Question 918117: Suppose that test scores in a math class are normally distributed with a mean of 72 and a standard deviation of 12.5. What score would a student need in order to be at the 90th percentile for that test?
I understood how to solve this problem until the 90th percentile, what does this mean exactly. can I not just find the z score from the (z-mean)/std?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you need to find the z-score associated with the 90th percentile.
then you need to translate that z-score into a raw score.
the 90th percentile means that 90% of the people who take the test will have scores less than you.
in the z-score table you look for a z-score of .90 because that is the area to the left of the z-score indicated.
looking into the z-score table at https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf, i find that an area of .90 equates to a z-score of somewhere between 1.28 and 1.29.
a z-score of 1.28 has an area of .89973 to the left of it.
a z-score of 1.29 has an area of .90147 to the left of it.
a z-score of 1.28 is closer to the area of .90 so we'll go with that.
you can use a calculator or you can interpolate to get closer, but it's usually not necessary to get that accurate.
for example, my T-84 calculator says that the z-score associated with an area of .90 to the left of it is equal to 1.281551567.
this is clearly closer to 1.28 than to 1.29.
normally, when you use the z-score tables, the z-scores will be 2 decimal places so, for practical purposes, you round the z-score to the nearest 2 decimal places and you should be ok unless greater accuracy is required.
now that you know the z-score is 1.28, you have to translate that z-score to a raw score.
the mean of the test is 72 and the standard deviation is 12.5
the formula for z-score is:
z-score = (raw score - mean) / standard deviation
put in what you know and solve for what you don't know.
you know:
1.28 = (raw score - 72) / 12.5
solve for raw score as follows:
multiply both sides of the equation by 12.5 to get:
12.5 * 1.28 = raw score - 72
add 72 to both sides of the equation to get:
12.5 * 1.28 + 72 = raw score.
this leads to a raw score of 88.
if the mean is 72 and the standard deviation is 12.5, then a raw score of 88 will be better than approximately 90% of the population taking the test.
to confirm you're right, calculate the z-score and then find the area on the normal distribution curve to the left of it.
z-score = (88 - 72) / 12.5 = 1.28
area tot the left of z score of 1.28 = .89973 which is about as close to .90 as you can get without interpolating or using a calculator.
graphically, this would look like this:
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