We are given that the time is normally distributed with mean \(\mu = 9.5\) and standard deviation \(\sigma = 1.3\). We need to find the value of \(t\) such that 90% of the time, Peter takes longer than \(t\) minutes. This means that 10% of the time, he takes less than \(t\) minutes.

Using the standard normal distribution table, we find the z-score corresponding to the 10th percentile, which is \(z = -1.282\).

We use the standardization formula:

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

Substituting the known values:

\(-1.282 = \frac{t - 9.5}{1.3}\)

Solving for \(t\):

\(t - 9.5 = -1.282 \times 1.3\)

\(t - 9.5 = -1.6666\)

\(t = 9.5 - 1.6666\)

\(t = 7.8334\)

Rounding to two decimal places, \(t = 7.83\).