A standardized test has a mean of 88 and a standard deviation of 12. What is the score at the 90th percentile? Assume a normal distribution.
how would I solve?
a. 90.00
b. 112.00
c. 103.36
d. 91.00
4 answers
Anyone?!
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.90) and its Z score. Insert Z score into equation above.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.90) and its Z score. Insert Z score into equation above.
.3159=(88-?)/12
For .90, Z value should be a number > +1.
Z = (score-88)/12
Z = (score-88)/12