Given that the times are normally distributed with a mean of 62 seconds and a standard deviation of 5 seconds, we need to find the time t such that 13% of the members take more than t minutes to swim 100 metres.

\(First, convert t minutes to seconds: t minutes = 60t seconds.\)

We know that 13% corresponds to a z-score of approximately 1.127 (from standard normal distribution tables).

Using the z-score formula:

\(\frac{60t - 62}{5} = 1.127\)

Solving for t:

\(60t - 62 = 5 \times 1.127 = 5.635\)

\(60t = 5.635 + 62 = 67.635\)

\(t = \frac{67.635}{60} \approx 1.13\)