(i) We know that 10% of cartons contain less than 440 millilitres of soup. This corresponds to a z-score of \(z = -1.282\) from the standard normal distribution table.

Using the standardization formula:

\(z = \frac{440 - \mu}{9}\)

Substitute \(z = -1.282\):

\(-1.282 = \frac{440 - \mu}{9}\)

Solving for \(\mu\):

\(440 - \mu = -1.282 \times 9\)

\(440 - \mu = -11.538\)

\(\mu = 440 + 11.538\)

\(\mu = 451.538\)

Rounding to the nearest whole number, \(\mu = 452\).

(ii) We need to find the probability that a carton has a volume more than 1.8 standard deviations above the mean. This corresponds to \(P(z > 1.8)\).

From the standard normal distribution table, \(P(z > 1.8) = 1 - 0.9641 = 0.0359\).

The expected number of cartons is:

\(0.0359 \times 150 = 5.385\)

Rounding to the nearest whole number, the number of cartons is 5.