Let the random variable \(X\) represent the number of students with blue eyes in a sample of 80 students. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.32\).

To approximate, we use a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) are given by:

\(\mu = np = 80 \times 0.32 = 25.6\)

\(\sigma^2 = np(1-p) = 80 \times 0.32 \times 0.68 = 17.408\)

Using the normal approximation, we apply continuity correction for \(P(X < 20)\):

\(P(X < 20) = P\left(Z < \frac{19.5 - 25.6}{\sqrt{17.408}}\right) = P(Z < -1.462)\)

Using standard normal distribution tables or a calculator, find:

\(P(Z < -1.462) = 1 - \Phi(1.462) = 1 - 0.9282 = 0.0718\)

Thus, the probability that fewer than 20 students have blue eyes is approximately \(0.0718 \leq p < 0.0719\).