(i) To find the probability that the profit is between $10,000 and $12,000, we standardize these values:

\(z_1 = \frac{12 - 6.4}{5.2} = 1.077\)

\(z_2 = \frac{10 - 6.4}{5.2} = 0.692\)

Using the standard normal distribution table, find:

\(\Phi(z_1) - \Phi(z_2) = 0.8593 - 0.7556 = 0.104\)

(ii) To find the probability of a loss, we standardize the value for a loss (profit less than $0):

\(P(\text{loss}) = P\left(z < \frac{0 - 6.4}{5.2}\right) = P(z < -1.231)\)

Using the standard normal distribution table, find:

\(P(\text{loss}) = 1 - 0.8909 = 0.109\)

The probability of exactly 1 loss in 4 days is given by the binomial distribution:

\(P(1) = (0.109)^1 (0.8909)^3 \times \binom{4}{1}\)

Calculate:

\(P(1) = 0.309 \text{ or } 0.308\)