Question 476297: SAT I scores around the nation tend to have a mean scale score around 500, a standard deviation of about 100 points and are approximately normally distributed. A person who scores (a perfect) 800 on the SAT I has approximately what percentile rank within the population?
Thanks
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! SAT I scores around the nation tend to have a mean scale score around 500, a standard deviation of about 100 points and are approximately normally distributed. A person who scores (a perfect) 800 on the SAT I has approximately what percentile rank within the population?
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z(800) = (800-500)/100 = 3
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P(z < 3) = 99.87%
So a score of 800 is the 99.87%ile score.
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Cheers,
Stan H.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! That person's percentile rank would be .998650 based on a normal distribution curve.
It assumes that .001350 of the population will score 800 or better on the test.
Since they can't score better, then the assumption is that .001350 of the population will score 800.
I used david m. lane's z-score calculator to derive this number.
the calculator can be found at the following link:
http://davidmlane.com/hyperstat/z_table.html
if you were to evaluate this using the z-tables, then you would need to do the following:
the mean is 500.
the standard distribution is 100.
the value you are interested in is 800.
the z-score would be (800-500)/100 = 3.0
you need to look up the z-score in the z-table.
one such z-table can be seen by clicking on the following:
http://lilt.ilstu.edu/dasacke/eco148/ZTable.htm
scan down the left side of that table for a z-score of 3.00 and you'll see that the z-score is equal .9987
this is very close to .998650 we derived from using david m. lane's z-score calculator.
in fact, .9987 would be equal to .998650 rounded to the nearest 4 decimal places.
in the z-table, the value shown there is the percent of the distribution curve that is to the left of the z-score.
using david m. lane's calculator, that would be equivalent to the area under the curve to the left of the value.
with david m. lane's calculator, you could enter a mean of 500 and a standard distribution of 100 and then enter a value of 800 in the below slot and then click on the below button to get the value.
alternately, you could enter a mean of 0 and a standard deviation of 1 and then enter a value of 3.0 in the below slot and thenc lick on the below button to get the value.
both cases should get you the same value of.897650.
i am talking about his top greph, not his bottom graph.
try it yourself and see how you do.
calculate the z-score yourself and look at the table and see if you can get the same answer i got.
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