We are given that the time is normally distributed with mean \(\mu = 3.5\) and standard deviation \(\sigma = 0.9\).

We need to find the value of \(t\) such that 90% of the time is greater than \(t\). This implies that 10% of the time is less than \(t\).

Using the standard normal distribution, we find the critical value for 10%, which corresponds to \(z = -1.282\).

We use the standardization formula:

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

Substitute the known values:

\(-1.282 = \frac{t - 3.5}{0.9}\)

Solving for \(t\):

\(t - 3.5 = -1.282 \times 0.9\)

\(t - 3.5 = -1.1538\)

\(t = 3.5 - 1.1538\)

\(t = 2.3462\)

Rounding to two decimal places, \(t = 2.35\).