(i) To find the mean \(\mu\) and standard deviation \(\sigma\), we use the standard normal distribution properties. Given that 96% of lengths are less than 34.1 cm, we find the z-score corresponding to 0.96, which is approximately 1.751. Thus,

\(\frac{34.1 - \mu}{\sigma} = 1.751\)

Similarly, 70% of lengths are more than 26.7 cm, meaning 30% are less than 26.7 cm. The z-score for 0.30 is approximately -0.524. Thus,

\(\frac{26.7 - \mu}{\sigma} = -0.524\)

Solving these two equations simultaneously, we find:

\(\mu = 28.4\), \(\sigma = 3.25\).

(iii) We need to find \(t\) such that \(P(31.8 < X < t) = 0.5\). The distribution of \(X\) is \(N(32.9, 2.4^2)\).

First, find the z-score for 31.8:

\(z = \frac{31.8 - 32.9}{2.4} = -0.4583\)

The cumulative probability for \(z = -0.4583\) is approximately 0.3235. Therefore, \(P(X > 31.8) = 1 - 0.3235 = 0.6765\).

We need \(P(31.8 < X < t) = 0.5\), so \(P(X < t) = 0.5 + 0.6765 = 0.8235\).

Find the z-score for 0.8235, which is approximately 0.929. Thus,

\(\frac{t - 32.9}{2.4} = 0.929\)

Solving for \(t\), we get:

\(t = 35.1\).