Given that the lengths of fish are normally distributed, we have:

\(P(X < 12) = \frac{42}{400} = 0.105\)

\(P(X > 19) = \frac{58}{400} = 0.145\)

Using the standard normal distribution table, we find the corresponding z-values:

\(\frac{12 - \mu}{\sigma} = -1.253\)

\(\frac{19 - \mu}{\sigma} = 1.058\)

We now have two equations:

\(12 - \mu = -1.253\sigma\)

\(19 - \mu = 1.058\sigma\)

Subtract the first equation from the second:

\(19 - 12 = 1.058\sigma + 1.253\sigma\)

\(7 = 2.311\sigma\)

Solving for \(\sigma\):

\(\sigma = \frac{7}{2.311} \approx 3.03\)

Substitute \(\sigma\) back into one of the equations to find \(\mu\):

\(12 - \mu = -1.253 \times 3.03\)

\(12 - \mu = -3.797\)

\(\mu = 12 + 3.797 \approx 15.8\)

Thus, the estimates for the mean and standard deviation are \(\mu = 15.8\) and \(\sigma = 3.03\).

(i) To find \(\sigma\), we use the standard normal distribution. The probability that \(X > 0\) is 0.25, which corresponds to a \(z\)-value of 0.674 (since the probability to the right of the mean is 0.25).

Using the standardization formula:

\(z = \frac{0 - (-3)}{\sigma} = 0.674\)

Solving for \(\sigma\):

\(\sigma = \frac{3}{0.674} \approx 4.45\)

(ii) The probability that a single value of \(X\) is positive is 0.25. We need to find the probability that fewer than 2 out of 8 values are positive. This is a binomial distribution problem with \(n = 8\) and \(p = 0.25\).

We calculate \(P(0) + P(1)\):

\(P(0) = (0.75)^8\)

\(P(1) = \binom{8}{1} (0.25)(0.75)^7\)

\(P(0) + P(1) = 0.1001 + 0.2670 = 0.367\)

(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.

Let the mean be \(\mu\) and the standard deviation be \(\sigma\).

For 36,800 km, the probability is 0.0082, which corresponds to a \(z\)-score of 2.4 (since \(P(Z > 2.4) = 0.0082\)).

For 31,000 km, the probability is 0.6915, which corresponds to a \(z\)-score of -0.5 (since \(P(Z > -0.5) = 0.6915\)).

Using the standardization formula:

\(z_1 = \frac{36800 - \mu}{\sigma} = 2.4\)

\(z_2 = \frac{31000 - \mu}{\sigma} = -0.5\)

We have two equations:

\(36800 - \mu = 2.4\sigma\)

\(31000 - \mu = -0.5\sigma\)

Subtract the second equation from the first:

\((36800 - \mu) - (31000 - \mu) = 2.4\sigma + 0.5\sigma\)

\(5800 = 2.9\sigma\)

\(\sigma = \frac{5800}{2.9} = 2000\)

Substitute \(\sigma = 2000\) back into one of the equations:

\(36800 - \mu = 2.4 \times 2000\)

\(36800 - \mu = 4800\)

\(\mu = 36800 - 4800 = 32000\)

Thus, the mean is 32,000 km and the standard deviation is 2,000 km.

(i) We know that \(P(X > 410) = \frac{225}{6000} = 0.0375\). Therefore, \(P\left( Z > \frac{410 - 400}{\sigma} \right) = 0.0375\), which implies \(P(Z > z) = 0.0375\). Thus, \(P(Z < z) = 0.9625\).

The corresponding \(z\) value for 0.9625 is approximately 1.78. Therefore, \(\frac{10}{\sigma} = 1.78\), giving \(\sigma = \frac{10}{1.78} = 5.62\) grams.

(ii) We need \(P(Z < -1.5)\) and \(P(Z > 1.5)\). This is \(\Phi(-1.5) + 1 - \Phi(1.5) = 2 - 2\Phi(1.5)\).

Using \(\Phi(1.5) \approx 0.9332\), we have \(2 - 2 \times 0.9332 = 0.1336\).

The expected number of packets is \(500 \times 0.1336 = 66.8\), so approximately 66 or 67 packets.