<PRE><B>Question 580970</B>
What would have to be the test score (rounded to once place to the right of a decimal) to qualify for admittance into a institution with a mean of normally distributed test scores of 82.5 and a standard deviation was 6.9, assuming admittance depended on being in the top 17 percent. 
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Find the z-score with a left tail of 0.83
invNorm(0.83) = 0.9542
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That tells you the score you are looking for is 0.9542
standard deviations above the mean.
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Find that score:
x = 0.9542*6.9+82.5 = 89.08
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Cheers,
Stan H.
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