Question 1132981
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.