(i) We know that the probability of the lunch break lasting more than 52 minutes is \(\frac{1}{4} = 0.25\). This corresponds to a standard normal value \(z\) such that \(P(Z > z) = 0.25\). From standard normal distribution tables, \(z \approx 0.674\).

Using the formula for the standard normal variable, \(z = \frac{52 - \mu}{5}\), we have:

\(\frac{52 - \mu}{5} = 0.674\)

Solving for \(\mu\):

\(52 - \mu = 5 \times 0.674\)

\(52 - \mu = 3.37\)

\(\mu = 52 - 3.37 = 48.63\)

Rounding to one decimal place, \(\mu = 48.6\).

(ii) We need to find the probability that the lunch break lasts between 40 and 46 minutes. First, standardize these values:

\(z_1 = \frac{40 - 48.63}{5} = -1.726\)

\(z_2 = \frac{46 - 48.63}{5} = -0.526\)

Using the standard normal distribution table, find the probabilities:

\(P(Z < -1.726) \approx 0.0422\)

\(P(Z < -0.526) \approx 0.2995\)

The probability that the lunch break lasts between 40 and 46 minutes is:

\(P(-1.726 < Z < -0.526) = 0.2995 - 0.0422 = 0.2573\)

The probability that this happens on every one of the next four days is:

\((0.2573)^4 = 0.00438\)