Let [tex]X[/tex] denote the random variable for scores on the exam. [tex]X[/tex] is normally distributed with mean 57 and standard deviation 9. Let [tex]Z[/tex] denote the random variable for scores following the standard normal distribution.
You have
[tex]\mathbb P(X<66)=\mathbb P\left(\dfrac{X-57}9<\dfrac{66-57}9\right)=\mathbb P(Z<1)[/tex]
You could use a table of z-scores to get a precise answer, or you could apply the empirical rule. You know 100% of the scores are contained within the distribution. The empirical rule states that approximately 68% of them will fall within one standard deviation of the mean, which means 32% fall without, with 16% lying to either side of this range.
[tex]\underbrace{100\%}_{\text{total}}=\underbrace{16\%}_{Z<-1}+\underbrace{68\%}_{-1<Z<1}+\underbrace{16\%}_{Z>1}[/tex]
[tex]\mathbb P(Z<1)[/tex] corresponds to the first two ranges, i.e.
[tex]\mathbb P(Z<1)=\mathbb P(Z<-1)+\mathbb P(-1<Z<1)=16\%+68\%=84\%[/tex]
So out of 4000 students, you can expect 84% of them, or about 3360, to score lower than a 66. This means the answer is 3366.
More precisely, [tex]\mathbb P(Z<1)\approx0.8413=84.13\%[/tex], which means closer to 3366 students would fall in this range.
