(i) Given that 5% of the fish are longer than 50 cm, we use the z-score for 95% which is 1.645. The equation is:

\(1.645 = \frac{50 - 38}{\sigma}\)

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

\(\sigma = \frac{50 - 38}{1.645} = 7.29\)

(ii) To find the proportion of fish shorter than 30 cm, we standardize:

\(z = \frac{30 - 38}{7.29} = -1.097\)

Using the standard normal distribution table, \(P(z < -1.097) = 1 - \Phi(1.097)\)

\(= 1 - 0.8637 = 0.136\)

(iii) The probability that a fish is longer than 50 cm is 0.05. For 9 fish, the probability that none are longer than 50 cm is \((0.95)^9\).

The probability that at least one is longer than 50 cm is:

\(1 - (0.95)^9 = 0.370\)