Given that the lengths are normally distributed, we use the standard normal distribution to find \(\mu\) and \(\sigma\).

For 4% longer than 12 cm, the z-score is:

\(z = \frac{12 - \mu}{\sigma}\)

From standard normal tables, \(z = 1.751\) for 4% in the upper tail.

Thus, \(1.751 = \frac{12 - \mu}{\sigma}\).

For 32% longer than 9 cm, the z-score is:

\(z = \frac{9 - \mu}{\sigma}\)

From standard normal tables, \(z = 0.468\) for 32% in the upper tail.

Thus, \(0.468 = \frac{9 - \mu}{\sigma}\).

We have two equations:

1. \(1.751 = \frac{12 - \mu}{\sigma}\)
2. \(0.468 = \frac{9 - \mu}{\sigma}\)

Subtract the second equation from the first:

\(1.751 - 0.468 = \frac{12 - \mu}{\sigma} - \frac{9 - \mu}{\sigma}\)

\(1.283 = \frac{3}{\sigma}\)

\(\sigma = \frac{3}{1.283} = 2.34\)

Substitute \(\sigma = 2.34\) into the first equation:

\(1.751 = \frac{12 - \mu}{2.34}\)

\(12 - \mu = 1.751 \times 2.34\)

\(12 - \mu = 4.1\)

\(\mu = 12 - 4.1 = 7.91\)