(i) Let \(X\) be the amount of fibre in a packet, \(X \sim N(160, \sigma^2)\). We know \(P(X > 190) = 0.19\). The z-score for 190 is given by:

\(z = \frac{190 - 160}{\sigma}\)

From the standard normal distribution table, \(P(Z > z) = 0.19\) corresponds to \(z = 0.878\).

Thus, \(\frac{30}{\sigma} = 0.878\).

Solving for \(\sigma\), we get:

\(\sigma = \frac{30}{0.878} = 34.2\)

(ii) Let \(Y\) be the number of packets with more than 190 grams of fibre. \(Y \sim \text{Binomial}(12, 0.19)\).

We need \(P(Y \geq 1) = 1 - P(Y = 0)\).

\(P(Y = 0) = (0.81)^{12}\)

Thus, \(P(Y \geq 1) = 1 - (0.81)^{12} = 0.920\).