(i) To find the probability that a randomly chosen can contains less than 440 ml of juice, we standardize the variable using the formula:

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

where \(x = 440\), \(\mu = 445\), and \(\sigma = 3.6\).

\(z = \frac{440 - 445}{3.6} = -1.389\)

The probability \(P(x < 440)\) is equivalent to \(P(z < -1.389)\).

Using the standard normal distribution table, \(P(z < -1.389) = 1 - \Phi(1.389) = 1 - 0.9176 = 0.0824\).

(ii) We know that 94% of the cans contain between \(445 - c\) ml and \(445 + c\) ml. This corresponds to the middle 94% of the normal distribution, leaving 3% in each tail.

The corresponding z-value for 3% in the lower tail is approximately \(z = 1.881\).

We use the equation:

\(\frac{c}{3.6} = 1.881\)

Solving for \(c\), we get:

\(c = 1.881 \times 3.6 = 6.77\).

Let the mean be \(\mu\) and the standard deviation be \(\sigma\). We are given that \(\mu = 5\sigma\).

The probability \(P(Y > 20) = 0.0732\) corresponds to a standard normal variable \(Z\) such that \(P(Z > z) = 0.0732\). From standard normal distribution tables, \(z \approx 1.452\).

Using the transformation for a normal distribution, \(Z = \frac{Y - \mu}{\sigma}\), we have:

\(1.452 = \frac{20 - \mu}{\mu / 5}\)

Solving for \(\mu\):

\(1.452 = \frac{20 - \mu}{\mu / 5}\)

\(1.452 \times \frac{\mu}{5} = 20 - \mu\)

\(0.2904\mu = 20 - \mu\)

\(1.2904\mu = 20\)

\(\mu = \frac{20}{1.2904} \approx 15.5\)

First, calculate the probability of a bag weighing more than 2.1 kg:

\(P(x > 2.1) = \frac{253}{8000} = 0.031625\)

Thus, the probability of a bag weighing less than 2.1 kg is:

\(P(x < 2.1) = 1 - 0.031625 = 0.968375\)

This corresponds to the cumulative distribution function \(\Phi(z)\), so:

\(\Phi(z) = 0.968375\)

Using standard normal distribution tables, find \(z\) such that \(\Phi(z) = 0.968375\). This gives:

\(z = 1.857 \text{ or } 1.858 \text{ or } 1.859\)

Using the z-score formula:

\(z = \frac{2.1 - 2.04}{\sigma}\)

Substitute \(z = 1.857\) (or similar values) to solve for \(\sigma\):

\(1.857 = \frac{2.1 - 2.04}{\sigma}\)

\(\sigma = \frac{0.06}{1.857} \approx 0.0323\)

(ii) To find the probability that at most one of the five observations is greater than 87, we first find \(P(X > 87)\).

Standardize: \(P(X > 87) = 1 - \Phi \left( \frac{87 - 82}{\sqrt{126}} \right)\)

\(= 1 - \Phi(0.445) = 1 - 0.6718 = 0.3282\)

Using the binomial distribution for 5 trials, \(P(0, 1) = (0.6718)^5 + \binom{5}{1} (0.6718)^4 (0.3282)\)

\(= 0.471\)

(iii) We need to find \(k\) such that \(P(87 < X < k) = 0.3\).

First, find \(P(X < 87) = 0.6718\).

Then, \(P(X < k) = 0.6718 + 0.3 = 0.9718\).

Find the corresponding \(z\)-value: \(z = 1.908\) or \(1.909\).

Use the equation: \(1.909 = \frac{k - 82}{\sqrt{126}}\)

Solve for \(k\): \(k = 103\).

Let the mean be \(\mu = 9.3\) and the standard deviation be \(\sigma\). We are given that \(P(X > 5.6) = 0.85\).

Using the standard normal distribution, we find \(P(Z > z) = 0.85\), which implies \(P(Z < z) = 0.15\).

From the standard normal distribution table, \(z \approx -1.036\).

Using the formula for the standard normal variable, \(z = \frac{X - \mu}{\sigma}\), we have:

\(z = \frac{5.6 - 9.3}{\sigma} = -1.036\)

Solving for \(\sigma\):

\(-1.036 = \frac{5.6 - 9.3}{\sigma}\)

\(\sigma = \frac{5.6 - 9.3}{-1.036}\)

\(\sigma = 3.57\)