Let the random variable \(X\) represent the number of students with blue eyes. Given that one in five students has blue eyes, the probability \(p\) is \(0.2\).

The expected number of students with blue eyes is \(\mu = 120 \times 0.2 = 24\).

The variance is \(\sigma^2 = 120 \times 0.2 \times 0.8 = 19.2\).

We approximate \(X\) by a normal distribution \(N(24, 19.2)\).

To find \(P(X < 33)\), we use a continuity correction: \(P(X < 33) \approx P(Z < \frac{32.5 - 24}{\sqrt{19.2}})\).

Calculate the standard score: \(Z = \frac{32.5 - 24}{\sqrt{19.2}} = 1.9398\).

Using standard normal distribution tables, \(P(Z < 1.9398) = 0.974\).