We are given that 75% of the time, Pia takes longer than t minutes. This implies that 25% of the time, she takes less than t minutes. We need to find the z-score corresponding to the 25th percentile of a standard normal distribution.

The z-score for the 25th percentile is approximately (-0.674).

Using the standardization formula:

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

Substitute the known values:

\(-0.674 = \frac{t - 10.1}{1.3}\)

Solve for \(t\):

\(t - 10.1 = -0.674 \times 1.3\)

\(t - 10.1 = -0.8762\)

\(t = 10.1 - 0.8762\)

\(t = 9.2238\)

Rounding to two decimal places, \(t = 9.22\).