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 = (81-66)/SD
Solve for SD.
(81-66)/SD = 0.67(z score of 0.75)