You're looking for \(k\) such that
\(\mathbb P(X
Transforming the standard normal distribution, you have
\(\mathbb P(X
where \(k^*\) is the cutoff score in terms of this new distribution. The z-score corresponding to a probability of 0.20 is approximately \(z=k^*=-0.8416\), which means the cutoff test score is
\(k^*=\dfrac{k-125}{12.51}\implies k=125+12.51k^*\approx114.47\)
Rounding to the nearest whole point, a score of about 114 is the bare minimum requirement to not have to take the remedial course.
