(i) To find the standard deviation \(\sigma\), use the standard normal distribution formula:

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

Given \(X = 4.2\), \(\mu = 3.9\), and \(P(X > 4.2) = 0.18\), find \(z\) such that \(P(Z > z) = 0.18\). From standard normal tables, \(z \approx 0.916\).

\(0.916 = \frac{4.2 - 3.9}{\sigma}\)

\(\sigma = \frac{0.3}{0.916} \approx 0.328\)

(ii) To find the probability that the length differs from the mean by less than half a minute, calculate:

\(z = \frac{4.4 - 3.9}{0.328} \approx 1.5267\)

\(z = \frac{3.4 - 3.9}{0.328} \approx -1.5267\)

Using standard normal tables, \(\Phi(1.5267) \approx 0.9364\).

Probability = \(2 \times 0.9364 - 1 = 0.873\)

(iii) Since the standard deviation is doubled, the spread is larger, resulting in a smaller \(z\)-value for the same difference from the mean. Therefore, \(p\) is less than the probability in part (ii).