First, we standardize the time of 20 minutes using the formula for the standard normal distribution:

\(Z = \frac{X - \mu}{\sigma}\)

where \(X = 20\), \(\mu = 14.6\), and \(\sigma = 5.2\).

\(Z = \frac{20 - 14.6}{5.2} = 1.03846\)

Next, we find the probability that \(Z > 1.03846\). Using standard normal distribution tables or a calculator, we find:

\(P(Z > 1.03846) = 1 - P(Z < 1.03846) = 1 - 0.8504 = 0.1496\)

This means approximately 14.96% of students take more than 20 minutes.

In a sample of 250 students, the expected number is:

\(250 \times 0.1496 = 37.4\)

Rounding to the nearest whole number, we expect about 37 or 38 students to take more than 20 minutes.