Question 1132981: Scores on a test have a mean of 73 and Q3 is 83. The scores have a distribution that is approximately normal.
F i n d P(90)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! not sure is this is correct, but here's my attempt at it.
a z-score associated with 25% of the area under the normal distribution curve to the right of it is equal to .6744897495.
that z-score is associated with a raw score of 83 and a mean of 73.
the formula for z-score is z = (x - m) / s
x is the raw score.
m is the mean
s is the standard deviation.
that formula becomes .6744897495 = (83 - 73) / s
solve for s to get s = 10 / .6744897495 = 14.8260222.
if the distribution is normal, that should be the standard deviation.
probability of getting a raw score greater than 83 should be .25 if this was done correctly.
solve for z to get z = (83 - 73) / 14.8260222.
you'll get z = .6744897495.
that z will get you a raw score of 83 if done correctly.
the z-score formula becomes .6744897495 = (x - 73) / 14.8260222.
solve for x to get 14.8268222 * .6744897495 + 73 = 83.
it looks ok.
probability of getting a score greater than 90 would them be done as follows.
z = (90 - 73) / 14.8268222 = 1.146632574.
probability of getting a z-score greater than 1.146632574 would be equal to .1257668062.
therefore, probability of getting a raw score greater than 90 will be .12576688062.
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