Given that the time to cook an egg is normally distributed with mean \(\mu = 4.2\) minutes and standard deviation \(\sigma = 0.6\) minutes. We need to find the value of \(t\) such that 12% of people take more than \(t\) minutes.

This implies that 88% of people take less than \(t\) minutes. We find the corresponding \(z\)-score for 0.88 in the standard normal distribution table, which is approximately \(z = 1.175\).

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute the known values:

\(1.175 = \frac{t - 4.2}{0.6}\)

Solving for \(t\):

\(t - 4.2 = 1.175 \times 0.6\)

\(t - 4.2 = 0.705\)

\(t = 4.2 + 0.705\)

\(t = 4.905\)

Rounding to two decimal places, \(t = 4.91\).