SOLUTION: A class is given an exam. The distribution of the scores is normal. The mean score is 80 and the standard deviation is 7. Determine the test score, c, such that the probability of
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Question 1158686: A class is given an exam. The distribution of the scores is normal. The mean score is 80 and the standard deviation is 7. Determine the test score, c, such that the probability of a student having a score less than c is 90 %.
P(x < c) = 0.9
Find c rounded to one decimal place. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
when z=1.645, have the 90th percentile score
1.645*7=x-mean
11.5=x-mean
x=91.5
Below that score will be 90% of the scores or P(x<90.5)=0.9
