Let the mean be \(\mu\) and the standard deviation be \(\sigma\).

For 36,800 km, the probability is 0.0082, which corresponds to a \(z\)-score of 2.4 (since \(P(Z > 2.4) = 0.0082\)).

For 31,000 km, the probability is 0.6915, which corresponds to a \(z\)-score of -0.5 (since \(P(Z > -0.5) = 0.6915\)).

Using the standardization formula:

\(z_1 = \frac{36800 - \mu}{\sigma} = 2.4\)

\(z_2 = \frac{31000 - \mu}{\sigma} = -0.5\)

We have two equations:

\(36800 - \mu = 2.4\sigma\)

\(31000 - \mu = -0.5\sigma\)

Subtract the second equation from the first:

\((36800 - \mu) - (31000 - \mu) = 2.4\sigma + 0.5\sigma\)

\(5800 = 2.9\sigma\)

\(\sigma = \frac{5800}{2.9} = 2000\)

Substitute \(\sigma = 2000\) back into one of the equations:

\(36800 - \mu = 2.4 \times 2000\)

\(36800 - \mu = 4800\)

\(\mu = 36800 - 4800 = 32000\)

Thus, the mean is 32,000 km and the standard deviation is 2,000 km.