Let the random variable \(X\) represent the travel time in minutes. \(X\) is normally distributed with mean \(\mu = 140\) and standard deviation \(\sigma = 12\).

We need to find \(P(X < 132)\).

First, standardize the variable using the formula:

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

Substitute the values:

\(Z = \frac{132 - 140}{12} = -0.6667\)

Now, find \(P(Z < -0.6667)\).

Using standard normal distribution tables or a calculator, \(P(Z < -0.6667) = 0.252\).