To find the least distance that must be thrown to qualify for a certificate, we need to find the 90th percentile of the normal distribution, as the top 10% of pupils qualify.

The mean \\(\mu \\\) is 35.0 m and the standard deviation \\(\sigma \\\) is 11.6 m. We need to find the value of \\(x \\\) such that the cumulative probability is 0.90.

Using the standard normal distribution table, the z-score corresponding to the 90th percentile is approximately \\(z = 1.282 \\\).

We use the z-score formula:

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

Substitute the known values:

\\(1.282 = \frac{x - 35.0}{11.6} \\\)

Solve for \\(x \\\):

\\(x - 35.0 = 1.282 \times 11.6 \\\)

\\(x - 35.0 = 14.8592 \\\)

\\(x = 35.0 + 14.8592 \\\)

\\(x \approx 49.9 \\\)

Therefore, the least distance that must be thrown to qualify for a certificate is 49.9 m.