We are given that 83.4% of adult male giraffes weigh more than 950 kg. This implies that the probability \(P(X > 950) = 0.834\).

Using the standard normal distribution table, we find that a probability of 0.834 corresponds to a \(z\)-score of approximately -0.97 (since we are looking for the left tail, we use the negative \(z\)-score).

The standardization formula is:

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

Substituting the known values:

\(-0.97 = \frac{950 - 1190}{\sigma}\)

Solving for \(\sigma\):

\(-0.97\sigma = 950 - 1190\)

\(-0.97\sigma = -240\)

\(\sigma = \frac{-240}{-0.97}\)

\(\sigma \approx 247\)