# Thread: percentiles... UGH

1. ## percentiles... UGH

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?

2. 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?
hey there,

i would resort to using normal distribution curve.
Normal distribution - Wikipedia, the free encyclopedia

percentile:
"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:

$\displaystyle z = \frac {score - mean}{Standard \text{ } Deviation}$
$\displaystyle z = \frac{700-500}{110}$
$\displaystyle z= 1.8182$

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;
$\displaystyle z = \frac {score - mean}{Standard \text{ } Deviation}$
$\displaystyle 0.615 = \frac {score - 500}{110}$
$\displaystyle score = (0.615 \times 110) + 500$
$\displaystyle score = 567.65$
This is the minimum score a person must obtain to be above the 73rd percentile.

hence score $\displaystyle \geqslant$ 567.65

c)
Given z = -0.55

using the formula;

$\displaystyle z = \frac {score - mean}{Standard \text{ } Deviation}$
$\displaystyle -0.55 = \frac {score - 500}{110}$
$\displaystyle score = (-0.55 \times 110) + 500$
$\displaystyle score = 439.5$

Hope this helps.