 0.67(z score of 0.75)
15/SD= 0.67
15/0.67
SD=22.3880
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