SOLUTION: In Statistics and Probability examination, the average grade was 74 and the standard deviation was 7. If 11.9% of the class is given X’s scores, and the grades follow a normal di

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Question 1179759: In Statistics and Probability examination, the average grade was 74 and the standard deviation was 7. If 11.9% of the class is given X’s scores, and the grades follow a normal distribution, what is the lowest possible score X and the highest possible score Y?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
Normal Distribution: µ = 74 and σ = 7
11.9% of the class is given X’s score

 .0595 above and below the mean: .5 + .0595 = .5595
invNorm(.5595,74,7) = 75

the highest possible score X is 75  (1 to the right of mean 74)
lowest is 74 - 1 = 73

Your Reference did not work for me.
If the intent was for 11.9% to get As...
invNorm(.881,74,7) = 82.26  Round Up to next whole number
or 
blue%28sigma%29%2Az+%2B+mu=blue+%28x%29   z%2A%28blue%28sigma%29%29+%2B+mu=blue+%28x%29
 7*1.18 + 74 = 82.26  Round Up to next whole number
 
For As:  lowest is 83  and highest, would be a possible 100.
Wish You the Best in your Studies.