Given that the times are normally distributed with mean \(\mu = 32.2\) and standard deviation \(\sigma = 9.6\), we need to find the value of \(t\) such that 20% of employees take longer than \(t\) minutes.

This implies that 80% of employees take less than \(t\) minutes. We need to find the corresponding \(z\)-value for 0.8 in the standard normal distribution table.

The \(z\)-value for 0.8 is approximately 0.842.

Using the standardization formula:

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

Substitute the known values:

\(0.842 = \frac{t - 32.2}{9.6}\)

Solve for \(t\):

\(t - 32.2 = 0.842 \times 9.6\)

\(t - 32.2 = 8.0832\)

\(t = 32.2 + 8.0832\)

\(t = 40.2832\)

Rounding to one decimal place, \(t = 40.3\).