Given that the running times are normally distributed with mean \(\mu = 41.2\) and standard deviation \(\sigma = 3.6\), we need to find the value of \(t\) such that 95% of the times are greater than \(t\).

This implies that 5% of the times are less than \(t\). The corresponding z-score for the 5th percentile is \(z = -1.645\).

Using the standardization formula:

\(\frac{t - 41.2}{3.6} = -1.645\)

Solving for \(t\):

\(t - 41.2 = -1.645 \times 3.6\)

\(t - 41.2 = -5.922\)

\(t = 41.2 - 5.922\)

\(t = 35.278\)

Rounding to one decimal place, \(t = 35.3\).