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The national average SAT score is 1028. Assume a normal distribution with standard deviation being 92. What is the 99th percentile score?
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The national average SAT score is 1028. Assume a normal distribution with standard deviation being 92.

What is the 99th percentile score?

User Chul
by
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Similar to your other question. Now, you're looking for the score
k such that


\mathbb P(X\le k)=0.99

This time, 0.99 corresponds to a z-score of approximately
\hat k=2.3264, which means


(k-1028)/(92)=\hat k\implies k\approx92(2.3264)+1028=1242.03\approx1242
User Zircon
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