Assuming scores are normally distributed, a score of 82% on Ms. Smith's test corresponds to the [tex]p[/tex]-th percentile, i.e.
[tex]P(X_S\le82)=p[/tex]
where [tex]X_S[/tex] is a random variable denoting scores on Ms. Smith's test.
Transform [tex]X_S[/tex] to [tex]Z[/tex], which follows the standard normal distribution:
[tex]P(X_S\le82)=P\left(\dfrac{X_S-75}2\le\dfrac{82-75}2\right)=P(Z\le3.5)\approx0.9998[/tex]
which means Amy scored at the 99.98th percentile.
This makes it so that Karina needs to score [tex]X_A=x[/tex] on Mr. Adams' test so that
[tex]P(X_A\le x)=0.9998[/tex]
Their test scores have the same [tex]z[/tex] score computed above, so
[tex]\dfrac{x-73}3=3.5\implies x=83.5[/tex]
so Karina needs to get a test score of at least 83.5%.
