hey there,A student's score on an SAT was 700 on a standardized test known to have a mean of 500 and a standard deviation of 110.
a) what percentile was the student in?
b) what score must a student get to be in the 73rd percentile?
c) another student had a z score equivalent of -0.55. what was her score on the test?
i would resort to using normal distribution curve.
Normal distribution - Wikipedia, the free encyclopedia
"A percentile is the value of a variable below which a certain percent of observations fall. So the 20th percentile is the value (or score) below which 20 percent of the observations may be found. "
You were given the mean and standard deviation and required to find percentile.
a) using probability on a normal curve with the information given to you,
percentile the student is in = probability (score < 700) x 100%
standardizing it using:
probability (score < 700) = probability ( Z < 1.8182 )
using the table for normal distribution, P(Z<1.8182) = 0.9654818
For percentile, 0.9654818 x 100% = 96.54818%
b) For part b) instead you will need to find the value of Z from the Z-table given the percentile.
First we convert the percentile to probability,
P ( Z< z) = 73/100 = 0.73
From the Z-table, 0.73 corresponds to z = 0.615.
Using the formula;
This is the minimum score a person must obtain to be above the 73rd percentile.
hence score 567.65
Given z = -0.55
using the formula;
Hope this helps.