(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:

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