The time, X, is normally distributed with mean \(\mu = 15.8\) hours and standard deviation \(\sigma = 4.2\) hours.

We need to find \(P(X < 21)\).

First, calculate the z-score for 21 hours:

\(z = \frac{21 - 15.8}{4.2} = 1.238\)

Using the standard normal distribution table, find \(\Phi(1.238)\), which gives the probability that a standard normal variable is less than 1.238.

\(\Phi(1.238) = 0.892\)

Thus, \(P(X < 21) = 0.892\).