(i) To find the probability that a randomly chosen packet weighs less than 1 kg, we standardize the variable:

\(P(X < 1) = P\left( Z < \frac{1 - 1.04}{0.017} \right) = P(Z < -2.353)\)

Using the standard normal distribution table, \(P(Z < -2.353) = 0.0093\).

(ii) The expected number of packets from 1000 kg is calculated by dividing the total weight by the mean weight per packet:

\(\frac{1000}{1.04} \approx 961 \text{ or } 962\)

(iii) Given that the probability of a packet weighing less than 1 kg is 0.0388, we find \(\mu\) by setting up the equation:

\(P(Z < -1.765) = 0.0388\)

Standardizing gives:

\(-1.765 = \frac{1 - \mu}{0.017}\)

Solving for \(\mu\):

\(1 - \mu = -1.765 \times 0.017\)

\(\mu = 1.03\)

(iv) With the new mean, the expected number of packets is:

\(\frac{1000}{1.03} \approx 971 \text{ or } 970\)