Question

The national average SAT score is 1028. The standard deviation is 92. What is the 78th percentile score?

Answer 1

You're looking for the score $k$ such that

$\mathbb P(X\le k)=0.78$

First transform $X$ to the standard normal random variable.

$\mathbb P(X\le k)=\mathbb P\left(\dfrac{X-1028}{92}\le\dfrac{k-1028}{92}\right)=\mathbb P(Z\le \hat k)=0.78$

Now 0.78 corresponds to a z-score of approximately $\hat k=0.7722$, which means

$\dfrac{k-1028}{92}=\hat k\implies k\approx92(0.7722)+1028=1099.04\approx1099$