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

percentile:

"Apercentileis 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:

Hence,

probability (score < 700) = probability ( Z < 1.8182 )

using the table for normal distribution, P(Z<1.8182) = 0.9654818

Z Table

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

c)

Given z = -0.55

using the formula;

Hope this helps.