SOLUTION: Students in a class take a test and the mean score is 95 and the standard deviation is 8. If the lowest possible score to pass the test is 72, what percentage of students did not p

Question 764633: Students in a class take a test and the mean score is 95 and the standard deviation is 8. If the lowest possible score to pass the test is 72, what percentage of students did not pass this exam?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean = 95
standard deviation = 8
lowest possible score to pass is 72.
normal distribution is assumed.
z = (72 - 95) / 8 = -2.875
since most tables don't go to the accuracy of more than 2 decimal places, round this off to -2.88.
anybody with a z-score less than -2.88 will have failed.
look up in the z-score tables to see that the probability of somebody getting a z-score less than -2.88 is equal to roughly .0020 which is equal to .2%.
a mean score of 95 with a standard deviation of 8 is suspicious if this is based on a normal distribution.
with a normal distribution, the probability that someone will have a score greater than 100 is calculated as follows:
z = (100 - 95) / 8 = .625
the probability of somebody getting a z-score greater than .625 is equal to .27 rounded to the nearest hundredth.
that's not likely if the top score is equal to 100 and nobody can get more than that.
this implies that the mean and / or standard deviation calculations are not correct or that the distribution is not normal and heavily skewed to the left (long tail on the left with most of the scores in the 95 range).

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