Given that 25% of the sheep weigh more than 67.5 kg, this corresponds to the 75th percentile of the normal distribution. The z-score for the 75th percentile is approximately 0.674.

Using the standardization formula:

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

Substitute the known values:

\(0.674 = \frac{67.5 - \mu}{4.92}\)

Solving for \(\mu\):

\(67.5 - \mu = 0.674 \times 4.92\)

\(67.5 - \mu = 3.31608\)

\(\mu = 67.5 - 3.31608\)

\(\mu = 64.18392\)

Rounding to one decimal place, \(\mu = 64.2\).